2.2: Quadratic Functions
- Refresher Topics
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Factoring Techniques
- Factoring Trinomials (of the form \( ax^2 + bx + c \))
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Solving Equations
- Solving Quadratic Equations (Quadratic Formula only)
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Linear Functions
- Slope of a Line
- Finding the Equation of a Line
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Quadratic Functions
- Vertex Form and Completing the Square (Completing the Square)
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Factoring Techniques
In this section, you will:
- Recognize characteristics of parabolas.
- Understand how the graph of a parabola is related to its quadratic function.
- Determine a quadratic function’s minimum or maximum value.
- Solve problems involving a quadratic function’s minimum or maximum value.
- Use the discriminant to determine the number of solutions to a quadratic equation.
- Solve equations that are quadratic-in-form.
- Solve a quadratic inequality using the graphical and sign chart methods.
- Solve an absolute value equation involving a quadratic.
- Graph the absolute value of a quadratic function.
Figure \( \PageIndex{ 1 } \) An array of satellite dishes. (credit: Matthew Colvin de Valle, Flickr)
Curved antennas, such as the ones shown in Figure \( \PageIndex{ 1 } \), are commonly used to focus microwaves and radio waves to transmit television and telephone signals, as well as satellite and spacecraft communication. The cross-section of the antenna is in the shape of a parabola, which can be described by a quadratic function.
In this section, we will investigate quadratic functions, which frequently model problems involving area and projectile motion. Working with quadratic functions can be less complex than working with higher degree functions, so they provide a good opportunity for a detailed study of function behavior.
Recognizing Characteristics of Parabolas
The graph of a quadratic function is a U-shaped curve called a parabola . One important feature of the graph is that it has an extreme point, called the vertex . If the parabola opens up, the vertex represents the lowest point on the graph, or the minimum value of the quadratic function. If the parabola opens down, the vertex represents the highest point on the graph, or the maximum value . In either case, the vertex is a turning point on the graph. The graph is also symmetric with a vertical line drawn through the vertex, called the axis of symmetry . These features are illustrated in Figure \( \PageIndex{ 2 } \).
Figure \( \PageIndex{ 2 } \)
The \( y \)-intercept is the point at which the parabola crosses the \( y \)-axis. The \( x \)-intercepts are the points at which the parabola crosses the \( x \)-axis. If they exist, the \( x \)-intercepts represent the zeros , or roots , of the quadratic function, the values of \(x\) at which \(y=0\).
Determine the vertex, axis of symmetry, zeros, and \( y \)-intercept of the parabola shown in Figure \( \PageIndex{ 3 } \).
Figure \( \PageIndex{ 3 } \)
- Solution
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The vertex is the turning point of the graph. We can see that the vertex is at \(( 3,1 )\). Because this parabola opens upward, the axis of symmetry is the vertical line that intersects the parabola at the vertex. So the axis of symmetry is \(x=3\). This parabola does not cross the \( x \)-axis, so it has no zeros. It crosses the \( y \)-axis at \(( 0,7 )\), so this is the \( y \)-intercept.
Understanding How the Graphs of Parabolas are Related to Their Quadratic Functions
The general form of a quadratic function presents the function in the form\[f(x)=a x^2 +bx+c, \nonumber \]where \(a\), \(b\), and \(c\) are real numbers and \(a \neq 0\). If \(a>0\), the parabola opens upward. If \(a<0\), the parabola opens downward. We can use the general form of a parabola to find the equation for the axis of symmetry.
The axis of symmetry is defined by \(x=-\frac{b}{2a} \). If we use the Quadratic Formula ,\[x= \dfrac{-b \pm \sqrt{ b^2 -4ac }}{ 2a }, \nonumber \]to solve \(a x^2 +bx+c=0\) for the \( x \)-intercepts, or zeros, we find the value of \(x\) halfway between them is always \(x=-\frac{b}{2a} \), the equation for the axis of symmetry.
Figure \( \PageIndex{ 4 } \) represents the graph of the quadratic function written in general form as \(y= x^2 +4x+3\). In this form, \(a=1\), \(b=4\), and \(c=3\). Because \(a>0\), the parabola opens upward. The axis of symmetry is \(x= -\frac{4}{2( 1 )} =-2\). This also makes sense, because we can see from the graph that the vertical line \(x=-2\) divides the graph in half. The vertex always occurs along the axis of symmetry. For a parabola that opens upward, the vertex occurs at the lowest point on the graph, in this instance, \((-2,-1)\). The \( x \)-intercepts, those points where the parabola crosses the \( x \)-axis, occur at \((-3,0)\) and \((-1,0)\).
Figure \( \PageIndex{ 4 } \)
The standard form of a quadratic function presents the function in the form\[f(x)=a (x-h)^2 +k, \nonumber \]where \(( h,k )\) is the vertex. Because the vertex appears in the standard form of the quadratic function, this form is also known as the vertex form of a quadratic function.
As with the general form, if \(a>0\), the parabola opens upward and the vertex is a minimum. If \(a<0\), the parabola opens downward, and the vertex is a maximum. Figure \( \PageIndex{ 5 } \) represents the graph of the quadratic function written in standard form as \(y=-3 ( x+2 )^2 +4\). Since \(x–h=x+2\) in this example, \(h=–2\). In this form, \(a=-3\), \(h=-2\), and \(k=4\). Because \(a<0\), the parabola opens downward. The vertex is at \(( -2, 4 )\).
Figure \( \PageIndex{ 5 } \)
The standard form is useful for determining how the graph is transformed from the graph of \(y= x^2 \). Figure \( \PageIndex{ 6 } \) is the graph of this basic function.
Figure \( \PageIndex{ 6 } \)
If \(k>0\), the graph shifts upward, whereas if \(k<0\), the graph shifts downward. In Figure \( \PageIndex{ 5 } \), \(k>0\), so the graph is shifted 4 units upward. If \(h>0\), the graph shifts toward the right and if \(h<0\), the graph shifts to the left. In Figure \( \PageIndex{ 5 } \), \(h<0\), so the graph is shifted 2 units to the left. The magnitude of \(a\) indicates the stretch of the graph. If \(| a |>1\), the point associated with a particular \( x \)-value shifts farther from the \( x \)-axis, so the graph appears to become narrower, and there is a vertical stretch. But if \(| a |<1\), the point associated with a particular \( x \)-value shifts closer to the \( x \)-axis, so the graph appears to become wider, but in fact there is a vertical compression. In Figure \( \PageIndex{ 5 } \), \(| a |>1\), so the graph becomes narrower.
The standard form and the general form are equivalent methods of describing the same function. We can see this by expanding out the general form and setting it equal to the standard form.\[ \begin{array}{rrcl}
& a (x-h)^2 +k & = & a x^2 +bx+c \\[6pt]
\implies & a x^2 -2ahx+(a h^2 +k) & = & a x^2 +bx+c \\[6pt]
\end{array} \nonumber \]For the linear terms to be equal, the coefficients must be equal.\[ –2ah = b \implies h=- \dfrac{b}{2a}. \nonumber \]This is the axis of symmetry we defined earlier. Setting the constant terms equal, we get\[ \begin{array}{rrcl}
& a h^2 +k & = & c \\[6pt]
\implies & k & = & c-a h^2 \\[6pt]
& & = & c-a\left(- \dfrac{b}{2a} \right)^2 \\[6pt]
& & = & c- \dfrac{b^2}{4a}. \\[6pt]
\end{array} \nonumber \]In practice, though, it is usually easier to remember that \( k \) is the output value of the function when the input is \(h\), so \(f(h)=k\).
