3.2: Describing Relationships with Quadratic Functions
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Goal: Describing Relationships with Quadratic Functions.
Let's graph the parabola \(y=f(x)=-2(x-1)^2 +3 \) using the transformations of \(y=x^2\) that we learned about in section 3.1. The graph of \(f\) is a parabola that is
- Shifted right 1 unit
- Vertically stretched by a factor of 2
- Reflected about the x-axis
- Shifted up 3 units
Notice, these transformations yield a graph with a similar shape to the parabola \(y=x^2\) just slightly altered. In general, we can conclude that the graph of the quadratic function \(y=a(x-h)^2+k\) is a parabola. In analyzing relationships that can described by a parabola similar to that of our introductory example, we will need to answer the following questions among others:
- Where is \(y=f(x)\) increasing? Where is \(y=f(x)\) decreasing?
- What is the maximum (or minimum) value of \(y=f(x)\)? Where does it occur?
- Where is \(y=f(x)\) positive? Where is \(y=f(x)\) negative?
In order, to address these questions we will need to identify these key features of the graph:
- The vertex
- The end behavior
- The zeros
The vertex of a parabola is the point where the graph changes direction.
The value \(x=a \) is a zero of the polynomial \(y=f(x)\) if \(f(a)=0 \).
The end behavior of a function describes what value(s) \(y\) approaches as x approaches \( \pm \infty \).
Focus: Identifying the End Behavior of a Quadratic Function in Vertex Form.
Notice that the transformed parabolas either open upward or open downward at the ends of the graph. Notice, when a parabola opens upward on both ends, as \(x \to \pm \infty \), \(y \to \infty \) as shown below. When a parabola opens upward on both ends, as \(x \to \pm \infty \), \(y \to -\infty \) as shown below.
Now the parabola \(y=x^2\) opens upward on both ends, but the final graph in our introductory example opened downward. Which transformation caused a change in the end behavior? The reflection changed the end behavior from opened upward to opening downward. Whereas \(y=2(x-1)^2 \) is a parabola opening upward on both ends with positive stretch factor \(a=2\), \(y=-2(x-1)^2 \) is a parabola that opens downward on both ends due to the reflection caused by the negative stretch factor \(a=-2\). In general, when the stretch factor \(a\) is positive, there is a vertical stretch which does not change the end behavior. When the stretch factor \(a\) is negative, there is a reflection about the x-axis which changes the end behavior so that the parabola opens downward on both ends.
The graph of the quadratic function \(y=a(x-h)^2+k\) is a parabola.
If \(a>0\), the parabola opens upward on both ends.
If \(a<0\), the parabola opens downward on both ends.
Describe the end behavior of the parabola \(f(x)=3(x+4)^{2}-1\).
Solution
The parabola opens upward since the stretch factor \(a=3>0\) is positive. There is no reflection.
Focus: Identifying the Vertex of a Quadratic Function in Vertex Form
The vertex of a parabola is where the graph changes direction from increasing to decreasing (or vice versa). In our introductory example where \(y=f(x)=-2(x-1)^2+3 \), notice only the horizontal and vertical shifts changed the location of the vertex of the parabola. The vertex is unaffected by vertical stretches, vertical compressions, and reflections. The shifts caused the vertex of \(y=x^2\) to move from the origin (0,0) right one and up three units to (1,3). In general, we say that the quadratic function \(y=a(x-h)^2 +k \) is in vertex form since we can identify the vertex directly from the horizontal and vertical shifts.
The vertex of the parabola \(y=a(x-h)^2+k \) is at \( (h,k) \).
Let \(y=-3(x+1)^2 +5 \).
- Identify the vertex.
- Describe the end behavior. Then graph the parabola.
- Where is \(f\) increasing? Where is \(f\) decreasing?
- What is the maximum value? Where does it occur?
Solution
- The vertex is at (-1,5). The shift left one and shift up five moves the vertex to (-1,5).
- The parabola opens down on both ends (as \(x \to \pm \infty , y \to -\infty \) ) since the stretch factor \(a=-3\) is negative.
- \(f\) increasing on \( (\infty,-1) \). \(f\) decreasing on \( [-1,\infty) \).
- The maximum value of \(f\) is \(y=5\) that occurs at \(x = -1 \).
A ball is thrown upward. The height in feet after \(t\) seconds is given by \(h=f(t)=-16(t-2)^2+64 \).
- What is the maximum height? When does the ball reach its maximum height?
