4.6 Graph Quadratic Functions Using Properties
Learning Objectives
By the end of this section, you will be able to:
- Recognize the graph of a quadratic function
- Find the axis of symmetry and vertex of a parabola
- Find the intercepts of a parabola
- Graph quadratic functions using properties
- Solve maximum and minimum applications
Try It
Before you get started, take this readiness quiz:
1) Graph the function [latex]f\left(x\right)={x}^{2}[/latex] by plotting points.
2) Solve: [latex]2{x}^{2}+3x-2=0[/latex].
3) Evaluate [latex]-\frac{b}{2a}[/latex] when [latex]a=3[/latex] and [latex]b=−6[/latex].
Recognize the Graph of a Quadratic Function
Previously we very briefly looked at the function [latex]f\left(x\right)={x}^{2}[/latex], which we called the square function. It was one of the first non-linear functions we looked at. Now we will graph functions of the form [latex]f\left(x\right)=a{x}^{2}+bx+c[/latex] if [latex]a\ne 0[/latex]. We call this kind of function a quadratic function.
Quadratic Function
A quadratic function, where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are real numbers and [latex]a\ne 0[/latex], is a function of the form
[latex]f\left(x\right)=a{x}^{2}+bx+c[/latex]
We graphed the quadratic function [latex]f\left(x\right)={x}^{2}[/latex] by plotting points.
| [latex]x[/latex] | [latex]f(x)=x^2[/latex] | [latex](x,f(x))[/latex] |
|---|---|---|
| [latex]-3[/latex] | [latex]9[/latex] | [latex](-3,9)[/latex] |
| [latex]-2[/latex] | [latex]4[/latex] | [latex](-2,4)[/latex] |
| [latex]-1[/latex] | [latex]1[/latex] | [latex](-1,1)[/latex] |
| [latex]0[/latex] | [latex]0[/latex] | [latex](0,0)[/latex] |
| [latex]1[/latex] | [latex]1[/latex] | [latex](1,1)[/latex] |
| [latex]2[/latex] | [latex]4[/latex] | [latex](2,4)[/latex] |
| [latex]3[/latex] | [latex]9[/latex] | [latex](3,9)[/latex] |
Every quadratic function has a graph that looks like this. We call this figure a parabola.
Let’s practice graphing a parabola by plotting a few points.
Example 4.6.1
Graph [latex]f\left(x\right)={x}^{2}-1[/latex].
Solution
We will graph the function by plotting points.
Step 1: Choose integer values for [latex]x[/latex], substitute them into the equation and simplify to find [latex]f\left(x\right)[/latex].
Record the values of the ordered pairs in the chart.
| [latex]f(x)=x^2-1[/latex] | |
|---|---|
| [latex]x[/latex] | [latex]f(x)[/latex] |
| [latex]0[/latex] | [latex]-1[/latex] |
| [latex]1[/latex] | [latex]0[/latex] |
| [latex]-1[/latex] | [latex]0[/latex] |
| [latex]2[/latex] | [latex]3[/latex] |
| [latex]-2[/latex] | [latex]3[/latex] |
Step 2: Plot the points, and then connect them with a smooth curve. The result will be the graph of the function [latex]f\left(x\right)={x}^{2}-1[/latex].
Try It
4) Graph [latex]f\left(x\right)=-{x}^{2}[/latex].
Solution
Try It
5) Graph [latex]f\left(x\right)={x}^{2}+1[/latex].
Solution
All graphs of quadratic functions of the form [latex]f(x)\;=\;ax^2\;+\;bx\;+\;c[/latex] are parabolas that open upward or downward. See Figure 4.6.5 below.
Notice that the only difference in the two functions is the negative sign before the quadratic term ([latex]x^2[/latex]) in the equation of the graph above. When the quadratic term, is positive, the parabola opens upward, and when the quadratic term is negative, the parabola opens downward.
Parabola Orientation
For the graph of the quadratic function [latex]f(x)=ax^2+bx+c[/latex], if:
- [latex]{a>0\text{, the parabola opens upward}}[/latex]
- [latex]{a<0\text{, the parabola opens downward}}[/latex]
Example 4.6.2
Determine whether each parabola opens upward or downward:
a. [latex]f\left(x\right)=-3{x}^{2}+2x-4[/latex]
b. [latex]f\left(x\right)=6{x}^{2}+7x-9[/latex]
Solution
a.
Step 1: Find the value of [latex]a[/latex].
[latex]\begin{eqnarray*}f(x)&=&{\color{red}{a}}x^2+bx+c\\f(x)&=&{\color{red}{-}}{\color{red}{3}}x^2+bx+c\\{\color{red}{a}}{\color{red}{}}&=&{\color{red}{-}}{\color{red}{3}}\end{eqnarray*}[/latex]
Step 2: Determine which direction the parabola will open.
Since the [latex]a[/latex] is negative, the parabola will open downward.
b.
Step 1: Find the value of [latex]a[/latex].
[latex]\begin{eqnarray*}f(x)&=&{\color{red}{a}}x^2+bx+c\\f(x)&=&{\color{red}{6}}x^2+7x-9\\{\color{red}{a}}{\color{red}{}}&=&{\color{red}{6}}\end{eqnarray*}[/latex]
Step 2: Determine which direction the parabola will open.