Let's group all this information together in one statement.
A quadratic function is a function of degree two. The graph of a quadratic function is a parabola . The general form of a quadratic function is \(f(x)=a x^2 +bx+c\) where \(a\), \(b\), and \(c\) are real numbers and \(a \neq 0\).
The standard form of a quadratic function is \(f(x)=a (x-h)^2 +k\).
The vertex \((h,k)\) is located at\[h=–\dfrac{b}{2a}, k = f(h) = f\left( -\dfrac{b}{2a}\right). \nonumber \]
Write an equation for the quadratic function \(g\) in Figure \( \PageIndex{ 7 } \) as a transformation of \(f(x)= x^2 \), and then expand the formula and simplify terms to write the equation in general form.
Figure \( \PageIndex{ 7 } \)
- Solution
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We can see the graph of \( g \) is the graph of \(f(x)= x^2\) shifted to the left 2 and down 3, giving a formula in the form \(g(x)=a (x+2)^2 –3\). Substituting the coordinates of any point on the curve, such as \((0,-1)\), we can solve for the stretch factor.\[ \begin{array}{rrcl}
& -1 & = & a (0+2)^2 -3 \\[6pt]
\implies & 2 & = & 4a \\[6pt]
\implies & \dfrac{1}{2} & = & a \\[6pt]
\end{array} \nonumber \]In standard form, the algebraic model for this graph is \(g(x)= \frac{1}{2} (x+2)^2 –3\).To write this in general polynomial form, we can expand the formula and simplify terms.\[ \begin{array}{rcl}
g(x) & = & \dfrac{1}{2} (x+2)^2 -3 \\[6pt]
& = & \dfrac{1}{2} (x+2)(x+2)-3 \\[6pt]
& = & \dfrac{1}{2} ( x^2 +4x+4)-3 \\[6pt]
& = & \dfrac{1}{2} x^2 +2x+2-3 \\[6pt]
& = & \dfrac{1}{2} x^2 +2x-1 \\[6pt]
\end{array} \nonumber \]Notice that the horizontal and vertical shifts of the basic graph of the quadratic function determine the location of the vertex of the parabola; the vertex is unaffected by stretches and compressions.
A coordinate grid has been superimposed over the quadratic path of a basketball in Figure \( \PageIndex{ 8 } \). Find an equation for the path of the ball. Does the shooter make the basket?
Figure \( \PageIndex{ 8 } \) (credit: modification of work by Dan Meyer)
Find the vertex of the quadratic function \(f(x)=2 x^2 –6x+7\). Rewrite the quadratic in standard form (vertex form).
- Solution
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The horizontal coordinate of the vertex will be at\[ h = –\dfrac{b}{2a} = –\dfrac{–6}{2(2)} = \dfrac{6}{4} = \dfrac{3}{2}. \nonumber \]The vertical coordinate of the vertex will be at\[k = f(h) = f\left( \dfrac{3}{2} \right) = 2 \left( \dfrac{3}{2} \right)^2 -6\left( \dfrac{3}{2} \right)+7 = \dfrac{5}{2}. \nonumber \]Rewriting into standard form, the stretch factor will be the same as the \(a\) in the original quadratic. Therefore, using the vertex to determine the shifts,\[f( x ) = 2 \left(x - \dfrac{3}{2} \right)^2 + \dfrac{5}{2}. \nonumber \]One reason we may want to identify the vertex of the parabola is that this point will inform us where the maximum or minimum value of the output occurs \(( k )\) and where it occurs \(( x )\).
If given a quadratic function in general form, most students are going to use the formula \( -\frac{b}{2a} \) to find the \( x \)-value of the vertex (and substitute this into the original quadratic function to find the corresponding \( y \)-value for the vertex). This is perfectly fine, and it will always lead you to the vertex of a given quadratic function; however, another approach, which is inevitably going to be necessary in this course, is to complete the square like you did in your Elementary Algebra course.
Rewrite the quadratic function \( f(x) = 2x^2 - 6x + 7 \) in standard (vertex) form by Completing the Square.
- Solution
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Since Completing the Square is an older skill, we will just use it here rather than explaining the entire process.\[ \begin{array}{rclcl}
f(x) & = & 2x^2 - 6x + 7 & & \\[6pt]
& = & 2(x^2 - 3x) + 7 & \quad & \left( \text{factoring the lead coefficient from both variable terms} \right) \\[6pt]
& = & 2\left(x^2 - 3x + {\color{red}\left( \dfrac{-3}{2} \right)^2} \right) + 7 {\color{red}-2 \left( \dfrac{-3}{2} \right)^2} & \quad & \left( \text{Completing the Square and offsetting the }2\left( \dfrac{-3}{2} \right)^2 \text{ term with }-2\left( \dfrac{-3}{2} \right)^2 \right) \\[6pt]
& = & 2\left(x - \dfrac{3}{2} \right)^2 + 7 -2 \left( \dfrac{9}{4} \right) & \quad & \left( \text{factoring and simplifying} \right) \\[6pt]
& = & 2\left(x - \dfrac{3}{2} \right)^2 + \dfrac{14}{2} - \dfrac{9}{2} & \quad & \left( \text{common denominator} \right) \\[6pt]
& = & 2\left(x - \dfrac{3}{2} \right)^2 + \dfrac{5}{2} & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]
The purpose of being introduced to the process of Completing the Square in your Algebra courses was to prove the Quadratic Formula. We now see that this same process can be used to rewrite a quadratic function in vertex form; however, these are not the only uses of Completing the Square. We will use this skill again in this course when working with conic sections. In Calculus II, you will find many opportunities to use this very same skill when manipulating a seemingly impossible-to-work-with expression into something more simple.
Given the equation \(g(x)=13+ x^2 -6x\), write the equation in general form and then in standard form.
Finding the Domain and Range of a Quadratic Function
Any number can be the input value of a quadratic function. Therefore, the domain of any quadratic function is all real numbers; however, because parabolas have a maximum or a minimum point, the range is restricted. Since the vertex of a parabola will be either a maximum or a minimum, the range will consist of all \( y \)-values greater than or equal to the \( y \)-coordinate at the turning point or less than or equal to the \( y \)-coordinate at the turning point, depending on whether the parabola opens up or down.
The domain of any quadratic function is all real numbers.
The range of a quadratic function written in general form, \(f(x)=a x^2 +bx+c\), with a positive \(a\) value is \(f(x) \geq f\left( -\frac{b}{2a} \right)\), or \(\left[ f\left( -\frac{b}{2a} \right), \infty \right)\).
The range of a quadratic function written in general form with a negative \(a\) value is \(f(x) \leq f\left( -\frac{b}{2a} \right)\), or \(\left( -\infty , f\left( -\frac{b}{2a} \right) \right]\).