- When is the ball going upward?
Solution
- The vertex of the parabola is at (2,64) since the shifts move the vertex right two and up 64. The parabola opens downward since the stretch factor \(a=-16\) is negative. The maximum height of the ball is 64 ft. It reaches the maximum height at \(t=2\) seconds.
- The ball going upward when \( 0<t<2 \).
Find a formula for a parabola with vertex at (3,7) that contains the point (1,0).
Solution
For the vertex to be at (3,7), the parabola must be shifted right three and up seven units. Therefore, the parabola must have the formula \[y=a(x-3)^2+7 \nonumber \]To find the stretch factor a, we can use the point (1,0) on the graph. If we plug in \(x=3\) into our formula, we get \(y=7\). \[0=a(1-3)^2+7 \nonumber \]Then evaluating and solving for \(a\) \[0=a(-2)^2+7 \nonumber \] \[0=a(4)+7 \nonumber \] \[-7=a(4) \nonumber \] \[\dfrac{-7}{4}=a \nonumber \] So the formula is \[y=\dfrac{-7}{4}(x-3)^2+7 \nonumber \]
Let \(y=f(x)=2(x-5)^2 -4 \).
- Identify the vertex.
- Describe the end behavior. Then graph the parabola.
- Where is \(f\) increasing? Where is \(f\) decreasing?
- What is the minimum value? Where does it occur?
- Answer
-
- The vertex is at (5,-4).
- The parabola opens up on both ends (as \(x \to \pm \infty , y \to \infty \) ) since the stretch factor \(a=2\) is positive.
- \(f\) increasing on \( [5,\infty) \). \(f\) decreasing on \( (-\infty,5) \).
- The minimum value of \(f\) is \(y=-4\) that occurs at \(x = 5 \).
Find a formula for a parabola with vertex at (-2,4) that contains the point (1,2).
- Answer
-
\(y=\dfrac{-2}{9}(x+2)^2+4 \)
Focus: The Zeros of a Parabola in Vertex Form
The zeros of a parabola are the values of \(x\) where \(y=f(x)=0\). The zeros are helpful in determining the sign of a quadratic function. If a quadratic function changes sign from positive to negative (or vice versa), it will do so at a zero.
Let \(f(x)=-2(x-1)^2+3\).
- Find the zeros.
- Determine the end behavior. Then graph the parabola.
- Where \(f\) is positive? Where is \(f\) negative?
Solution
- To find the zeros, let \(y=0\). So \[-2(x-1)^2+3=0 \nonumber \]Since there is only one x-term in the equation, we can solve by isolating \(x\). \[-2(x-1)^2=-3 \nonumber \] \[(x-1)^2=\dfrac{-3}{-2} \nonumber \] \[x-1 = \pm \sqrt{\dfrac{3}{2}} \nonumber \] \[x=1 \pm \sqrt{\dfrac{3}{2}} \nonumber \] \[x \approx 2.225 \text{ and } x \approx -0.225 \nonumber \]
- The parabola opens downward on both ends (as \(x \to \pm \infty, y \to \pm -\infty \)) since \(a=-2\) is negative.
- \(f\) is positive on \( (-0.225,2.225) \) since the graph is above the x-axis on this interval. \(f\) is negative on \( (-\infty,-0.225) \cup (2.225,\infty) \) since the graph is below the x-axis on these intervals.
Let \(f(x)=3(x+2)^2-4\).
- Find the zeros.
- Determine the end behavior. Then graph the parabola.
- Where \(f\) is positive? Where is \(f\) negative?
- Answer
-
- The zeros are \(x=-2 \pm \dfrac{2}{\sqrt{3}} \). So \(x \approx -0.845 \) and \(x \approx -3.155 \)
- The parabola opens upward on both ends since \(a=3\) is positive.
- \(f\) is positive on \( (-\infty,-3.155) \cup (-0.845,\infty) \). \(f\) is negative on \( (-3.155, -0.845) \) .
Focus: Quadratic Functions in Standard Form
As we learned earlier in this section, the quadratic function \(y=2(x+1)^2-3 \) in vertex form has an end behavior that opens up on both ends (\(x \to \pm \infty, y \to \pm \infty \)) since the stretch factor \(a=2\) is positive. However, if we perform the operations in the quadratic function \[y=2(x+1)^2-3=2(x+1)(x+1)-3\nonumber \]\[y=2[x^2+2x+1]-3\nonumber \]\[y=2x^2+4x+2-3\nonumber \]\[y=2x^2+4x-1\nonumber \]First, a quadratic function can also be written in the form \(y=ax^2+bx+c\). We call this a quadratic function in standard form.
