1.2: Basic Classes of Functions
- Page ID
- 139030
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\( \newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\)
( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\id}{\mathrm{id}}\)
\( \newcommand{\Span}{\mathrm{span}}\)
\( \newcommand{\kernel}{\mathrm{null}\,}\)
\( \newcommand{\range}{\mathrm{range}\,}\)
\( \newcommand{\RealPart}{\mathrm{Re}}\)
\( \newcommand{\ImaginaryPart}{\mathrm{Im}}\)
\( \newcommand{\Argument}{\mathrm{Arg}}\)
\( \newcommand{\norm}[1]{\| #1 \|}\)
\( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\)
\( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\AA}{\unicode[.8,0]{x212B}}\)
\( \newcommand{\vectorA}[1]{\vec{#1}} % arrow\)
\( \newcommand{\vectorAt}[1]{\vec{\text{#1}}} % arrow\)
\( \newcommand{\vectorB}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vectorC}[1]{\textbf{#1}} \)
\( \newcommand{\vectorD}[1]{\overrightarrow{#1}} \)
\( \newcommand{\vectorDt}[1]{\overrightarrow{\text{#1}}} \)
\( \newcommand{\vectE}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash{\mathbf {#1}}}} \)
\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)
\( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)
\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Calculate the slope of a linear function and interpret its meaning.
- Sketch the graph of a function that has been shifted, stretched, or reflected from its initial graph position.
- Evaluate a piecewise-defined function.
- Graph a quadratic function by finding the vertex and intercepts.
We have studied the general characteristics of functions, so now let’s examine some specific classes of functions. We begin by reviewing the basic properties of linear and quadratic functions, and then generalize to include higher-degree polynomials. By combining root functions with polynomials, we can define general algebraic functions and distinguish them from the transcendental functions we examine later in this chapter. We finish the section with examples of piecewise-defined functions and take a look at how to sketch the graph of a function that has been shifted, stretched, or reflected from its initial form.
Linear Functions and Slope
The easiest type of function to consider is a linear function. Linear functions have the form \(f(x)=ax+b\), where \(a\) and \(b\) are constants. In Figure \(\PageIndex{1}\), we see examples of linear functions when a is positive, negative, and zero. Note that if \(a>0\), the graph of the line rises as \(x\) increases. In other words, \(f(x)=ax+b\) is increasing on \((−∞, ∞)\). If \(a<0\), the graph of the line falls as \(x\) increases. In this case, \(f(x)=ax+b\) is decreasing on \((−∞, ∞)\). If \(a=0\), the line is horizontal.
As suggested by Figure \(\PageIndex{1}\), the graph of any linear function is a line. One of the distinguishing features of a line is its slope. The slope is the change in \(y\) for each unit change in \(x\). The slope measures both the steepness and the direction of a line. If the slope is positive, the line points upward when moving from left to right. If the slope is negative, the line points downward when moving from left to right. If the slope is zero, the line is horizontal. To calculate the slope of a line, we need to determine the ratio of the change in \(y\) versus the change in \(x\). To do so, we choose any two points \((x_1,y_1)\) and \((x_2,y_2)\) on the line and calculate \(\dfrac{y_2−y_1}{x_2−x_1}\). In Figure \(\PageIndex{2}\), we see this ratio is independent of the points chosen.
Consider line \(L\) passing through points \((x_1,y_1)\) and \((x_2,y_2)\). Let \(Δy=y_2−y_1\) and \(Δx=x_2−x_1\) denote the changes in \(y\) and \(x\),respectively. The slope of the line is
\[m=\dfrac{y_2−y_1}{x_2−x_1}=\dfrac{Δy}{Δx} \nonumber \]
We now examine the relationship between slope and the formula for a linear function. Consider the linear function given by the formula \(f(x)=ax+b\). As discussed earlier, we know the graph of a linear function is given by a line. We can use our definition of slope to calculate the slope of this line. As shown, we can determine the slope by calculating \((y_2−y_1)/(x_2−x_1)\) for any points \((x_1,y_1)\) and \((x_2,y_2)\) on the line. Evaluating the function \(f\) at \(x=0\), we see that \((0,b)\) is a point on this line. Evaluating this function at \(x=1\), we see that \((1,a+b)\) is also a point on this line. Therefore, the slope of this line is
\[\dfrac{(a+b)−b}{1−0}=a. \nonumber \]
We have shown that the coefficient \(a\) is the slope of the line. We can conclude that the formula \(f(x)=ax+b\) describes a line with slope \(a\). Furthermore, because this line intersects the \(y\)-axis at the point \((0,b)\), we see that the \(y\)-intercept for this linear function is \((0,b)\). We conclude that the formula \(f(x)=ax+b\) tells us the slope, \(a\), and the \(y\)-intercept, \((0,b)\), for this line. Since we often use the symbol \(m\) to denote the slope of a line, we can write
\[\underbrace{f(x)=mx+b}_{\text{slope-intercept form}} \nonumber \]
to denote the slope-intercept form of a linear function.