The range of a quadratic function written in standard form, \(f(x)=a (x-h)^2 +k\), with a positive \(a\) value is \(f(x) \geq k\); the range of a quadratic function written in standard form with a negative \(a\) value is \(f(x) \leq k\).
The theorem above is only provided for completeness - it should not be memorized.
Find the domain and range of \(f(x)=-5 x^2 +9x-1\).
- Solution
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As with any quadratic function, the domain is all real numbers.
Because \(a\) is negative, the parabola opens downward and has a maximum value. We need to determine the maximum value. We can begin by finding the \( x \)-value of the vertex.\[h=-\dfrac{b}{2a} =-\dfrac{9}{2(-5)} = \dfrac{9}{10}. \nonumber \]The maximum value is given by \(f(h)\).\[f\left( \dfrac{9}{10} \right) = -5 \left( \dfrac{9}{10} \right)^2 +9\left( \dfrac{9}{10} \right)-1 = \dfrac{61}{20}. \nonumber \]The range is \(\left( -\infty , \frac{61}{20} \right]\).
Find the domain and range of \(f(x)=2 \left(x - \frac{4}{7} \right)^2 + \frac{8}{11} \).
Determining the Maximum and Minimum Values of Quadratic Functions
The output of the quadratic function at the vertex is the maximum or minimum value of the function, depending on the orientation of the parabola. We can see the maximum and minimum values in Figure \( \PageIndex{ 9 } \).
Figure \( \PageIndex{ 9 } \)
There are many real-world scenarios that involve finding the maximum or minimum value of a quadratic function, such as applications involving area and revenue.
A backyard farmer wants to enclose a rectangular space for a new garden within her fenced backyard. She has purchased 80 feet of wire fencing to enclose three sides, and she will use a section of the backyard fence as the fourth side.
- Find a formula for the area enclosed by the fence if the sides of fencing perpendicular to the existing fence have length \(L\).
- What dimensions should she make her garden to maximize the enclosed area?
- Solutions
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Let’s use a diagram such as Figure \( \PageIndex{ 10 } \) to record the given information. It is also helpful to introduce a temporary variable, \(W\), to represent the width of the garden and the length of the fence section parallel to the backyard fence.
Figure \( \PageIndex{ 10 } \)- We know we have only 80 feet of fence available, and \(L+W+L=80\), or more simply, \(2L+W=80\). This allows us to represent the width, \( W \), in terms of \( L \):\[W=80-2L. \nonumber \]Now we are ready to write an equation for the area the fence encloses. We know the area of a rectangle is length multiplied by width, so\[ A = LW =L (80 - 2L) = 80L - 2 L^2. \ nonumber \] This formula represents the area of the fence in terms of the variable length \(L\). The function, written in general form, is\[A(L)= - 2 L^2 + 80L . \ nonumber \]
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The quadratic has a negative leading coefficient, so the graph will open downward, and the vertex will be the maximum value for the area. In finding the vertex, we must be careful because the equation is not written in standard polynomial form with decreasing powers. This is why we rewrote the function in general form above. Since \(a\) is the coefficient of the squared term, \(a=-2\), \(b=80\), and \(c=0\).
To find the vertex:\[ \begin{array}{rclcrcl}
h & = & - \dfrac{80}{2(-2)} & \quad & k & = & A(20) \\[6pt]
& = & 20 & \text{and} & & = & 80(20) -2(20)^2 \\[6pt]
& & & & & = & 800 \\[6pt]
\end{array} \nonumber \]The maximum value of the function is an area of 800 squa re feet, wh ich occurs when \(L=20\) feet. When the shorter sides are 20 feet, there is 40 feet of fencing left for the longer side. To maximize the are a, she sho uld enclose the garden so the two shorter sides have length 20 feet and the longer side parallel to the existing fence has length 40 feet.
Example \( 6 \) also could be solved by graphing the quadratic function. We can see where the maximum area occurs on a graph of the quadratic function in Figure \( \PageIndex{ 11 } \).
Figure \( \PageIndex{ 11 } \)
The unit price of an item affects its supply and demand. That is, if the unit price goes up, the demand for the item will usually decrease. For example, a local newspaper currently has 84,000 subscribers at a quarterly charge of $30. Market research has suggested that if the owners raise the price to $32, they would lose 5,000 subscribers. Assuming that subscriptions are linearly related to the price, what price should the newspaper charge for a quarterly subscription to maximize their revenue?
- Solution
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Revenue is the amount of money a company brings in. In this case, the revenue can be found by multiplying the price per subscription times the number of subscribers, or quantity. We can introduce variables, \(p\) for price per subscription and \(Q\) for quantity, giving us the equation\[\text{Revenue}=pQ. \nonumber \]Because the number of subscribers changes with the price, we need to find a relationship between the variables. We know that currently \(p=30\) and \(Q=84,000\). We also know that if the price rises to $32, the newspaper would lose 5,000 subscribers, giving a second pair of values, \(p=32\) and \(Q=79,000\). From this we can find a linear equation relating the two quantities. The slope will be\[ m= \dfrac{79,000-84,000}{32-30} = \dfrac{-5,000}{2} = -2,500. \nonumber \]This tells us the paper will lose 2,500 subscribers for each dollar they raise the price. We can then solve for the \( y \)-intercept.\[ \begin{array}{rrclcl}
& Q & = & -2500p+b & & \\[6pt]
\implies & 84,000 & =& -2500(30)+b & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & b & = & 159,000 & \quad & \left( \text{isolating the variable} \right) \\[6pt]
\end{array} \nonumber \]This gives us the linear equation \(Q=-2,500p+159,000\) relating cost and subscribers. We now return to our revenue equation.\[ \begin{array}{rrcl}
& \text{Revenue} & = & pQ \\[6pt]
\implies & \text{Revenue} & = & p(-2,500p+159,000) \\[6pt]
\implies & \text{Revenue} & = & -2,500 p^2 +159,000p \\[6pt]
\end{array} \nonumber \]We now have a quadratic function for revenue as a function of the subscription charge. To find the price that will maximize revenue for the newspaper, we can find the vertex.\[h =-\dfrac{159,000}{2(-2,500)} = 31.8. \nonumber \]The model tells us that the maximum revenue will occur if the newspaper charges $31.80 for a subscription. To find what the maximum revenue is, we evaluate the revenue function.\[\text{maximum revenue} = -2,500 (31.8)^2 +159,000(31.8) = 2,528,100. \nonumber \]
As before, Example \( \PageIndex{ 7 } \) could also be solved by graphing the quadratic as in Figure \( \PageIndex{ 12 } \). We can see the maximum revenue on a graph of the quadratic function.
Figure \( \PageIndex{ 12 } \)
Finding the \( x \)- and \( y \)-Intercepts of a Quadratic Function
Much as we did in the application problems above, we also need to find intercepts of quadratic equations for graphing parabolas. Recall that we find the \( y \)-intercept of a quadratic by evaluating the function at an input of zero, and we find the \( x \)-intercepts at locations where the output is zero. Notice in Figure \( \PageIndex{ 13 } \) that the number of \( x \)-intercepts can vary depending upon the location of the graph.