A function written in form \(y=ax^2+bx+c\) is called a quadratic function in standard form.
Secondly, notice the stretch factor \(a=2\) of the quadratic function in vertex form was distributed to the \(x^2\) term of our quadratic function in standard form. Essentially, the stretch factor \(a=2\) in vertex form is the coefficient \(a=2\) of \(x^2\) in standard form. We used the stretch factor to determine the end behavior of the parabola. Since the stretch factor \(a=2\) is positive, whether the quadratic function in vertex form as \(y=2(x+1)^2-3 \) or in standard form as \(y=2x^2+4x-1\), the parabola opens upward on both sides (\(x \to \pm \infty, y \to \pm \infty \)).
For a quadratic function in standard form. \(y=ax^2+bx+c\), if \(a>0\), the parabola opens up on both ends. If \(a<0\), the parabola opens down on both ends.
Determine the end behavior of the quadratic function \(y=-3x^2+4x+1 \)
Solution
Since the stretch factor \(a=-3\) is negative, the parabola opens down on both ends.
Focus: The Zeros of a Quadratic Function in Standard Form
To find the zeros of a quadratic function like \(y=x^2-5x+4\), we let \(y=0\) and solve the equation \[x^{2} -5x+4=0\nonumber\]We can't solve by isolating \(x\) like we did in example 3.2.5 since there are two unlike variable terms \(x^2\) and \(-5x\). However, this is an equation that we learned how to solve in algebra. We have two options
Option 1: Solve by factoring
Factoring the right side \[x^{2} -5x+4=0\nonumber\]\[(x-4)(x-1)=0\nonumber\]Then setting each factor to zero, using the zero product property (if two factors multiply two zero, one of them must be zero) \[x-4=0 \text{ and } x-1=0\nonumber\]Then solving for \(x\) in each equation, we have \[x=4 \text{ and } x=1\nonumber\]
Option 2: Solve using the quadratic formula
Recall, the quadratic formula from algebra allows us to solve these types of quadratic equations.
The solutions to the equation \(ax^{2} +bx+c\) are of the form \(x=\dfrac{-b\pm \sqrt{b^{2} -4ac} }{2a}\)
In our example, we need to first identity the coefficients, \(a=1\), \(b=-5\), and \(c = 4\). Then substitute them into the quadratic formula \[x=\dfrac{-(-5)\pm \sqrt{(-5)^{2} -4(1)(4)} }{2(1)} \nonumber \]\[x=\dfrac{5\pm \sqrt{25 -19}}{2}\nonumber \]\[x=\dfrac{5\pm \sqrt{9}}{2} \nonumber \]\[x=\dfrac{5\pm 3}{2} \nonumber \]\[x=\dfrac{5+ 3}{2}=\dfrac{2}{2}=1 \text{ and } x=\dfrac{5+ 3}{2}=\dfrac{8}{2}=4\nonumber \]
In general, we can use either strategy to solve a quadratic equation like this and find the zeros of a quadratic function.
Let \(f(x)=2x^{2} +5x-3\)
- Find the zeros
- Identify the end behavior. Then graph the parabola
- Determine where \(f\) is positive. Where is \(f\) is negative?
Solution
- To find the zeros, let \(y=0\). So \[2x^{2} +5x-3=0\nonumber\]The solving by factoring \[(2x-1)(x+3)=0\nonumber\]Setting each factor to zero and solving, \[2x-1=0 \text{ and } x+3=0\nonumber\]Then solving for \(x\) in each equation, we have zeros at \[x=\dfrac{1}{2} \text{ and } x=-3\nonumber\]
- Since the stretch factor \(a=2\) is postive, the parabola opens up on both ends ( as \(x \to \pm \infty, y \to \infty \) ).
- \(f\) positive on \( (-\infty,-3) \cup (\dfrac{1}{2},\infty) \). \(f\) negative on \( (-3),\dfrac{1}{2}) \).