Sometimes it is convenient to express a linear function in different ways. For example, suppose the graph of a linear function passes through the point \((x_1,y_1)\) and the slope of the line is \(m\). Since any other point \((x,f(x))\) on the graph of \(f\) must satisfy the equation
\[m=\dfrac{f(x)−y_1}{x−x_1}, \nonumber \]
this linear function can be expressed by writing
\[\underbrace{f(x)−y_1=m(x−x_1)}_{\text{point-slope equation}}. \nonumber \]
We call this equation the point-slope equation for that linear function.
Since every nonvertical line is the graph of a linear function, the points on a nonvertical line can be described using the slope-intercept or point-slope equations. However, a vertical line does not represent the graph of a function and cannot be expressed in either of these forms. Instead, a vertical line is described by the equation \(x=k\) for some constant \(k\). Since neither the slope-intercept form nor the point-slope form allows for vertical lines, we use the notation
\[\underbrace{ax+by=c}_{\text{standard form}}, \nonumber \]
where \(a,b\) are both not zero, to denote the standard form of a line.
Consider a line passing through the point \((x_1,y_1)\) with slope \(m\). The equation
\[y−y_1=m(x−x_1) \nonumber \]
is the point-slope equation for that line.
Consider a line with slope \(m\) and \(y\)-intercept \((0,b).\) The equation
\[y=mx+b \nonumber \]
is an equation for that line in slope-intercept form.
The standard form of a line is given by the equation
\[ax+by=c, \nonumber \]
where \(a\) and \(b\) are both not zero. This form is more general because it allows for a vertical line, \(x=k\).
Consider the line passing through the points \((11,−4)\) and \((−4,5)\), as shown in Figure \(\PageIndex{3}\).
- Find the slope of the line.
- Find an equation for this linear function in point-slope form.
- Find an equation for this linear function in slope-intercept form.
Solution
1. The slope of the line is
\[m=\dfrac{y_2−y_1}{x_2−x_1}=\dfrac{5−(−4)}{−4−11}=−\dfrac{9}{15}=−\dfrac{3}{5}. \nonumber \]
2. To find an equation for the linear function in point-slope form, use the slope \(m=−3/5\) and choose any point on the line. If we choose the point \((11,−4)\), we get the equation
\[f(x)+4=−\dfrac{3}{5}(x−11). \nonumber \]
3. To find an equation for the linear function in slope-intercept form, solve the equation in part b. for \(f(x)\). When we do this, we get the equation
\[f(x)=−\dfrac{3}{5}x+\dfrac{13}{5}. \nonumber \]
Consider the line passing through points \((−3,2)\) and \((1,4)\).
- Find the slope of the line.
- Find an equation of that line in point-slope form.
- Find an equation of that line in slope-intercept form.
- Hint
-
The slope \(m=Δy/Δx\).
- Answer a
-
\(m=1/2\).
- Answer b
-
The point-slope form is \(y−4=\dfrac{1}{2}(x−1)\).
- Answer c
-
The slope-intercept form is \(y=\dfrac{1}{2}x+\dfrac{7}{2}\).
Jessica leaves her house at 5:50 a.m. and goes for a 9-mile run. She returns to her house at 7:08 a.m. Answer the following questions, assuming Jessica runs at a constant pace.
- Describe the distance \(D\) (in miles) Jessica runs as a linear function of her run time \(t\) (in minutes).
- Sketch a graph of \(D\).