Figure \( \PageIndex{ 13 } \) Number of \( x \)-intercepts of a parabola
Find the \( y \)- and \( x \)-intercepts of the quadratic \(f(x)=3 x^2 +5x-2\).
- Solution
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We find the \( y \)-intercept by evaluating \(f( 0 )\).\[f(0) =3 (0)^2 +5(0)-2 =-2. \nonumber \]So the \( y \)-intercept is at \(( 0,-2 )\).
For the \( x \)-intercepts, we find all solutions of \(f( x )=0\). That is, we want to solve\[0=3 x^2 +5x-2. \nonumber \]In this case, the quadratic can be factored easily, providing the simplest method for solution.\[0=(3x-1)(x+2). \nonumber \]Thus, \(x= \frac{1}{3} \) or \(x=-2\). So the \( x \)-intercepts are at \(\left( \frac{1}{3}, 0 \right)\) and \(( -2,0 )\).
By graphing the function, we can confirm that the graph crosses the \( y \)-axis at \((0,-2)\). We can also confirm that the graph crosses the \( x \)-axis at \(\left( \frac{1}{3}, 0 \right)\) and \((-2,0)\). See Figure \( \PageIndex{ 14 } \)
Figure \( \PageIndex{ 14 } \)
Example \( 8 \) illustrates a strength of having the general form of a quadratic function - the \( y \)-intercept is the constant term. That is, given\[ f(x) = ax^2 + bx + c, \nonumber \]the \( y \)-intercept is \( f(0) = c \). This is only true, however, when the quadratic is given in general form.
Another lesson to be learned from Example \( \PageIndex{ 8 } \) is that to solve \( ax^2 + bx + c = 0 \), we don't have to always resort to the Quadratic Formula. In fact, whenever you need to solve a quadratic equation, you should spend about 30 seconds or so trying to factor the given quadratic before plunging into the algorithm called the Quadratic Formula.
As Figure \( \PageIndex{ 13 } \) illustrates, there are some quadratic functions that do not have \( x \)-intercepts. Algebraically, this occurs when the given quadratic function is prime . Just like a prime number, a prime quadratic is not factorable (across the real number system). That is, no matter how ingenius we are, we will never be able to write a prime quadratic in the form \( \left( c_1 x + d_1 \right)\left( c_2 x + d_2 \right) \) (where \( c_1 \), \( c_2 \), \( d_1 \), and \( d_2 \) are real numbers).
In Mathematics, we always strive to develop methods to reduce our workload. It would be nice to know ahead of time that the quadratic function\[ f(x) = ax^2 + bx + c \nonumber \]is prime. Knowing this would allow us to skip trying to find any \( x \)-intercepts. Luckily, we have such a method.
The \( x \)-intercepts of a quadratic function occur when\[ ax^2 + bx + c = 0. \nonumber \]We know from the Quadratic Formula that the solutions to this equation are\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. \nonumber \]However, the \( x \)-axis (where our \( x \)-intercepts lay) is a real number axis. That is, only real numbers exist along the \( x \)-axis. Hence, our \( x \)-intercepts cannot be imaginary numbers.
From our previous work, we know we have an imaginary number if our expression contains the square root of a negative number. Therefore, we don't want the radicand in the numerator of the Quadratic Formula to be negative. That is, we require\[ b^2 - 4ac \geq 0. \nonumber \]This expression is so important in Mathematics, it deserves its own name.
The discriminant of the quadratic function\[ f(x) = ax^2 + bx + c \nonumber \]is the value \( b^2 - 4ac \).
From our discussion above, if the discriminant is negative, the quadratic function will not have any \( x \)-intercepts. Likewise, if the discriminant is negative, the equation\[ ax^2 + bx + c = 0 \nonumber \]will not have real solutions.
If the discriminant is \( 0 \), then (from the Quadratic Formula) the \( x \)-intercepts for the quadratic function would occur at\[ x = \dfrac{-b \pm\sqrt{0}}{2a} = -\dfrac{b}{2a}. \nonumber \]Since this is a single value, a discriminant of \( 0 \) implies that the graph of the quadratic function has only one \( x \)-intercept (as seen in the middle frame of Figure \( \PageIndex{ 13 } \)).
When the discriminant is positive, the quadratic function has two real roots, which means its graph crosses the \( x \)-axis twice.
Let\[ f(x) = ax^2 + bx + c. \nonumber \]The graph of \( f \) has
- two \( x \)-intercepts if the discriminant is positive,
- one \( x \)-intercept if the discriminant is \( 0 \), and
- no \( x \)-intercepts if the discriminant is negative.
I titled this theorem, "Zeros and the Discriminant" because the zeros of a function occur precisely at the \( x \) (or horizontal) intercepts.
Find the \( y \)- and \( x \)-intercepts of the quadratic \(f(x)=-3 x^2 +4x-2\).
- Solution
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Since the quadratic function is already in general form, finding the \( y \)-intercept is as simple as looking for the constant term. In this case, \( \left( 0,-2 \right) \) is the \( y \)-intercept.
As for any \( x \)-intercepts, per my advice earlier, we spend roughly 30 seconds trying to factor the function, but have no success. Before running straight to the full Quadratic Formula, let's only compute one part - the discriminant (heck, we need to anyway if we are eventually going to use the Quadratic Formula).\[ b^2 - 4ac = 4^2 - 4(-3)(-2) = 16 - 24 = -8 < 0. \nonumber \]Since the discriminant is negative, the function doesn't have any \( x \)-intercepts.
It is important to clarify that finding the \( x \)-intercepts (or roots) of a quadratic function and solving a quadratic equation are often, but not always, the same thing. How are these processes different? Since the horizontal intercepts of a graph can only be real numbers, we stop our search for such intercepts if the discriminant of the quadratic function is negative. In contrast , the solutions to a random quadratic equation can be imaginary, and therefore the discriminant can be negative.
Find all the solutions of the quadratic equation.\[0=-3 x^2 +4x-2 \nonumber \]
- Solution
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From Example \( \PageIndex{ 9 } \), we know the discriminant is \( -8 \). Thus, the solutions of this equation are going to be imaginary. An
incredibly
important piece of advice is to start our solution process by multiplying both sides of the given equation by \( -1 \). We do this because the original lead coefficient is negative, which only serves to complicate using the Quadratic Formula. Therefore, let's focus on solving\[ 0 = 3x^2 - 4x + 2. \nonumber \]Note that the discriminant,\[ b^2 - 4ac = (-4)^2 - 4(3)(2) = 16 - 24 = -8 \nonumber \]doesn't change when multiplying both sides of the quadratic equation by \( -1 \). Now we can solve.\[ \begin{array}{rclcl}
x & = & \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} & \quad & \left( \text{Quadratic Formula} \right) \\[6pt]
& = & \dfrac{-(-4) \pm \sqrt{\text{discriminant}}}{2(3)} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{4 \pm \sqrt{-8}}{6} & \quad & \left( \text{substituting and simplifying} \right) \\[6pt]
& = & \dfrac{4 \pm \sqrt{4 \cdot -2}}{6} & & \\[6pt]
& = & \dfrac{4 \pm 2 \sqrt{-2}}{6} & \quad & \left( \text{simplifying the radical} \right) \\[6pt]
& = & \dfrac{4 \pm 2 \sqrt{-1} \sqrt{2}}{6} & & \\[6pt]
& = & \dfrac{4 \pm 2 i \sqrt{2}}{6} & \quad & \left( \sqrt{-1} = i \right) \\[6pt]
& = & \dfrac{4}{6} \pm \dfrac{2 i \sqrt{2}}{6} & \quad & \left( \text{splitting the fraction} \right) \\[6pt]
& = & \dfrac{2}{3} \pm \dfrac{i \sqrt{2}}{3} & \quad & \left( \text{simplifying fractions} \right) \\[6pt]
& = & \dfrac{2}{3} \pm \dfrac{\sqrt{2}}{3}i & \quad & \left( \text{rewriting in the standard form of a complex number} \right) \\[6pt]
\end{array} \nonumber \]
Find the \( x \)-intercepts of the quadratic function \(f(x)=2 x^2 +4x-4\).