Notice how each zero is related to a corresponding factor. In intoductory example in this part, notice we got the zero \(x=4\) of \(y=x^2-5x+4\) by setting the factor \( (x-4) \) to zero and solving. Similarly, we found the zero \(x=1\) by setting the factor \( (x-1) \) to zero and solving. In example 3.2.7, we found the zero \(x=-3\) of \(f(x)=2x^{2} +5x-3\) by setting the factor \( (x+3) \) to zero and solving. If \(x=2\) was a zero of another polynomial \(f(x) \), then the polynomial must have a factor of \( (x-2) \). This idea is the basis of The Factor Theorem.
A polynomial \(y=f(x) \) has a zero \(x=k\) if and only if \( (x-k) \) is a factor of \(f(x)\).
Given that \(x=2\) and \(x=-4\) are zeros of a quadratic function.
- Find a possible formula
- Find a formula for a parabola with these zeros that passes through the point (0,3).
Solution
- Since \(x=2\) is a zero, \( (x-2) \) is a factor. Similarly, Since \(x=-4\) is a zero, \( (x+4) \) is a factor. So a possible formula for a quadratic function with these zeros is \(f(x)=(x-2)(x+4)=x^2-2x+8 \).
- Notice all of the functions in the graph below have zeros at \(x=2\) and \(x=-4\), but only one of them passes through the point (0,3). All of them are vertical stretches, vertical compressions, and/or reflections of the graph in part (a). In addition to the factors \( (x-2) \) and \( (x+4) \), the formula for the quadratic function that passes through the point (0,3) could have a constant stretch factor \(a\). So \[f(x)=a(x-2)(x+4) \nonumber\]To find the stretch factor \(a\), we can use the known point \(x=0\), \(y=3\). Substituting these values into our formula. \[3=a(0-2)(0+4) \nonumber\]\[3=a(-2)(4) \nonumber\]\[3=a(-8) \nonumber\]Then solving for \(a\), we have \[a=\dfrac{3}{-8} \nonumber\]So our formula is \[f(x)=-\dfrac{3}{8}(x-2)(x+4) \nonumber\]
Find a formula for a polynomial with zeros \(x=0\), \(x=1\), and \(x=-2\) that passes through the point (3,4).
Solution
Since \(x=1\) is a zero, \( (x-1) \) is a factor. Since \(x=-2\) is a zero, \( (x+2) \) is also a factor. Since \(x=0\) is also a zero, \( (x) \) is a factor. So a possible formula \(f\) is \[f(x)=ax(x-1)(x+2) \nonumber\]To find the strech factor \(a\), we can plug in the known point \(x=3\), \(y=4\) and solve for \(a\)\[4=a(3)(3-1)(3+2) \nonumber\]\[4=a(3)(2)(5) \nonumber\]\[4=a(30) \nonumber\]\[a=\dfrac{4}{30}=\dfrac{2}{15} \nonumber\]So our formula is \[f(x)=\dfrac{2}{15}x(x-1)(x+2) \nonumber\]
The polynomial in example 3.2.9 had three distinct zeros \(x=0\), \(x=1\), and \(x=-2\) and therefore three corresponding linear factors \(f(x)=\dfrac{2}{15}x(x-1)(x+2) \). When we multiply the three linear factors, we get \(f(x)=\dfrac{2}{15}x(x-1)(x+2)= \dfrac{2}{15}x(x^2+x-2)=\dfrac{2}{15}x^3+\dfrac{2}{15}x^2-\dfrac{4}{15}x\). This is not a quadratic function as we can see from its graph below.
The polynomial in example 3.2.8 has two zeros and therefore two corresponding linear factors \(f(x)=-\dfrac{3}{8}(x-2)(x+4) \). When we multiply these two linear factors, we get \(f(x)=-\dfrac{3}{8}(x-2)(x+4) = -\dfrac{3}{8}(x^2-2x-8)= -\dfrac{3}{8}x^2+\dfrac{3}{4}+3 \), a quadratic function. This really highlights the relationship between the size of the polynomial, the number of linear factors, and the number of zeros.
The degree of a polynomial is the largest exponent that occurs that occurs in each of the individual terms
The quadratic polynomial in example 3.2.8, \(f(x)=-\dfrac{3}{8}(x-2)(x+4) = -\dfrac{3}{8}x^2+\dfrac{3}{4}+3 \), has degree two, two linear factors, and two distinct zeros. The polynomial in example 3.2.9, \(f(x)=\dfrac{2}{15}x(x-1)(x+2)=\dfrac{2}{15}x^3+\dfrac{2}{15}x^2-\dfrac{4}{15}x\) has degree three, linear factors, and three distinct zeros. Be extension, a polynomial of degree five, like \(f(x)=x^5 -5x^3 +4x \), has five linear factors, and at most five distinct zeros.