Since the [latex]a[/latex] is positive, the parabola will open upward.
Try It
6) Determine whether the graph of each function is a parabola that opens upward or downward:
a. [latex]f\left(x\right)=2{x}^{2}+5x-2[/latex]
b. [latex]f\left(x\right)=-3{x}^{2}-4x+7[/latex]
Solution
a. up
b. down
Try It
7) Determine whether the graph of each function is a parabola that opens upward or downward:
a. [latex]f\left(x\right)=-2{x}^{2}-2x-3[/latex]
b. [latex]f\left(x\right)=5{x}^{2}-2x-1[/latex]
Solution
a. down
b. up
Find the Axis of Symmetry and Vertex of a Parabola
Look again at Figure 4.6.5. Do you see that we could fold each parabola in half and then one side would lie on top of the other? The ‘fold line’ is a line of symmetry. We call it the axis of symmetry of the parabola.
We show the same two graphs again with the axis of symmetry. See Figure 4.6.6 below.
The equation of the axis of symmetry can be derived by using the Quadratic Formula. We will omit the derivation here and proceed directly to using the result. The equation of the axis of symmetry of the graph of
[latex]f(x)=ax^2+bx+c[/latex] is [latex]x=-\frac{b}{2a}[/latex]
So to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula [latex]x=-\frac{b}{2a}[/latex].
| [latex]\begin{eqnarray*}{\color{red}{f}}{\color{red}{(}}{\color{red}{x}}{\color{red}{)}}{\color{red}{}}&=&{\color{red}{a}}{\color{red}{x}}^{\color{red}{2}}{\color{red}{+}}{\color{red}{b}}{\color{red}{x}}{\color{red}{+}}{\color{red}{c}}\\f(x)&=&x^2+4x+3\\\text{Axis of Symmetry:}\\x&=&-\frac b{2a}\\x&=&-\frac4{2\times1}\\x&=&-2\end{eqnarray*}[/latex] | [latex]\begin{eqnarray*}{\color{red}{f}}{\color{red}{(}}{\color{red}{x}}{\color{red}{)}}{\color{red}{}}&=&{\color{red}{a}}{\color{red}{x}}^{\color{red}{2}}{\color{red}{+}}{\color{red}{b}}{\color{red}{x}}{\color{red}{+}}{\color{red}{c}}\\f(x)&=&-x^2+4x+3\\\text{Axis of Symmetry:}\\x&=&-\frac b{2a}\\x&=&-\frac4{2(-1)}\\x&=&2\end{eqnarray*}[/latex] |
Notice that these are the equations of the dashed blue lines on the graphs.
The point on the parabola that is the lowest (parabola opens up), or the highest (parabola opens down), lies on the axis of symmetry. This point is called the vertex of the parabola.
We can easily find the coordinates of the vertex, because we know it is on the axis of symmetry. This means its [latex]x[/latex]-coordinate is [latex]-\frac{b}{2a}[/latex]. To find the [latex]y[/latex]-coordinate of the vertex we substitute the value of the [latex]x[/latex]-coordinate into the quadratic function.
| [latex]\begin{eqnarray*}f(x)&=&x^2+4x+3\\\text{Axis of Symmetry:}\;{\color{red}{x}}{\color{red}{}}&=&{\color{red}{-}}{\color{red}{2}}\\\text{Vertex is}\;({\color{red}{-}}{\color{red}{2}},\_\_)\\f(x)&=&{\color{red}{x}}^2+4{\color{red}{x}}+3\\f(x)&=&{({\color{red}{-}}{\color{red}{2}})}^2+4({\color{red}{-}}{\color{red}{2}})+3\\f(x)&=&-1\\\text{Vertex is}\;({\color{red}{-}}{\color{red}{2}},-1)\end{eqnarray*}[/latex] | [latex]\begin{eqnarray*}f(x)&=&-x^2+4x+3\\\text{Axis of Symmetry:}\;{\color{red}{x}}{\color{red}{}}&=&{\color{red}{2}}\\\text{Vertex is}\;({\color{red}{2}},\_\_)\\f(x)&=&-{\color{red}{x}}^2+4{\color{red}{x}}+3\\f(x)&=&-{({\color{red}{2}})}^2+4({\color{red}{2}})+3\\f(x)&=&7\\\text{Vertex is}\;({\color{red}{2}},7)\end{eqnarray*}[/latex] |
Axis of Symmetry and Vertex of a Parabola
The graph of the function [latex]f(x)={a}{x}^2+{b}{x}+c[/latex] is a parabola where:
- the axis of symmetry is the vertical line [latex]x=-\frac{b}{2a}[/latex].
- the vertex is a point on the axis of symmetry, so its [latex]x[/latex]-coordinate is [latex]-\frac{b}{2a}[/latex].
- the [latex]y[/latex]-coordinate of the vertex is found by substituting [latex]x=-\frac{b}{2a}[/latex] into the quadratic equation.