- Solution
-
We begin by solving for when the output will be zero; however, because this quadratic is not easily factorable, we start by checking the discriminant to see if there are any \( x \)-intercepts at all.\[ b^2 - 4ac = 4^2 - 4(2)(-4) = 16 + 32 = 48 > 0. \nonumber \]Thus, there will be two \( x \)-intercepts. Our time computing the discriminant is not wasted since we will need its value when solving the equation \( 2 x^2 + 4x - 4 = 0 \). Use the Quadratic Formula, we get the following:\[ \begin{array}{rclcl}
x & = & \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} & \quad & \left( \text{Quadratic Formula} \right) \\[6pt]
& = & \dfrac{-4 \pm \sqrt{\text{discriminant}}}{2(2)} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{-4 \pm \sqrt{48}}{4} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{-4 \pm \sqrt{16 \cdot 3}}{4} & & \\[6pt]
& = & \dfrac{-4 \pm 4\sqrt{3}}{4} & \quad & \left( \text{simplifying the radical expression} \right) \\[6pt]
& = & \dfrac{4(-1 \pm \sqrt{3})}{4} & \quad & \left( \text{factoring out the GCF from the numerator} \right) \\[6pt]
& = & \dfrac{\cancelto{1}{4}(-1 \pm \sqrt{3})}{\cancelto{1}{4}} & \quad & \left( \text{canceling like factors} \right) \\[6pt]
& = & -1 \pm \sqrt{3} & & \\[6pt]
\end{array} \nonumber \]The graph has \( x \)-intercepts at \((-1-\sqrt{ 3 } ,0)\) and \((-1+\sqrt{ 3 } ,0)\).
Again, we can check our work for Example \( \PageIndex{ 11 } \) by graphing the given function on a graphing utility and observing the \( x \)-intercepts. See Figure \( \PageIndex{ 15 } \).
Figure \( \PageIndex{ 15 } \)
In a separate Checkpoint, we found the standard and general form for the function \(g(x)=13+ x^2 -6x\). Now find the \( y \)- and \( x \)-intercepts (if any).
Focusing on Calculus - Solving Equations that are Quadratic-in-Form
You have already solved quite a few quadratic equations throughout your academic journey in Mathematics; however, as you approach Calculus, you will need to focus your attention on pattern recognition instead of expecting equations of the form \( ax^2 = bx + c = 0 \). It turns out that any equation of the form\[ a\left[ f(x) \right]^2 + b f(x) + c = 0 \nonumber \]can be solved using methods similar to those from our Elementary and Intermediate Algebra courses. Let's illustrate the process with a few examples.
Solve \(3(3x - 1)^2 + 13(3x - 1) -10=0\).
- Solution
-
Many students would immediately distribute things out to get\[ 27x^2 + 21x - 20 = 0 \nonumber \]and try to factor; however, this is not taking advantage of the form of the original equation.
Notice that there are some common expressions within the original equation - namely \( 3x - 1 \). If we let \( u = 3x - 1 \), we could rewrite the equation as\[ 3u^2 + 13u - 10 = 0. \nonumber \]Hopefully, you agree that this equation looks much easier to solve. This type of substitution is formally called a \( u \)-substitution . The choice of the variable \( u \) is arbitrary, and you are encouraged to use whatever variable you want (other than the variable in the original equation).
We can factor the left side of this new equation to get\[ (3u - 2)(u + 5) = 0. \nonumber \]Therefore, \( u = \frac{2}{3} \) or \( u = -5 \).
Since the original problem involved the variable \( x \), we need to resubstitute \( u = 3x - 1 \) to arrive at\[ \begin{array}{rrclcrclcl}
& u & = & \dfrac{3}{2} & \text{ and } & u & = & -5 & & \\[6pt]
\implies & 3x - 1 & = & \dfrac{3}{2} & \text{ and } & 3x - 1 & = & -5 & \quad & \left( \text{resubstituting} \right) \\[6pt]
\implies & 3x & = & \dfrac{3}{2} + 1 & \text{ and } & 3x & = & -5 + 1 & \quad & \left( \text{adding }1\text{ to both sides} \right) \\[6pt]
\implies & 3x & = & \dfrac{5}{2} & \text{ and } & 3x & = & -4 & \quad & \left( \text{simplifying} \right) \\[6pt]
\implies & x & = & \dfrac{5}{6} & \text{ and } & x & = & -\dfrac{4}{3} & \quad & \left( \text{dividing both sides by }3 \right) \\[6pt]
\end{array} \nonumber \]Thus, the solutions to the original equation are \( x = \frac{5}{6} \) and \( x = -\frac{4}{3} \).
You will be using the \( u \)-substitution process illustrated in Example \( \PageIndex{ 12 } \) often throughout Calculus. As such, it's best to start looking for patterns like this from this point forward.
Solve \( x^4 - 3x^2 = 45 + x^2 \).
- Solution
-
Since this equation has several terms involving powers of \( x \), it is best to collect all information on one side (this tends to "reveal" the type of equation we are working with). Subtracting \( 45 \) and \( x^2 \) from both sides, we get\[ x^4 - 4x^2 - 45 = 0. \nonumber \]The key observation here is that the exponent on \( x \) in the first term is
twice
that of the exponent on \( x \) in the second term. That is, this has the same form as a quadratic equation. Letting \( u = x^2 \), we get\[ \begin{array}{rclcl}
x^4 - 4x^2 - 45 & = & \left( x^2 \right)^2 - 4x^2 - 45 & \quad & \left( \text{Laws of Exponents: Power Law} \right) \\[6pt]
& = & u^2 - 4u - 45 & \quad & \left( \text{substituting } u = x^2 \right) \\[6pt]
\end{array} \nonumber \]Thus, we instead focus on solving\[ \begin{array}{rrclcl}
& u^2 - 4u - 45 & = & 0 & & \\[6pt]
\implies & (u - 9)(u + 5) & = & 0 & \quad & \left( \text{factoring} \right) \\[6pt]
\end{array} \nonumber \]Hence, \( u = 9 \) or \( u = -5 \). However, we are supposed to be solving for \( x \), so we resubstitute \( u = x^2 \).\[ \implies x^2 = 9 \quad \text{or} \quad x^2 = -5. \nonumber \]While we can easily solve the first equation to get \( x = \pm 3 \), the second equation doesn't have a solution (over the real number system) because there isn't a real number whose square becomes a negative number. Thus, the solutions to the original equation are \( x = -3 \) and \( x = 3 \).