- Interpret the meaning of the slope.
Solution
a. At time \(t=0\), Jessica is at her house, so \(D(0)=0\). At time \(t=78\) minutes, Jessica has finished running \(9\) mi, so \(D(78)=9\). The slope of the linear function is
\[m=\dfrac{9−0}{78−0}=\dfrac{3}{26}.\nonumber \]
The \(y\)-intercept is \((0,0)\), so the equation for this linear function is
\[D(t)=\dfrac{3}{26}t. \nonumber \]
b. To graph \(D\), use the fact that the graph passes through the origin and has slope \(m=3/26.\)
c. The slope \(m=3/26≈0.115\) describes the distance (in miles) Jessica runs per minute, or her average velocity.
Piecewise-Defined Functions
Sometimes a function is defined by different formulas on different parts of its domain. A function with this property is known as a piecewise-defined function. The absolute value function is an example of a piecewise-defined function because the formula changes with the sign of \(x\):
\[f(x)=\begin{cases}−x, & \text{if } x<0\\x, & \text{if } x≥0\end{cases}. \nonumber \]
Other piecewise-defined functions may be represented by completely different formulas, depending on the part of the domain in which a point falls. To graph a piecewise-defined function, we graph each part of the function in its respective domain, on the same coordinate system. If the formula for a function is different for \(x<a\) and \(x>a\), we need to pay special attention to what happens at \(x=a\) when we graph the function. Sometimes the graph needs to include an open or closed circle to indicate the value of the function at \(x=a\). We examine this in the next example.
Sketch a graph of the following piecewise-defined function:
\[f(x)=\begin{cases}x+3, & \text{if } x<1\\(x−2)^2, & \text{if } x≥1\end{cases} \nonumber \]
Solution
Graph the linear function \(y=x+3\) on the interval \((−∞,1)\) and graph the quadratic function \(y=(x−2)^2\) on the interval \([1,∞)\). Since the value of the function at \(x=1\) is given by the formula \(f(x)=(x−2)^2\), we see that \(f(1)=1\). To indicate this on the graph, we draw a closed circle at the point \((1,1)\). The value of the function is given by \(f(x)=x+3\) for all \(x<1\), but not at \(x=1\). To indicate this on the graph, we draw an open circle at \((1,4)\).
2) Sketch a graph of the function
\(f(x)=\begin{cases}2−x, & \text{if } x≤2\\x+2, & \text{if } x>2\end{cases}.\)
Solution:
In a big city, drivers are charged variable rates for parking in a parking garage. They are charged $10 for the first hour or any part of the first hour and an additional $2 for each hour or part thereof up to a maximum of $30 for the day. The parking garage is open from 6 a.m. to 12 midnight.
- Write a piecewise-defined function that describes the cost \(C\) to park in the parking garage as a function of hours parked \(x\).
- Sketch a graph of this function \(C(x).\)
Solution
1.Since the parking garage is open 18 hours each day, the domain for this function is \(\{x\,|\,0<x≤18\}\). The cost to park a car at this parking garage can be described piecewise by the function
\[C(x)=\begin{cases}10, & \text{for } 0<x≤1\\12, & \text{for } 1<x≤2\\14, & \text{for } 2<x≤3\\16, & \text{for } 3<x≤4\\ ⋮\\30, & \text{for } 10<x≤18\end{cases}. \nonumber \]
2.The graph of the function consists of several horizontal line segments.
The cost of mailing a letter is a function of the weight of the letter. Suppose the cost of mailing a letter is \(49¢\) for the first ounce and \(21¢\) for each additional ounce. Write a piecewise-defined function describing the cost \(C\) as a function of the weight \(x\) for \(0<x≤3\), where \(C\) is measured in cents and \(x\) is measured in ounces.
- Hint
-
The piecewise-defined function is constant on the intervals \((0,1],\,(1,2],\,….\)
- Answer
-
\[C(x)=\begin{cases}49, 0<x≤1\\70, 1<x≤2\\91, 2<x≤3\end{cases} \nonumber \]
Transformations of Functions
We have seen several cases in which we have added, subtracted, or multiplied constants to form variations of simple functions. In the previous example, for instance, we subtracted 2 from the argument of the function \(y=x^2\) to get the function \(f(x)=(x−2)^2\). This subtraction represents a shift of the function \(y=x^2\) two units to the right. A shift, horizontally or vertically, is a type of transformation of a function. Other transformations include horizontal and vertical scalings, and reflections about the axes.