A polynomial of degree n has at most n distinct zeros.
Let \(f(x)=3x^2-14x-5\)
- Find the zeros
- Identify the end behavior. Then graph the parabola
- Determine where \(f\) is positive. Where is \(f\) is negative?
- Answer
-
- The zeros are \(x = -\dfrac{1}{3} \) and \(x = 5 \)
- The parabola opens up on both ends since \(a=3>0\).
- \(f\) positive on \( (-\infty,-\dfrac{1}{3}) \cup (5,\infty) \). \(f\) negative on \( (-\dfrac{1}{3},5) \).
Find a formula for a polynomial with zeros \(x=-1\) and \(x=3\) that passes through the point (0,7).
- Answer
-
\(f(x)=-\dfrac{7}{3}(x+1)(x-3) \)
Let \(f(x)=x^2+x+1\)
- Find the zeros
- Identify the end behavior. Then graph the parabola
- Determine where \(f\) is positive. Where is \(f\) is negative?
- Answer
-
- There are no real zeros. The zeros are \(x=\dfrac{-1 \pm \sqrt{-3}}{2} \) which are complex numbers.
- The parabola opens up on both ends since \(a=1>0\).
- \(f\) is positive on \( -\infty,\infty) \). \(f\) is never negative.
Notice, only real zeros are x-intercepts. When we graph points on the coordinate axis, they have real components. In exercise 3.2.6, we encountered only complex zeros \(x=\dfrac{-1 \pm \sqrt{-3}}{2} \). This meant that the quadratic function had no x-intercepts where the function could change sign. Since the graph opened up on both ends where \(y \to \infty \) as \( x \to +\pm \infty \), the parabola must always be above the x-axis. Therefore, \(f\) is positive for all real values of \(x\). We will explore complex zeros in more detail in section 3.3.
Secondly, notice that not all quadratic functions have two distinct zeros. For example, \(y=(x-1)^2\) has only one zero \(x=1\). However, notice that there are two linear factors of \( (x-1) \) that correspond to the zero \(x=1\). The graph of \(y=(x-1)^2\) is a horizontal shift right one of the graph of \(y=x^2\) moving the x-intercept to \(x=1\) as shown in the graph below. With only one distinct real zero, there is only one x-intercept on the graph.
Focus: The Vertex of a Quadratic Function in Standard Form
In addition to the zeros and end behavior, we need to be able to identify the vertex of a quadratic function in standard form in order to analyze any relationship described by it. First, notice that the graph of \(y=ax^2+bx+c\) is a vertical shift c units of the graph of \(y=ax^2+bx\) as shown in the graph below. In the graph, we have assumed \(a<0\) and \(c>0\).
Notice the vertical shift c units changed the y-coordinate of the vertex of \(y=ax^2+bx\) but not the x-coordinate of the vertex. Both \(y=ax^2+bx+c\) and \(y=ax^2+bx\) have the same x-coordinate of their vertices. If we can identify the x-coordinate of the vertex of \(y=ax^2+bx\), we have the x-coordinate of the vertex of a parabola in standard form. To do so, let's first find the zeros of \(y=ax^2+bx\). Letting \(y=0\) and factoring we have, \[ax^2+bx=0 \nonumber\]\[x(ax+b)=0 \nonumber\]Setting each factor to zero and solving \[x=0 \text{ and } ax+b=0 \nonumber\]\[x=0 \text{ and } x=\dfrac{-b}{a} \nonumber\]Notice a parabola is symmetric about a vertical line through the vertex (called the axis of symmetry). So the distance from the x-coordinate of the vertex to each zero on either side is the same. Therefore, the distance from the zero at \(x=0\) to the x-coordinate of the vertex is half of the distance between the two zeros. Therefore, the x-coordinate of the vertex is \(x_{vertex}=\dfrac{1}{2}(\dfrac{-b}{a})=\dfrac{-b}{2a} \) as illustrated in the graph below.
The x-coordinate of the vertex of a quadratic function \(y=ax^2+bx+c\) in standard form is located at \[x_{vertex}=-\dfrac{b}{2a} \nonumber\]
Let \(f(x)=x^{2} -6x+1\).
- Find the vertex
- Identify the end behavior. Then sketch a graph.
- Determine where \(f\) is increasing. Where is \(f\) decreasing?