Solving Quadratic Inequalities
We now turn our attention to quadratic inequalities . If we consider \(f(x) = x^2-x-6\), then solving \(x^2 - x -6 < 0\) corresponds graphically to finding the values of \(x\) for which the graph of \(y=f(x)=x^2-x-6\) (the parabola) is below the graph of \(y=0\) (the \(x\)-axis). See Figure \( \PageIndex{ 16 } \).
Figure \( \PageIndex{16} \)
We can see that the graph of \(f\) dips below the \(x\)-axis between its two \(x\)-intercepts. The zeros of \(f\) are \(x=-2\) and \(x=3\) in this case and they divide the domain (the \(x\)-axis) into three intervals: \((-\infty, -2)\), \((-2,3)\) and \((3, \infty)\). For every number in \((-\infty, -2)\), the graph of \(f\) is above the \(x\)-axis; in other words, \(f(x) > 0\) for all \(x\) in \((-\infty, -2)\). Similarly, \(f(x) < 0\) for all \(x\) in \((-2,3)\), and \(f(x) > 0\) for all \(x\) in \((3, \infty)\). Thus, the solution to \(x^2 - x -6 < 0\) is all \( x \) in the interval \( \left( -2,3 \right) \).
The method introduced above (graphing the quadratic function and determining when the graph satisfies the inequality) is called the graphical method of solving a quadratic inequality. This is the method I prefer to teach in my courses because it simultaneously reinforces graphing and it is much faster (in the long run) than the alternative method. Before discussing the alternative method, let's look at a couple examples.
Solve the following inequalities using the graphing method.
- \(2x^2 \leq 3-x\)
- \(x^2 - 2x > 1\)
- Solutions
-
-
To solve \(2x^2 \leq 3-x\), we first get \(0\) on one side of the inequality which yields \(2x^2+x-3 \leq 0\). Therefore, we want to know when the function \( f(x) = 2x^2 + x - 3 \) is at or below the \( x \)-axis. Since the lead coefficient is positive, we know the parabola representing \( f \) opens upward. We just need to find the zeros of \(f\) by solving \(2x^2 + x - 3 = 0\) for \(x\). Factoring gives \((2x+3)(x-1)=0\), so \(x = -\frac{3}{2}\) or \(x = 1\). A simple and quick sketch yields the following:
By our sketch, we can see \( f(x) \) is at or below the \( x \)-axis on the interval \( \left[ -\frac{3}{2}, 1 \right] \). This is the solution to our inequality. -
Once again, we rewrite \(x^2-2x > 1\) as \(x^2-2x-1>0\) and we identify \(f(x)=x^2-2x-1\). When we go to find the zeros of \(f\), we find, to our chagrin, that the quadratic \(x^2-2x-1\) doesn’t factor nicely. Hence, we resort to the Quadratic Formula to solve \(x^2-2x-1=0\), and arrive at \(x=1 \pm \sqrt{2}\). As before, our parabola is opening upward. We spend a moment sketching the graph of \( f \) and asking ourselves when the parabola is strictly above the \( x \)-axis (because we are solving \( f(x) > 0 \)).
So, we see that the interval notation \(\left(-\infty, 1-\sqrt{2}\right) \cup \left(1+\sqrt{2},\infty\right)\) solves our inequality.
-
To solve \(2x^2 \leq 3-x\), we first get \(0\) on one side of the inequality which yields \(2x^2+x-3 \leq 0\). Therefore, we want to know when the function \( f(x) = 2x^2 + x - 3 \) is at or below the \( x \)-axis. Since the lead coefficient is positive, we know the parabola representing \( f \) opens upward. We just need to find the zeros of \(f\) by solving \(2x^2 + x - 3 = 0\) for \(x\). Factoring gives \((2x+3)(x-1)=0\), so \(x = -\frac{3}{2}\) or \(x = 1\). A simple and quick sketch yields the following:
The alternative method is called the sign chart method of solving a quadratic inequality. Like the graphical method, we start by finding the \( x \)-intercepts. We then mark those intercepts on a number line. For example, considering our original inequality \( x^2 - x - 6< 0 \), we would create the sign chart in Figure \( \PageIndex{17} \).
Figure \( \PageIndex{17} \)
Here, the \((+)\) above a portion of the number line indicates \(f(x) > 0\) for those values of \(x\); the \((-)\) indicates \(f(x) < 0\) there. The numbers labeled on the number line are the zeros of \(f\), so we place \(0\) above them. We see at once that the solution to \(f(x) < 0\) is \((-2,3)\).
An important property of quadratic functions is that if the function is positive at one point and negative at another, the function must have at least one zero in between. Graphically, this means that a parabola can’t be above the \(x\)-axis at one point and below the \(x\)-axis at another point without crossing the \(x\)-axis. This allows us to determine the sign of all of the function values on a given interval by testing the function at just one value in the interval. We use this idea in the following example.
Solve the following inequalities analytically using sign diagrams. Verify your answer graphically.
- \(2x^2 \leq 3-x\)
- \(x^2 - 2x > 1\)
- Solutions
-
-
To solve \(2x^2 \leq 3-x\), we first get \(0\) on one side of the inequality which yields \(2x^2+x-3 \leq 0\). We find the zeros of \(f(x) = 2x^2 + x - 3\) by solving \(2x^2 + x - 3 = 0\) for \(x\). Factoring gives \((2x+3)(x-1)=0\), so \(x = -\frac{3}{2}\) or \(x = 1\). We place these values on the number line with \(0\) above them and choose test values in the intervals \(\left(-\infty, -\frac{3}{2}\right)\), \(\left(-\frac{3}{2},1\right)\) and \((1,\infty)\). For the interval \(\left(-\infty, -\frac{3}{2}\right)\), we choose \(x=-2\); for \(\left(-\frac{3}{2},1\right)\), we pick \(x=0\); and for \((1,\infty)\), \(x=2\). Evaluating the function at the three test values gives us \(f(-2) = 3 > 0\), so we place \((+)\) above \(\left(-\infty, -\frac{3}{2}\right)\); \(f(0)=-3 < 0\), so \((-)\) goes above the interval \(\left(-\frac{3}{2},1\right)\); and, \(f(2) = 7\), which means \((+)\) is placed above \((1,\infty)\).
Since we are solving \(2x^2+x-3 \leq 0\), we look for solutions to \(2x^2+x-3 < 0\) as well as solutions for \(2x^2+x-3 =0\). For \(2x^2+x-3 < 0\), we need the intervals which we have a \((-)\). Checking the sign diagram, we see this is \(\left(-\frac{3}{2},1\right)\). We know \(2x^2+x-3 =0\) when \(x=-\frac{3}{2}\) and \(x=1\), so our final answer is \(\left[-\frac{3}{2},1\right]\).