A vertical shift of a function occurs if we add or subtract the same constant to each output \(y\). For \(c>0\), the graph of \(f(x)+c\) is a shift of the graph of \(f(x)\) up \(c\) units, whereas the graph of \(f(x)−c\) is a shift of the graph of \(f(x)\) down \(c\) units. For example, the graph of the function \(f(x)=x^3+4\) is the graph of \(y=x^3\) shifted up \(4\) units; the graph of the function \(f(x)=x^3−4\) is the graph of \(y=x^3\) shifted down \(4\) units (Figure \(\PageIndex{9}\)).
A horizontal shift of a function occurs if we add or subtract the same constant to each input \(x\). For \(c>0\), the graph of \(f(x+c)\) is a shift of the graph of \(f(x)\) to the left \(c\) units; the graph of \(f(x−c)\) is a shift of the graph of \(f(x)\) to the right \(c\) units. Why does the graph shift left when adding a constant and shift right when subtracting a constant? To answer this question, let’s look at an example.
Consider the function \(f(x)=|x+3|\) and evaluate this function at \(x−3\). Since \(f(x−3)=|x|\) and \(x−3<x\), the graph of \(f(x)=|x+3|\) is the graph of \(y=|x|\) shifted left \(3\) units. Similarly, the graph of \(f(x)=|x−3|\) is the graph of \(y=|x|\) shifted right \(3\) units (Figure \(\PageIndex{10}\)).
A vertical scaling of a graph occurs if we multiply all outputs \(y\) of a function by the same positive constant. For \(c>0\), the graph of the function \(cf(x)\) is the graph of \(f(x)\) scaled vertically by a factor of \(c\). If \(c>1\), the values of the outputs for the function \(cf(x)\) are larger than the values of the outputs for the function \(f(x)\); therefore, the graph has been stretched vertically. If \(0<c<1\), then the outputs of the function \(cf(x)\) are smaller, so the graph has been compressed. For example, the graph of the function \(f(x)=3x^2\) is the graph of \(y=x^2\) stretched vertically by a factor of 3, whereas the graph of \(f(x)=x^2/3\) is the graph of \(y=x^2\) compressed vertically by a factor of \(3\) (Figure \(\PageIndex{11b}\)).
The horizontal scaling of a function occurs if we multiply the inputs \(x\) by the same positive constant. For \(c>0\), the graph of the function \(f(cx)\) is the graph of \(f(x)\) scaled horizontally by a factor of \(c\). If \(c>1\), the graph of \(f(cx)\) is the graph of \(f(x)\) compressed horizontally. If \(0<c<1\), the graph of \(f(cx)\) is the graph of \(f(x)\) stretched horizontally. For example, consider the function \(f(x)=\sqrt{2x}\) and evaluate \(f\) at \(x/2\). Since \(f(x/2)=\sqrt{x}\), the graph of \(f(x)=\sqrt{2x}\) is the graph of \(y=\sqrt{x}\) compressed horizontally. The graph of \(y=\sqrt{x/2}\) is a horizontal stretch of the graph of \(y=\sqrt{x}\) (Figure \(\PageIndex{12}\)).
We have explored what happens to the graph of a function \(f\) when we multiply \(f\) by a constant \(c>0\) to get a new function \(cf(x)\). We have also discussed what happens to the graph of a function \(f\)when we multiply the independent variable \(x\) by \(c>0\) to get a new function \(f(cx)\). However, we have not addressed what happens to the graph of the function if the constant \(c\) is negative. If we have a constant \(c<0\), we can write \(c\) as a positive number multiplied by \(−1\); but, what kind of transformation do we get when we multiply the function or its argument by \(−1?\) When we multiply all the outputs by \(−1\), we get a reflection about the \(x\)-axis. When we multiply all inputs by \(−1\), we get a reflection about the \(y\)-axis. For example, the graph of \(f(x)=−(x^3+1)\) is the graph of \(y=(x^3+1)\) reflected about the \(x\)-axis. The graph of \(f(x)=(−x)^3+1\) is the graph of \(y=x^3+1\) reflected about the \(y\)-axis (Figure \(\PageIndex{13}\)).