Example 4.6.3
For the graph of [latex]f\left(x\right)=3{x}^{2}-6x+2[/latex] find:
a. the axis of symmetry
b. the vertex
Solution
a.
[latex]\begin{eqnarray*}{\color{black}{f}}{\color{black}{(}}{\color{black}{x}}{\color{black}{)}}{\color{black}{}}&=&{\color{red}{a}}{\color{black}{x}}^{\color{black}{2}}{\color{black}{+}}{\color{blue}{b}}{\color{black}{x}}{\color{black}{+}}{\color{black}{c}}\\f(x)&=&{\color{red}{3}}x^2{\color{blue}{-}}{\color{blue}{6}}x+2\end{eqnarray*}[/latex]
Step 1: Find the axis of symmetry.
The axis of symmetry is the vertical line [latex]{x=-}\frac{{\color{blue}{b}}}{{2}{\color{red}{a}}}[/latex].
Step 2: Substitute the values of [latex]a,b[/latex] into the equation.
[latex]{x=-}\frac{{\color{blue}{-6}}}{{2\times}{\color{red}{3}}}[/latex]
Step 3: Simplify.
[latex]x=1[/latex]
Step 4: Write in a statement.
The axis of symmetry is the line [latex]x=1[/latex].
b.
Step 1: Find the vertex.
The vertex is a point on the line of symmetry, so its [latex]x[/latex]-coordinate will be [latex]x=1[/latex].
Step 2: Find [latex]f\left(1\right)[/latex].
[latex]{f(1)=3{(}{\color{red}{1}}{)}^2-6(}{\color{red}{1}}{)+2}[/latex]
Step 3: Simplify.
[latex]{f(1)=3\times}{\color{red}{1}}{-6+2}[/latex]
Step 4: The result is the [latex]y[/latex]-coordinate.
[latex]f(1)=-1[/latex]
Step 5: Write as a statement.
The vertex is [latex]\left(1,-1\right)[/latex].
Try It
8) For the graph of [latex]f\left(x\right)=2{x}^{2}-8x+1[/latex] find:
a. the axis of symmetry
b. the vertex
Solution
a. [latex]x=2[/latex]
b. [latex]\left(2,-7\right)[/latex]
Try It
9) For the graph of [latex]f\left(x\right)=2{x}^{2}-4x-3[/latex] find:
a. the axis of symmetry
b. the vertex
Solution
a. [latex]x=1[/latex]
b. [latex]\left(1,-5\right)[/latex]
Find the Intercepts of a Parabola
When we graphed linear equations, we often used the [latex]x[/latex]– and [latex]y[/latex]-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.
Remember, at the [latex]y[/latex]-intercept the value of [latex]x[/latex] is zero. So to find the [latex]y[/latex]-intercept, we substitute [latex]x=0[/latex] into the function.
Let’s find the [latex]y[/latex]-intercepts of the two parabolas shown in Figure 4.6.7.
An [latex]x[/latex]-intercept results when the value of [latex]f(x)[/latex] is zero. To find an [latex]x[/latex]-intercept, we let [latex]f(x)=0[/latex]. In other words, we will need to solve the equation [latex]0=ax^2+bx+c[/latex] for [latex]x[/latex].
[latex]\begin{eqnarray}\hfill f\left(x\right)& =\hfill & a{x}^{2}+bx+c\hfill \\ \hfill 0& =\hfill & a{x}^{2}+bx+c\hfill\end{eqnarray}[/latex]
Solving quadratic equations like this is exactly what we have done earlier in this chapter!
We can now find the [latex]x[/latex]-intercepts of the two parabolas we looked at. First we will find the [latex]x[/latex]-intercepts of the parabola whose function is:
[latex]f(x)=x^2+4x+3[/latex].
Step 1: Let [latex]f(x)=0[/latex].
[latex]{\color{red}{0}}{=x^2+4x+3}[/latex]
Step 2: Factor.
[latex]0=(x+1)(x+3)[/latex]
Step 3: Solve using the Zero Product Property.
[latex]\begin{eqnarray*}x+1&=&0\;\;\;\;\\x&=&-1\\\\x+3&=&0\\x&=&-3\end{eqnarray*}[/latex]
Step 4: Write as a statement.
The [latex]x[/latex]-intercepts are [latex]\left(-1,0\right)[/latex] and [latex]\left(-3,0\right)[/latex].
Now we will find the [latex]x[/latex]-intercepts of the parabola whose function is:
[latex]f(x)=−x^2+4x+3[/latex]
Step 1: Let [latex]f(x)=0[/latex].
This quadratic does not factor, so we use the Quadratic Formula.
[latex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/latex]
Step 2: Substitute in known values.
[latex]\begin{eqnarray*}a&=&{\color{red}{-}}{\color{red}{1}},\;b={\color{blue}{4}},\;c={\color{green}{3}}\\x&=&\frac{-{\color{blue}{4}}\pm\sqrt{{\color{blue}{4}}^2-4({\color{red}{-}}{\color{red}{1}})({\color{green}{3}})}}{2({\color{red}{-}}{\color{red}{1}})}\\\end{eqnarray*}[/latex]
Step 3: Simplify.