To verify our solution graphically, we refer to the original inequality, \(2x^2 \leq 3-x\). We let \(g(x) = 2x^2\) and \(h(x)=3-x\). We are looking for the \(x\) values where the graph of \(g\) is below that of \(h\) (the solution to \(g(x) < h(x)\)) as well as the points of intersection (the solutions to \(g(x)=h(x)\)). -
Once again, we re-write \(x^2-2x > 1\) as \(x^2-2x-1>0\) and we identify \(f(x)=x^2-2x-1\). When we go to find the zeros of \(f\), we find, to our chagrin, that the quadratic \(x^2-2x-1\) doesn’t factor nicely. Hence, we resort to the Quadratic Formula to solve \(x^2-2x-1=0\), and arrive at \(x=1 \pm \sqrt{2}\). As before, these zeros divide the number line into three pieces. To help us decide on test values, we approximate \(1 - \sqrt{2} \approx -0.4\) and \(1 + \sqrt{2} \approx 2.4\). We choose \(x=-1\), \(x=0\) and \(x=3\) as our test values and find \(f(-1)= 2\), which is \((+)\); \(f(0)=-1\) which is \((-)\); and \(f(3)=2\) which is \((+)\) again. Our solution to \(x^2-2x-1>0\) is where we have \((+)\), so, in interval notation \(\left(-\infty, 1-\sqrt{2}\right) \cup \left(1+\sqrt{2},\infty\right)\).
To check the inequality \(x^2 - 2x > 1\) graphically, we set \(g(x) = x^2-2x\) and \(h(x)=1\). We are looking for the \(x\) values where the graph of \(g\) is above the graph of \(h\).
-
To solve \(2x^2 \leq 3-x\), we first get \(0\) on one side of the inequality which yields \(2x^2+x-3 \leq 0\). We find the zeros of \(f(x) = 2x^2 + x - 3\) by solving \(2x^2 + x - 3 = 0\) for \(x\). Factoring gives \((2x+3)(x-1)=0\), so \(x = -\frac{3}{2}\) or \(x = 1\). We place these values on the number line with \(0\) above them and choose test values in the intervals \(\left(-\infty, -\frac{3}{2}\right)\), \(\left(-\frac{3}{2},1\right)\) and \((1,\infty)\). For the interval \(\left(-\infty, -\frac{3}{2}\right)\), we choose \(x=-2\); for \(\left(-\frac{3}{2},1\right)\), we pick \(x=0\); and for \((1,\infty)\), \(x=2\). Evaluating the function at the three test values gives us \(f(-2) = 3 > 0\), so we place \((+)\) above \(\left(-\infty, -\frac{3}{2}\right)\); \(f(0)=-3 < 0\), so \((-)\) goes above the interval \(\left(-\frac{3}{2},1\right)\); and, \(f(2) = 7\), which means \((+)\) is placed above \((1,\infty)\).
Applications Involving Quadratic Functions
A ball is thrown upward from the top of a 40 foot high building at a speed of 80 feet per second. The ball’s height above ground can be modeled by the equation \(H(t)=-16 t^2 +80t+40\).
- When does the ball reach the maximum height?
- What is the maximum height of the ball?
- When does the ball hit the ground?
- When will the ball be above 104 feet?
- Solutions
-
- The ball reaches the maximum height at the vertex of the parabola.\[h= -\dfrac{80}{2(-16)} = \dfrac{80}{32} = \dfrac{5}{2} =2.5. \nonumber \]The ball reaches a maximum height after 2.5 seconds.
- To find the maximum height, find the \( y \)-coordinate of the vertex of the parabola.\[ k = H\left( -\dfrac{b}{2a} \right) = H( 2.5 ) =-16 ( 2.5 )^2 +80( 2.5 )+40 =140. \nonumber \]The ball reaches a maximum height of 140 feet.
-
We need to solve\[ 0 = -16t^2 + 80t + 40 \nonumber \]which has the same solution as\[ 0 = 16t^2 - 80t - 40. \nonumber \]Since the coefficients are all divisible by \( 8 \), we divide both sides by \( 8 \) to get\[ 0 = 2t^2 - 10t - 5. \nonumber \]Try as you might, the quadratic on the right side doesn't factor. Therefore, we use the Quadratic Formula. Before going any further, let's compute the discriminant (if it's negative, then we can stop because this would mean the ball would never touch the ground).\[ \text{discriminant} = b^2 - 4ac = (-10)^2 - 4(2)(-5) = 100 +40 = 140 > 0. \nonumber \]Thus, we continue with the Quadratic Formula.\[ \begin{array}{rclcl}
t & = & \dfrac{-b \pm \sqrt{\text{discriminant}}}{2a} & \quad & \left( \text{Quadratic Formula} \right) \\[6pt]
& = & \dfrac{10 \pm \sqrt{140}}{2(2)} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{10 \pm \sqrt{140}}{4} & \quad & \left( \text{simplifying} \right) \\[6pt]
\end{array} \nonumber \]Because this is an application and the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.\[t \approx 5.458 \quad \text{or} \quad t \approx -0.458 \nonumber \]The second answer is outside the reasonable domain of our model, so we conclude the ball will hit the ground after about 5.458 seconds. See Figure \( \PageIndex{ 18 } \).
Figure \( \PageIndex{ 18 } \) -
This question is akin to asking us to solve \( -16t^2 + 80t + 40 > 104 \). We starting by getting \( 0 \) on one side:\[ -16t^2 + 80t - 64 > 0. \nonumber \]Again, it would be better to get that lead coefficient to be positive, so we multiply both sides by \( -1 \) (remember that the inequality flips):\[ 16t^2 - 80t + 64 < 0. \nonumber \]We then divide both sides by the GCF of each of the coefficients (which is 16):\[ t^2 - 5t + 4 < 0. \nonumber \]We want to know when the outputs of this parabola are below the \( t \)-axis. We know this parabola is opening upward and the left side factors to \( (t - 4)(t - 1) \). This implies the roots are \( t =1 \) and \( t = 4 \). This means that our solution is the interval \( (1,4) \) (see Figure \( \PageIndex{ 19 } \)).
Figure \( \PageIndex{ 19 } \)
It's important to note that Figure \( \PageIndex{ 19 } \) in Example \( \PageIndex{ 15 } \) doesn't represent the path of the ball. We arrived at the graph in Figure \( \PageIndex{ 19 } \) by manipulating an inequality based on the function for the path of the ball's travel.
A rock is thrown upward from the top of a 112-foot high cliff overlooking the ocean at a speed of 96 feet per second. The rock’s height above ocean can be modeled by the equation \(H(t)=-16 t^2 +96t+112\).
- When does the rock reach the maximum height?
- What is the maximum height of the rock?
- When does the rock hit the ocean?
Solving Absolute Value Equations Involving Quadratics
We finish this section with two synthesis synthesis topics:
- Solving equations involving the composition of the absolute value function and a quadratic function, and
- Graphing the absolute value of a quadratic function.
Both of these topics will seem more complex than our previous work because you are combining previously-learned skills (called synthesizing skills). The crucial thing to remember is the piecewise definition of the absolute value\[ | \blacksquare | = \begin{cases}
\blacksquare , & \text{ if } \blacksquare \geq 0 \\[6pt]
-\blacksquare , & \text{ if } \blacksquare < 0 \\[6pt]
\end{cases} \nonumber \]We will be combining this knowledge with our graphing skills for quadratic functions.