If the graph of a function consists of more than one transformation of another graph, it is important to transform the graph in the correct order. Given a function \(f(x)\), the graph of the related function \(y=cf(a(x+b))+d\) can be obtained from the graph of \(y=f(x)\)by performing the transformations in the following order.
- Horizontal shift of the graph of \(y=f(x)\). If \(b>0\), shift left. If \(b<0\) shift right.
- Horizontal scaling of the graph of \(y=f(x+b)\) by a factor of \(|a|\). If \(a<0\), reflect the graph about the \(y\)-axis.
- Vertical scaling of the graph of \(y=f(a(x+b))\) by a factor of \(|c|\). If \(c<0\), reflect the graph about the \(x\) -axis.
- Vertical shift of the graph of \(y=cf(a(x+b))\). If \(d>0\), shift up. If \(d<0\), shift down.
We can summarize the different transformations and their related effects on the graph of a function in the following table.
| Transformation of \(f (c>0)\) | Effect of the graph of \(f\) |
|---|---|
| \(f(x)+c\) | Vertical shift up \(c\) units |
| \(f(x)-c\) | Vertical shift down \(c\) units |
| \(f(x+c)\) | Shift left by \(c\) units |
| \(f(x-c)\) | Shift right by \(c\) units |
| \(cf(x)\) |
Vertical stretch if \(c>1\); vertical compression if \(0<c<1\) |
| \(f(cx)\) |
Horizontal stretch if \(0<c<1\); horizontal compression if \(c>1\) |
| \(-f(x)\) | Reflection about the \(x\)-axis |
| \(f(-x)\) | Reflection about the \(y\)-axis |
For each of the following functions, a. and b., sketch a graph by using a sequence of transformations of a well-known function.
- \(f(x)=−|x+2|−3\)
- \(f(x)=3\sqrt{-x}+1\)
Solution
1.Starting with the graph of \(y=|x|\), shift \(2\) units to the left, reflect about the \(x\)-axis, and then shift down \(3\) units.
2. Starting with the graph of \(y=\sqrt{x},\) reflect about the \(y\)-axis, stretch the graph vertically by a factor of 3, and move up 1 unit.
Describe how the function \(f(x)=−(x+1)^2−4\) can be graphed using the graph of \(y=x^2\) and a sequence of transformations
- Answer
-
Shift the graph \(y=x^2\) to the left 1 unit, reflect about the \(x\)-axis, then shift down 4 units.
Analyzing 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 \(\PageIndex{2}\).
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 \(y=0\).
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 \(f(x)=ax^2+bx+c\) 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 \[h=–\dfrac{b}{2a},\;k=f(h)=f(\dfrac{−b}{2a}).\]
Given a quadratic function in general form, find the vertex of the parabola.
- Identify \(a\), \(b\), and \(c\).
- Find \(h\), the x-coordinate of the vertex, by substituting \(a\) and \(b\) into \(h=–\frac{b}{2a}\).
- Find \(k\), the y-coordinate of the vertex, by evaluating \(k=f(h)=f\Big(−\frac{b}{2a}\Big)\).
Find the vertex of the quadratic function \(f(x)=2x^2–6x+7\). Rewrite the quadratic in standard form (vertex form).
Solution
The horizontal coordinate of the vertex will be at
\[\begin{align} h&=–\dfrac{b}{2a} \\ &=-\dfrac{-6}{2(2)} \\ &=\dfrac{6}{4} \\ &=\dfrac{3}{2}\end{align}\]
The vertical coordinate of the vertex will be at
\[\begin{align} k&=f(h) \\ &=f\Big(\dfrac{3}{2}\Big) \\ &=2\Big(\dfrac{3}{2}\Big)^2−6\Big(\dfrac{3}{2}\Big)+7 \\ &=\dfrac{5}{2} \end{align}\]
Rewriting into standard form, the stretch factor will be the same as the \(a\) in the original quadratic.
\[f(x)=ax^2+bx+c \\ f(x)=2x^2−6x+7\]
Using the vertex to determine the shifts,
\[f(x)=2\Big(x–\dfrac{3}{2}\Big)^2+\dfrac{5}{2}\]
Analysis
One reason we may want to identify the vertex of the parabola is that this point will inform us what the maximum or minimum value of the function is, \((k)\),and where it occurs, \((h)\).