[latex]\begin{eqnarray*}x&=&\frac{-4\pm\sqrt{16+12}}{-2}\\x&=&\frac{-4\pm\sqrt{28}}{-2}\\x&=&\frac{-4\pm2\sqrt7}{-2}\\x&=&\frac{-2(\pm2\sqrt7)}{-2}\\x&=&2\pm\sqrt7\end{eqnarray*}[/latex]
Step 4: Write as a statement.
The [latex]x[/latex]-intercepts are [latex]\left(2+\sqrt7,0\right)[/latex] and [latex]\left(2-\sqrt7,0\right)[/latex].
We will use the decimal approximations of the [latex]x[/latex]-intercepts, so that we can locate these points on the graph:
[latex]\begin{array}{cccccc}\left(2+\sqrt{7},0\right)\approx \left(4.6,0\right)\hfill & & & & & \left(2-\sqrt{7},0\right)\approx \left(-0.6,0\right)\hfill \end{array}[/latex]
Do these results agree with our graphs? See Figure 4.6.8.
Find the Intercepts of a Parabola
To find the intercepts of a parabola whose function is [latex]f\left(x\right)=a{x}^{2}+bx+c[/latex]:
[latex]\begin{array}{cccccc}y\textbf{-intercept}&&&&&x\textbf{-intercepts}\\\text{Let}\;x=0\;\text{and solve for}\;f\left(x\right)&&&&&\text{Let}\;f\left(x\right)=0\;\text{and solve for}\;x\end{array}[/latex]
Example 4.6.3
Find the intercepts of the parabola whose function is [latex]f\left(x\right)={x}^{2}-2x-8[/latex].
Solution
Step 1: To find the [latex]y[/latex]-intercept
Let [latex]x=0[/latex] and solve for [latex]f\left(x\right)[/latex].
[latex]\begin{eqnarray*}f(x)&=&x^2-2x-8\\f(0)&=&{\color{red}{0}}^2-2\times{\color{red}{0}}-8\\f(0)&=&-8\\\end{eqnarray*}[/latex]
When [latex]x=0[/latex], then [latex]f\left(0\right)=-8[/latex].
The [latex]y[/latex]-intercept is the point [latex]\left(0,-8\right)[/latex].
Step 2: To find the [latex]x[/latex]-intercept
Let [latex]f\left(x\right)=0[/latex] and solve for [latex]x[/latex].
[latex]\begin{eqnarray*}f(x)=x^2-2x-8\\0=x^2-2x-8\end{eqnarray*}[/latex]
Step 3: Solve by factoring.
[latex]\begin{eqnarray*}(x-4)(x+2)&=&0\\\\x-4&=&0\\x&=&4\\\\x+2&=&0\\x&=&-2\end{eqnarray*}[/latex]
When [latex]f\left(x\right)=0[/latex], then [latex]x=4\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=-2[/latex].
The [latex]x[/latex]–intercepts are the points [latex]\left(4,0\right)[/latex] and [latex]\left(-2,0\right)[/latex].
Try It
10) Find the intercepts of the parabola whose function is [latex]f\left(x\right)={x}^{2}+2x-8[/latex].
Solution
[latex]y[/latex]-intercept: [latex]\left(0,-8\right)[/latex]
[latex]x[/latex]-intercepts: [latex]\left(-4,0\right),\left(2,0\right)[/latex]
Try It
11) Find the intercepts of the parabola whose function is [latex]f\left(x\right)={x}^{2}-4x-12[/latex].
Solution
[latex]y[/latex]-intercept: [latex]\left(0,-12\right)[/latex]
[latex]x[/latex]-intercepts: [latex]\left(-2,0\right),\left(6,0\right)[/latex]
In this chapter, we have been solving quadratic equations of the form [latex]ax^2+bx+c=0[/latex]. We solved for [latex]x[/latex] and the results were the solutions to the equation.
We are now looking at quadratic functions of the form [latex]f(x)=ax^2+bx+c[/latex]. The graphs of these functions are parabolas. The [latex]x[/latex]-intercepts of the parabolas occur where [latex]f(x)=0[/latex].
For example:
[latex]\begin{array}{ccccccccc}\hfill \mathbf{\text{Quadratic equation}}\hfill & & & & & & & & \hfill \phantom{\rule{4em}{0ex}}\mathbf{\text{Quadratic function}}\hfill \\ \hfill \begin{array}{}\\ \hfill {x}^{2}-2x-15& =\hfill & 0\hfill \\ \hfill \left(x-5\right)\left(x+3\right)& =\hfill & 0\hfill \\ \hfill x-5=0\phantom{\rule{0.8em}{0ex}}x+3& =\hfill & 0\hfill \\ \hfill x=5\phantom{\rule{2.3em}{0ex}}x& =\hfill & -3\hfill \end{array}\hfill & & & & \hfill \begin{array}{c}\hfill \text{Let}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=0.\hfill \\ \\ \end{array}\hfill & & & & \hfill \phantom{\rule{4em}{0ex}}\begin{array}{ccc}\hfill f\left(x\right)& =\hfill & {x}^{2}-2x-15\hfill \\ \hfill 0& =\hfill & {x}^{2}-2x-15\hfill \\ \hfill 0& =\hfill & \left(x-5\right)\left(x+3\right)\hfill \\ \hfill x-5& =\hfill & 0\phantom{\rule{0.8em}{0ex}}x+3=0\hfill \\ \hfill x& =\hfill & 5\phantom{\rule{2.3em}{0ex}}x=-3\hfill \end{array}\hfill \\ & & & & & & & & \hfill \phantom{\rule{4em}{0ex}}\left(5,0\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(-3,0\right)\hfill \\ & & & & & & & & \hfill \phantom{\rule{4em}{0ex}}x\text{-intercepts}\hfill \end{array}[/latex]
The solutions of the quadratic function are the [latex]x[/latex] values of the [latex]x[/latex]–intercepts.