Solve each absolute value equation.
- \( \left|2x^2 + x - 3\right| = 0 \)
- \( \left|2x^2 + x - 3\right| = -6 \)
- \( \left|x^2 - 5x \right| = 6 \)
- Solution
-
- As with any synthesis question, you need to think critical more than with basic questions. In this case, you are being asked when the left side is zero. The only time the absolute value is zero is when its argument is zero. So this question boils down to solving\[ 2x^2 + x - 3 = 0. \nonumber \]In Example \( \PageIndex{ 13 } \), we found the zeros of this function to be \( x = -\frac{3}{2} \) and \( x = 1 \). Thus, the solutions to this absolute value equation are \( x = -\frac{3}{2} \) and \( x = 1 \).
- Again, we have a synthesis question where thinking critically ahead of time saves time. We are being asked, "When is an absolute value equal to a negative number?" We know that this cannot happen. Thus, this absolute value equation does not have a solution. In Mathematics, we often say the solution DNE (does not exist).
-
We finally arrive at an example which requires more than superficial thinking. Using the definition of the absolute value, we see that the left side has the equivalent form\[ | x^2 - 5x | = \begin{cases}
x^2 - 5x , & \text{ if } x^2 - 5x \geq 0 \\[6pt]
-(x^2 - 5x) , & \text{ if } x^2 - 5x < 0 \\[6pt]
\end{cases} \nonumber \]So, we need to determine when \( x^2 - 5x = 0 \) because this is where the "pivot point" of the piecewise definition occurs. We factor the equation to \( x(x-5) = 0 \) and note that the solutions are \( x = 0 \) and \( x = 5 \). Since \( y = x^2 - 5x \) is a parabola opening upward, and we just found out that the zeros of this function are \( 0 \) and \( 5 \), this function is at or above the horizontal axis on the interval \( \left( -\infty, 0 \right] \cup \left[ 5, \infty \right) \). Thus, it is below the horizontal axis on \( \left( 0, 5 \right) \). We modify our piecewise definition accordingly.\[ | x^2 - 5x | = \begin{cases}
x^2 - 5x , & \text{ if } x \in \left( -\infty, 0 \right] \cup \left[ 5, \infty \right) \\[6pt]
-(x^2 - 5x) , & \text{ if } x \in \left( 0,5 \right) \\[6pt]
\end{cases} \nonumber \]Thus, our original equation becomes the following:\[ \begin{array}{rcc|cc}
& \textbf{Case #1} & \quad & \quad & \textbf{Case #2} \\[6pt]
& x \in \left( -\infty, 0 \right] \cup \left[ 5, \infty \right) & \quad & \quad & x \in \left( 0,5 \right) \\[6pt]
\hline & x^2 - 5x = 6 & \quad & \quad & -(x^2 - 5x) = 6 \\[6pt]
\implies & x^2 - 5x - 6 = 0 & \quad & \quad & -x^2 + 5x - 6 = 0 \\[6pt]
\implies & (x - 6)(x + 1) = 0 & \quad & \quad & x^2 - 5x + 6 = 0 \\[6pt]
\implies & x = 6 \text{ or } x = -1 & \quad & \quad & (x - 2)(x - 3) = 0 \\[6pt]
\implies & & \quad & \quad & x = 2 \text{ or } x = 3 \\[6pt]
\end{array} \nonumber \]Hence, our solutions to the original absolute value equation are \( x = -1, 2, 3 \) and \( 6 \).
It is very important to check that the solutions from Example \( \PageIndex{ 17} \) occur in the proper intervals. That is, we arrives at the solutions \( x = -1 \) and \( x = 6 \) whenever \( x \in \left( -\infty, 0 \right] \cup \left[ 5, \infty \right) \). Both of these solutions exist in these intervals, so we are fine. I leave it to you to check that the second solution set exists within the interval \( (0,5) \).
We conclude this section with a graph of a more complicated absolute value function.
Graph \(f(x) = |x^2 - x - 6|\).
- Solution
-
Using the definition of absolute value, we have\[ f(x) = \left\{ \begin{array}{rcl}
\left(x^2 - x - 6\right), & \text{ if } & x^2 - x - 6 \geq 0 \\
-(x^2 - x - 6), & \text{ if } & x^2 - x - 6 \lt 0 \\
\end{array} \right.\nonumber \]We begin by graphing \(y = g(x) = x^2 - x - 6\) using the intercepts and the vertex. To find the \(x\)-intercepts, we solve \(x^2 - x - 6 = 0\). Factoring gives \((x-3)(x+2)=0\) so \(x=-2\) or \(x=3\). Hence, \((-2,0)\) and \((3,0)\) are \(x\)-intercepts. The \(y\)-intercept \((0,-6)\) is found by setting \(x=0\). To plot the vertex, we find\[x = -\frac{b}{2a} = -\frac{-1}{2(1)} = \frac{1}{2}\), and \(y = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)-6 = -\frac{25}{4} = -6.25.\nonumber \]Plotting, we get the parabola on the left side of Figure \( \PageIndex{20} \).To obtain points on the graph of \(y = f(x) = |x^2-x-6|\), we can take points on the graph of \(g(x) = x^2-x-6\) and apply the absolute value to each of the \(y\) values on the parabola. We see from the graph of \(g\) that for \(x \leq -2\) or \(x \geq 3\), the \(y\)-values on the parabola are greater than or equal to zero (since the graph is on or above the \(x\)-axis), so the absolute value leaves these portions of the graph alone. For \(x\) between \(-2\) and \(3\), however, the \(y\)-values on the parabola are negative. For example, the point \((0,-6)\) on \(y = x^2-x-6\) would result in the point \((0,|-6|) = (0,-(-6))= (0,6)\) on the graph of \(f(x) = |x^2-x-6|\). Proceeding in this manner for all points with \(x\)-coordinates between \(-2\) and \(3\) results in the graph on the right side of Figure \( \PageIndex{20} \).
Figure \( \PageIndex{20} \)
If we take a step back and look at the graphs of \(g\) and \(f\) in the Example \( \PageIndex{ 18} \), we notice that to obtain the graph of \(f\) from the graph of \(g\), we reflect a
portion
of the graph of \(g\) about the \(x\)-axis. We can see this analytically by substituting \(g(x) = x^2-x-6\) into the formula for \(f(x)\) and calling to mind the concept of reflections from transformations of functions.\[f(x) = \left\{ \begin{array}{rcl}
g(x), & \text{ if } & g(x) \geq 0 \\
-g(x), & \text{ if } & g(x) \lt 0 \\
\end{array} \right.\nonumber \]The function \(f\) is defined so that when \(g(x)\) is negative (i.e., when its graph is below the \(x\)-axis), the graph of \(f\) is its refection across the \(x\)-axis. This is a general template to graph functions of the form \(f(x) = |g(x)|\). From this perspective, the graph of \(f(x) = |x|\) can be obtained by reflecting the portion of the line \(g(x) =x\) which is below the \(x\)-axis back above the \(x\)-axis creating the characteristic "\(vee\)" shape.
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