Given the equation \(g(x)=13+x^2−6x\), write the equation in general form and then in standard form.
- Answer
-
\(g(x)=x^2−6x+13\) in general form; \(g(x)=(x−3)^2+4\) 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. 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 \(f(x)=ax^2+bx+c\) with a positive \(a\) value is \(f(x){\geq}f ( −\frac{b}{2a}\Big)\), or \([ f(−\frac{b}{2a}),∞ ) \); the range of a quadratic function written in general form with a negative a value is \(f(x) \leq f(−\frac{b}{2a})\), or \((−∞,f(−\frac{b}{2a})]\).
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 \(f(x) \leq k\).
Given a quadratic function, find the domain and range.
- Identify the domain of any quadratic function as all real numbers.
- Determine whether \(a\) is positive or negative. If \(a\) is positive, the parabola has a minimum. If \(a\) is negative, the parabola has a maximum.
- Determine the maximum or minimum value of the parabola, \(k\).
- If the parabola has a minimum, the range is given by \(f(x){\geq}k\), or \(\left[k,\infty\right)\). If the parabola has a maximum, the range is given by \(f(x){\leq}k\), or \(\left(−\infty,k\right]\).
Find the domain and range of \(f(x)=−5x^2+9x−1\).
Solution
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.
\[\begin{align} h&=−\dfrac{b}{2a} \\ &=−\dfrac{9}{2(-5)} \\ &=\dfrac{9}{10} \end{align}\]
The maximum value is given by \(f(h)\).
\[\begin{align} f(\dfrac{9}{10})&=5(\dfrac{9}{10})^2+9(\dfrac{9}{10})-1 \\&= \dfrac{61}{20}\end{align}\]
The range is \(f(x){\leq}\frac{61}{20}\), or \(\left(−\infty,\frac{61}{20}\right]\).
Find the domain and range of \(f(x)=2\Big(x−\frac{4}{7}\Big)^2+\frac{8}{11}\).
- Answer
-
The domain is all real numbers. The range is \(f(x){\geq}\frac{8}{11}\), or \(\left[\frac{8}{11},\infty\right)\).
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.
Given a quadratic function \(f(x)\), find the y- and x-intercepts.
- Evaluate \(f(0)\) to find the y-intercept.
- Solve the quadratic equation \(f(x)=0\) to find the x-intercepts.
Find the y- and x-intercepts of the quadratic \(f(x)=3x^2+5x−2\).
Solution
We find the y-intercept by evaluating \(f(0)\).
\[\begin{align} f(0)&=3(0)^2+5(0)−2 \\ &=−2 \end{align}\]
So the y-intercept is at \((0,−2)\).
For the x-intercepts, we find all solutions of \(f(x)=0\).
\[0=3x^2+5x−2\]
In this case, the quadratic can be factored easily, providing the simplest method for solution.
\[0=(3x−1)(x+2)\]
\[\begin{align} 0&=3x−1 & 0&=x+2 \\ x&= \frac{1}{3} &\text{or} \;\;\;\;\;\;\;\; x&=−2 \end{align}\]
So the x-intercepts are at \((\frac{1}{3},0)\) and \((−2,0)\).
Analysis
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 \(\Big(\frac{1}{3},0\Big)\) and \((−2,0)\). See Figure \(\PageIndex{14}\).
Rewriting Quadratics in Standard Form
In Example \(\PageIndex{7}\), the quadratic was easily solved by factoring. However, there are many quadratics that cannot be factored. We can solve these quadratics by first rewriting them in standard form.
Given a quadratic function, find the x-intercepts by rewriting in standard form.
- Substitute a and \(b\) into \(h=−\frac{b}{2a}\).
- Substitute \(x=h\) into the general form of the quadratic function to find \(k\).
- Rewrite the quadratic in standard form using \(h\) and \(k\).
- Solve for when the output of the function will be zero to find the x-intercepts.
Find the x-intercepts of the quadratic function \(f(x)=2x^2+4x−4\).