Earlier, we saw that quadratic equations have [latex]2[/latex], [latex]1[/latex], or [latex]0[/latex] solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the functions give the [latex]x[/latex]–intercepts of the graphs, the number of [latex]x[/latex]–intercepts is the same as the number of solutions.
Previously, we used the discriminant to determine the number of solutions of a quadratic function of the form [latex]a{x}^{2}+bx+c=0[/latex]. Now we can use the discriminant to tell us how many [latex]x[/latex]-intercepts there are on the graph.
Before you to find the values of the [latex]x[/latex]-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.
Example 4.6.4
Find the intercepts of the parabola for the function [latex]f\left(x\right)=5{x}^{2}+x+4[/latex].
Solution
Step 1: To find the [latex]y[/latex]-intercept
Let [latex]x=0[/latex] and solve for [latex]f\left(x\right)[/latex].
[latex]\begin{eqnarray*}f(0)&=&5\times{\color{red}{0}}^2+{\color{red}{0}}+4\\f(0)&=&4\end{eqnarray*}[/latex]
When [latex]x=0[/latex], then [latex]f\left(0\right)=4[/latex].
The [latex]y[/latex]-intercept is the point [latex]\left(0,4\right)[/latex].
Step 2: To find the [latex]x[/latex]-intercept
Let [latex]f\left(x\right)=0[/latex] and solve for [latex]x[/latex].
[latex]\begin{eqnarray*}f(x)&=&5x^2+x+4\\0&=&5x^2+x+4\end{eqnarray*}[/latex]
Step 3: Find the value of the discriminant to predict the number of solutions which is also the number of [latex]x[/latex]-intercepts.
[latex]\begin{eqnarray*}&=&b^2-4ac\\&=&1^2-4\times5\times4\\&=&1-80\\&=&-79\end{eqnarray*}[/latex]
Since the value of the discriminant is negative, there is no real solution to the equation. There are no [latex]x[/latex]-intercepts.
Try It
12) Find the intercepts of the parabola whose function is [latex]f\left(x\right)=3{x}^{2}+4x+4[/latex].
Solution
[latex]y[/latex]-intercept: [latex]\left(0,4\right)[/latex] no [latex]x[/latex]-intercept
Try It
13) Find the intercepts of the parabola whose function is [latex]f\left(x\right)={x}^{2}-4x-5[/latex].
Solution
[latex]y[/latex]-intercept: [latex]\left(0,-5\right)[/latex]
[latex]x[/latex]-intercepts: [latex]\left(-1,0\right),\left(5,0\right)[/latex]
Graph Quadratic Functions Using Properties
Now we have all the pieces we need in order to graph a quadratic function. We just need to put them together. In the next example we will see how to do this.
Example 4.6.5
Graph [latex]f(x)=x^2−6x+8[/latex] by using its properties.
Solution
Step 1: Determine whether the parabola opens upward or downward.
Look at [latex]a[/latex] in the equation.
[latex]{\color{red}{a}}{{\color{red}{=}}{\color{red}{1}}{\color{red}{,}}{\color{red}{\;}}{\color{red}{=}}{\color{red}{-}}{\color{red}{6}}{\color{red}{,}}{\color{red}{\;}}{\color{red}{c}}{\color{red}{=}}}{\color{red}{8}}{}[/latex]
Since [latex]a[/latex] is positive, the parabola opens upward.
Step 2: Find the axis of symmetry
[latex]f(x)=x^2-6x+8[/latex]
The axis of symmetry is the line [latex]x=-\frac b{2a}[/latex].
Axis of Symmetry
[latex]\begin{eqnarray*}x&=&-\frac b{2a}\\x&=&-\frac{\left(-6\right)}{2\times1}\\x&=&3\end{eqnarray*}[/latex]
The axis of symmetry is the line [latex]x=3[/latex].
Step 3: Find the vertex.
The vertex is on the axis of symmetry. Substitute [latex]x=3[/latex] into the function.
Vertex
[latex]\begin{eqnarray*}f(x)&=&x^2-6x+8\\f(3)&=&\left({\color{red}{3}}\right)^2-6\left({\color{red}{3}}\right)+8\\f(3)&=&-1\end{eqnarray*}[/latex]
The vertex is [latex](3,-1)[/latex].
Step 4: Find the [latex]y[/latex]-intercept.