- What is the minimum value of \(f\)? Where does it occur?
Solution
- The x-coordinate of the vertex using \(x_{vertex}=-\dfrac{b}{2a} \). Since \(a=1\) and \(b=-6\), the x-coordinate of the vertex is \[x_{vertex}=-\dfrac{b}{2a} =-\dfrac{-6}{2(1)} =\dfrac{6}{2} =3 \nonumber\]To find the y-coordinate of the vertex now that we know the x-coordinate of the vertex, we can evaluate \(f\) at \(x=3\), so \[y_{vertex}=f(3)=(3)^2-6(3)+1 = -8\nonumber\]The vertex is at (3,-8).
- The parabola opens upward since \(a=1>0\).
- \(f\) is increasing on \( [3,\infty) \) and decreasing on \( (-\infty,3) \).
- The minimum occurs at the vertex. The minimum value is \(y=-8\). It occurs at \(x=3\).
Let \(g(x)=2x^{2}+4x+1\).
- Find the vertex
- Identify the end behavior. Then sketch a graph.
- Determine where \(g\) is increasing. Where is \(g\) decreasing?
- What is the minimum value of \(g\)? Where does it occur?
Solution
- Since \(a=2\) and \(b=4\), the x-coordinate of the vertex is \[x_{vertex}=-\dfrac{b}{2a} =-\dfrac{4}{2(2)} =\dfrac{-4}{4} =-1 \nonumber\]Then we can evaluate \(g\) at \(x=-1\) to find y-coordinate of the vertex \(g\), so \[y_{vertex}=g(-1)=2(-1)^2+4(-1)+1 = -1\nonumber\]The vertex is at (-1,-1).
- The parabola opens upward since \(a=1>0\).
- \(f\) is increasing on \( [3,\infty) \) and decreasing on \( (-\infty,3) \).
- The minimum occurs at the vertex. The minimum value is \(y=-8\). It occurs at \(x=3\).
The profit of a company that produces and sells \(x\) units of a product can be modeled by the function \(P=-0.01x^2+120x-1000\). What is the company's maximum profit? What production level will maximize profit?
Solution
The profit is modeled by a quadratic function. The maximum will occur at the vertex. To find the vertex, the x-coordinate of the vertex is \[x_{vertex}=-\dfrac{b}{2a} =-\dfrac{120}{2(-.01)} =6000 \nonumber\] since \(a=-0.01\) and \(b=120\). The P-coordinate of the vertex can be found by evaluating \[P_{vertex}=-0.01(6000)^2+120(6000)-1000 = 359,000 \nonumber\]The vertex is at \(x=6000\) and \(P=359,000\). The parabola opens downward since \(a=-0.01<0\) as shown in the graph below. So the maximum profit is P=$359,000 when 6000 units are produced and sold.
Let \(f(x)=-2x^{2}+8x+7\).
- Find the vertex
- Identify the end behavior. Then sketch a graph.
- Determine where \(g\) is increasing. Where is \(g\) decreasing?
- What is the maximum value of \(g\)? Where does it occur?
- Answer
-
- The x-coordinate of the vertex is \[x_{vertex}=-\dfrac{b}{2a} =-\dfrac{8}{2(-2)} =2 \nonumber\]The y-coordinate of the vertex is \[y_{vertex}=-2(2)^2+8(2)+7 = 15 \nonumber\]So the vertex is at (2,15).
- The parabola opens upward since \(a=-2<0\).
- \(f\) is increasing on \( -\infty,2) \) and decreasing on \( [2,\infty) \).
- The maximum value is \(y=15\). It occurs at \(x=2\).
A ball is thrown upward from the ground at \(64 \dfrac{ft.}{s} \). The height of the ball in feet after \(t\) seconds is given by \(h=-16t^2+64t\) while it is in flight. What is the maximum height of the ball? When does it reach its maximum height.
- Answer
-
The t-coordinate of the vertex is \[t_{vertex}=-\dfrac{b}{2a} =-\dfrac{64}{2(-16)} =2 \nonumber\]The h-coordinate of the vertex is \[h_{vertex}=-16(2)^2+64(2) = 64 \nonumber\]Since the parabola opens downward, the ball reaches a maximumm height of 64 ft. at \(t=2\) seconds.
Important Topics of this Section
- Quadratic functions
- Standard form
- Vertex form
- Vertex as a maximum / vertex as a minimum
- Zeros of quadratic functions
- The end behavior of quadratic functions