Solution
We begin by solving for when the output will be zero.
\[0=2x^2+4x−4 \nonumber\]
Because the quadratic is not easily factorable in this case, we solve for the intercepts by first rewriting the quadratic in standard form.
\[f(x)=a(x−h)^2+k\nonumber\]
We know that \(a=2\). Then we solve for \(h\) and \(k\).
\[\begin{align*} h&=−\dfrac{b}{2a} & k&=f(−1) \\ &=−\dfrac{4}{2(2)} & &=2(−1)^2+4(−1)−4 \\ &=−1 & &=−6 \end{align*}\]
So now we can rewrite in standard form.
\[f(x)=2(x+1)^2−6\nonumber\]
We can now solve for when the output will be zero.
\[\begin{align*} 0&=2(x+1)^2−6 \\ 6&=2(x+1)^2 \\ 3&=(x+1)^2 \\ x+1&={\pm}\sqrt{3} \\ x&=−1{\pm}\sqrt{3} \end{align*}\]
The graph has x-intercepts at \((−1−\sqrt{3},0)\) and \((−1+\sqrt{3},0)\).
Analysis
We can check our work by graphing the given function on a graphing utility and observing the x-intercepts. See Figure \(\PageIndex{15}\).
In Try It \(\PageIndex{1}\), 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).
- Answer
-
y-intercept at \((0, 13)\), No x-intercepts
Solve \(x^2+x+2=0\).
Solution
Let’s begin by writing the quadratic formula: \(x=\frac{−b{\pm}\sqrt{b^2−4ac}}{2a}\).
When applying the quadratic formula, we identify the coefficients \(a\), \(b\) and \(c\). For the equation \(x^2+x+2=0\), we have \(a=1\), \(b=1\), and \(c=2\). Substituting these values into the formula we have:
\[\begin{align*} x&=\dfrac{−b{\pm}\sqrt{b^2−4ac}}{2a} \\ &=\dfrac{−1{\pm}\sqrt{1^2−4⋅1⋅(2)}}{2⋅1} \\ &=\dfrac{−1{\pm}\sqrt{1−8}}{2} \\ &=\dfrac{−1{\pm}\sqrt{−7}}{2} \\ &=\dfrac{−1{\pm}i\sqrt{7}}{2} \end{align*}\]
The solutions to the equation are \(x=\frac{−1+i\sqrt{7}}{2}\) and \(x=\frac{−1-i\sqrt{7}}{2}\) or \(x=−\frac{1}{2}+\frac{i\sqrt{7}}{2}\) and \(x=\frac{-1}{2}−\frac{i\sqrt{7}}{2}\).
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 \(H(t)=−16t^2+80t+40\).
When does the ball reach the maximum height?
What is the maximum height of the ball?
When does the ball hit the ground?
The ball reaches the maximum height at the vertex of the parabola.
\[\begin{align} h &= −\dfrac{80}{2(−16)} \\ &=\dfrac{80}{32} \\ &=\dfrac{5}{2} \\ & =2.5 \end{align}\]
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.
\[\begin{align} k &=H(−\dfrac{b}{2a}) \\ &=H(2.5) \\ &=−16(2.5)^2+80(2.5)+40 \\ &=140 \end{align}\]
The ball reaches a maximum height of 140 feet.
To find when the ball hits the ground, we need to determine when the height is zero, \(H(t)=0\).
We use the quadratic formula.
\[\begin{align} t & =\dfrac{−80±\sqrt{80^2−4(−16)(40)}}{2(−16)} \\ & = \dfrac{−80±\sqrt{8960}}{−32} \end{align} \]
Because the square root does not simplify nicely, we can use a calculator to approximate the values of the solutions.
\[t=\dfrac{−80-\sqrt{8960}}{−32} ≈5.458 \text{ or }t=\dfrac{−80+\sqrt{8960}}{−32} ≈−0.458 \]
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 \(\PageIndex{16}\).
\(\PageIndex{13}\): 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 \(H(t)=−16t^2+96t+112\).
- When does the rock reach the maximum height?
- What is the maximum height of the rock?
- When does the rock hit the ocean?
Solution
a. 3 seconds b. 256 feet c. 7 seconds