Find the point symmetric to the [latex]y[/latex]-intercept across the axis of symmetry.
We find [latex]f(0)[/latex].
[latex]y[/latex]-intercept
[latex]\begin{eqnarray*}f(x)&=&x^2-6x+8\\f(0)&=&\left({\color{red}{0}}\right)^2-6\left({\color{red}{0}}\right)+8\\f(0)&=&8\end{eqnarray*}[/latex]
The [latex]y[/latex]-intercept is [latex](0,8)[/latex].
We use the axis of symmetry to find a point symmetric to the [latex]y[/latex]-intercept. The [latex]y[/latex]-intercept is [latex]3[/latex] units left of the axis of symmetry, [latex]x=3[/latex].
A point [latex]3[/latex] units to the right of the axis of symmetry has [latex]x=6[/latex].
Point symmetric to [latex]y[/latex]-intercept: [latex](6,8)[/latex].
Step 5: Find the [latex]x[/latex]-intercepts.
Find additional points if needed.
We solve [latex]f(x)=0[/latex].
We can solve this quadratic equation by factoring.
[latex]x[/latex]-intercepts
[latex]\begin{eqnarray*}f(x)&=&x^2-6x+8\\{\color{red}{0}}&=&x^2-6x+8\\{\color{red}{0}}&=&\left(x-2\right)\left(x-4\right)\\x&=&2\;\text{or}\;x=4\end{eqnarray*}[/latex]
The [latex]x[/latex]-intercepts are [latex](2,0)[/latex] and [latex](4,0)[/latex].
Step 6: Graph the parabola.
We graph the vertex, intercepts, and the point symmetric to the [latex]y[/latex]-intercept.
We connect these [latex]5[/latex] points to sketch the parabola.
Try It
14) Graph [latex]f\;(x)\;=\;x^2+\;2x-\;8[/latex] by using its properties.
Solution
Try It
15) Graph [latex]f\;(x)\;=\;x^2\;-\;8x\;+\;12[/latex] by using its properties.
Solution
We list the steps to take in order to graph a quadratic function here.
HOW TO
To graph a quadratic function using properties.
- Determine whether the parabola opens upward or downward.
- Find the equation of the axis of symmetry.
- Find the vertex.
- Find the [latex]y[/latex]-intercept. Find the point symmetric to the [latex]y[/latex]-intercept across the axis of symmetry.
- Find the [latex]x[/latex]-intercepts. Find additional points if needed.
- Graph the parabola.
We were able to find the [latex]x[/latex]-intercepts in the last example by factoring. We find the [latex]x[/latex]-intercepts in the next example by factoring, too.
Example 4.6.7
Graph [latex]f(x)=x^2+6x−9[/latex] by using its properties.
Solution
[latex]\begin{array}{l}\begin{array}{c}{\color{red}{f}}{\color{red}{(}}{\color{red}{x}}{\color{red}{)}}{\color{red}{=}}{\color{red}{a}}{\color{red}{x}}^{\color{red}{2}}{\color{red}{+}}{\color{red}{b}}{\color{red}{x}}{\color{red}{+}}{\color{red}{c}}\\f(x)=-x^2+6x-9\end{array}\end{array}[/latex]
Step 1: Find the direction of the parabola.
Since a is [latex]-1[/latex], the parabola opens downward.
Step 2: Find the axis of symmetry and vertex.
To find the equation of the axis of symmetry, use [latex]x=-\frac{b}{2a}[/latex].
[latex]\begin{eqnarray*}x&=&-\frac b{2a}\\x&=&-\frac6{2(-1)}\\x&=&3\end{eqnarray*}[/latex]
The axis of symmetry is [latex]x=3[/latex].
The vertex is on the line [latex]x=3[/latex].
Step 3: Graph the axis of symmetry and vertex.
Step 4: Find [latex]f\left(3\right)[/latex].
[latex]\begin{eqnarray*}f(x)&=&-x^2+6x-9\\f(3)&=&-{\color{red}{3}}^2+6\times{\color{red}{3}}-9\\f(3)&=&-9+18-9\\f(3)&=&0\end{eqnarray*}[/latex]
Step 5: Graph [latex]f\left(3\right)[/latex].
The vertex is [latex]\left(3,0\right)[/latex].
Step 6: Find [latex]f\left(0\right)[/latex]. The [latex]y[/latex]-intercept occurs when [latex]x=0[/latex].
[latex]f(x)=-x^2+6x-9[/latex]
Step 7: Substitute [latex]x=0[/latex].
[latex]{f(0)=-}{\color{red}{0}}{^2+6\times}{\color{red}{0}}{-9}[/latex]
Step 8: Simplify.
[latex]f(0)=-9[/latex]
The [latex]y[/latex]-intercept is [latex]\left(0,-9\right)[/latex].
Step 9: Graph the points of symmetry.
The point [latex]\left(0,-9\right)[/latex] is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is [latex]\left(6,-9\right)[/latex].
Point symmetric to the [latex]y[/latex]-intercept is [latex]\left(6,-9\right)[/latex]
The [latex]x[/latex]-intercept occurs when [latex]f\left(x\right)=0[/latex].
[latex]f(x)=-x^2+6x-9[/latex]
Step 10: Find [latex]f\left(x\right)=0[/latex].
[latex]{\color{darkred}{0}}{=-x^2+6x-9}[/latex]
Step 11: Factor the GCF.
[latex]0=-(x^2-6x+9)[/latex]
Step 12: Factor the trinomial.
[latex]0=-{(x-3)}^2[/latex]
Step 13: Solve for [latex]x[/latex].
[latex]0=3[/latex]
Step 14: Connect the points to graph the parabola.
Try It
16) Graph [latex]f\;(x)\;=\;3x^2\;+\;12x\;-\;12[/latex] by using its properties.
Solution
Try It
17) Graph [latex]f\;(x)\;=\;4x^2\;+\;24x\;+\;36[/latex] by using its properties.
Solution
For the graph of [latex]f(x)=−x^2+6x−9[/latex], the vertex and the [latex]x[/latex]-intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation [latex]0=−x^2+6x−9[/latex] is [latex]0[/latex], so there is only one solution. That means there is only one [latex]x[/latex]-intercept, and it is the vertex of the parabola.
How many [latex]x[/latex]-intercepts would you expect to see on the graph of [latex]f(x)=x^2+4x+5[/latex]?
Example 4.6.8
Graph [latex]f(x)=x^2+4x+5[/latex] by using its properties.
Solution
[latex]\begin{array}{c}{\color{red}{f}}{\color{red}{(}}{\color{red}{x}}{\color{red}{)}}{\color{red}{=}}{\color{red}{a}}{\color{red}{x}}^{\color{red}{2}}{\color{red}{+}}{\color{red}{b}}{\color{red}{x}}{\color{red}{+}}{\color{red}{x}}\\f(x)=x^2+4x+5\end{array}[/latex]
Step 1: Find the direction of the parabola.
Since [latex]a[/latex] is [latex]1[/latex], the parabola opens upward.
Step 2: Find the axis of symmetry.
To find the axis of symmetry, find [latex]x=-\frac{b}{2a}[/latex].
[latex]\begin{eqnarray*}x&=&-\frac b{2a}\\x&=&-\frac4{(2)1}\\x&=&-2\end{eqnarray*}[/latex]
Step 3: Graph the axis of symmetry.
The equation of the axis of symmetry is [latex]x=-2[/latex].
The vertex is on the line [latex]x=-2[/latex].
Step 4: Find [latex]f\left(x\right)[/latex] when [latex]x=-2[/latex].
[latex]\begin{eqnarray*}f(x)&=&x^2+4x+5\\f(-2)&=&{(-2)}^2+4(-2)+5\\f(-2)&=&4-8+5\\f(-2)&=&1\end{eqnarray*}[/latex]
Step 5: Graph the vertex.
The vertex is [latex]\left(-2,1\right)[/latex].
The [latex]y[/latex]-intercept occurs when [latex]x=0[/latex].
[latex]f(x)=x^2+4x-5[/latex]
Step 6: Find [latex]f\left(0\right).[/latex]
[latex]f(0)=5[/latex]
Step 7: Simplify.
[latex]f(0)=5[/latex]
The [latex]y[/latex]-intercept is [latex]\left(0,5\right)[/latex].
Step 8: Graph the points.
The point [latex]\left(-4,5\right)[/latex] is two units to the left of the line of symmetry.
The point two units to the right of the line of symmetry is [latex]\left(0,5\right)[/latex].
Step 9: State the point that is symmetric to the [latex]y[/latex]-intercept.
Point symmetric to the [latex]y[/latex]-intercept is [latex]\left(-4,5\right)[/latex].
The [latex]x[/latex]-intercept occurs when [latex]f\left(x\right)=0[/latex].
Step 10: Find [latex]f\left(x\right)=0[/latex].
[latex]{\color{darkred}{0}}{=x^2+4x+5}[/latex]
Step 11: Test the discriminant.
[latex]\begin{eqnarray*}&=&b^2-4ac\\&=&4^2-4\times1\times5\\&=&16-20\\&=&-4\end{eqnarray*}[/latex]
Since the value of the discriminant is negative, there is no real solution and so no [latex]x[/latex]-intercept.
Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.
Try It
18) Graph [latex]f\;(x)\;=\;x^2-\;2x\;+\;3[/latex] by using its properties.
Solution
Try It
19) Graph [latex]f\;(x)\;=\;-3x^2\;-\;6x\;-\;4[/latex] by using its properties.
Solution
Finding the [latex]y[/latex]-intercept by finding [latex]f (0)[/latex] is easy, isn’t it? Sometimes we need to use the Quadratic Formula to find the [latex]x[/latex]-intercepts.
Example 4.6.9
Graph [latex]f(x)=2x^2−4x−3[/latex] by using its properties.
Solution
[latex]\begin{array}{c}{\color{red}{f}}{\color{red}{(}}{\color{red}{x}}{\color{red}{)}}{\color{red}{=}}{\color{red}{a}}{\color{red}{x}}^{\color{red}{2}}{\color{red}{+}}{\color{red}{b}}{\color{red}{x}}{\color{red}{+}}{\color{red}{x}}\\f(x)=2x^2-4x-3\end{array}[/latex]
Step 1: Find the direction of the parabola.
Since [latex]a[/latex] is [latex]2[/latex], the parabola opens upward.
Step 2: Find the axis of symmetry.
To find the equation of the axis of symmetry, use [latex]x=-\frac{b}{2a}[/latex].
[latex]\begin{eqnarray*}x&=&-\frac b{2a}\\x&=&-\frac{-4}{2\times2}\\x&=&1\end{eqnarray*}[/latex]
The equation of the axis of symmetry is [latex]x=1.[/latex]
The vertex is on the line [latex]x=1[/latex].
[latex]f(x)=2x^2-4x-3[/latex]
Step 3: Find [latex]f\left(1\right)[/latex].
[latex]\begin{eqnarray*}f(x)&=&2{({\color{red}{1}})}^2-4({\color{red}{1}})-3\\f(1)&=&2-4-3\\f(1)&=&-5\end{eqnarray*}[/latex]
The vertex is [latex]\left(1,-5\right)[/latex].
The [latex]y[/latex]-intercept occurs when [latex]x=0[/latex].
[latex]f(x)=2x^2-4x-3[/latex]
Step 4: Find [latex]f\left(0\right)[/latex].
[latex]{f(x)=2{(}{\color{darkred}{0}}{)}^2-4(}{\color{darkred}{0}}{)-3}[/latex]
Step 5: Simplify.
[latex]f(0)=-3[/latex]
The [latex]y[/latex]-intercept is [latex]\left(0,-3\right).[/latex]
Step 6: State the point that is symmetric to the y-intercept.
The point [latex]\left(0,-3\right)[/latex] is one unit to the left of the line of symmetry.
Point symmetric to the [latex]y[/latex]-intercept is[latex]\left(2,-3\right)[/latex]
The point one unit to the right of the line of symmetry is [latex]\left(2,-3\right)[/latex].
The [latex]x[/latex]-intercept occurs when [latex]y=0[/latex].
[latex]f(x)=2x^2-4x-3[/latex]
Step 7: Find [latex]f\left(x\right)=0[/latex].
[latex]{\color{darkred}{0}}{=2x^2-4x-3}[/latex]
Step 8: Use the Quadratic Formula.
[latex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/latex]
Step 9: Substitute in the values of [latex]a,b,[/latex] and [latex]c[/latex].
[latex]\begin{eqnarray*}{\color{red}{a}}{\color{red}{}}&=&{\color{red}{2}},\;{\color{blue}{b}}{\color{blue}{=}}{\color{blue}{-}}{\color{blue}{4}},\;{\color{green}{c}}{\color{green}{=}}{\color{green}{3}}\\x&=&\frac{-({\color{blue}{-}}{\color{blue}{4}})\pm\sqrt{{({\color{blue}{-}}{\color{blue}{4}})}^2-4({\color{red}{2}})({\color{green}{3}})}}{2({\color{red}{2}})}\end{eqnarray*}[/latex]
Step 10: Simplify.
[latex]x=\frac{4\pm\sqrt{16+24}}4[/latex]
Step 11: Simplify inside the radical.
[latex]x=\frac{4\pm\sqrt{40}}4[/latex]
Step 12: Simplify the radical.
[latex]x=\frac{4\pm2\sqrt{10}}4[/latex]
Step 13: Factor the GCF.
[latex]x=\frac{2(2\pm\sqrt{10})}4[/latex]
Step 14: Remove common factors.
[latex]x=\frac{2\pm\sqrt{10}}2[/latex]
Step 15: Write as two equations.
[latex]x=\frac{2+\sqrt{10}}2,\;x=\frac{2-\sqrt{10}}2[/latex]
Step 16: Approximate the values.
[latex]x\approx2.5,\;x\approx-0.6[/latex]
The approximate values of the [latex]x[/latex]-intercepts are [latex]\left(2.5,0\right)[/latex] and [latex]\left(-0.6,0\right).[/latex]
Step 17: Graph the parabola using the points found.
Try It
20) Graph [latex]f\;(x)\;=\;5x^2\;+\;10x\;+\;3[/latex] by using its properties.
Solution
Try It
21) Graph [latex]f\;(x)\;=\;-3x^2\;-\;6x\;+\;5[/latex] by using its properties.
Solution
Solve Maximum and Minimum Applications
Knowing that the vertex of a parabola is the lowest or highest point of the parabola gives us an easy way to determine the minimum or maximum value of a quadratic function. The [latex]y[/latex]-coordinate of the vertex is the minimum value of a parabola that opens upward. It is the maximum value of a parabola that opens downward. See 4.6.29.
Minimum or Maximum Values of a Quadratic Function
The [latex]y[/latex]-coordinate of the vertex of the graph of a quadratic function is the:
-
- minimum value of the quadratic equation if the parabola opens upward.
- maximum value of the quadratic equation if the parabola opens downward.
Example 4.6.10
Find the minimum or maximum value of the quadratic function [latex]f\left(x\right)={x}^{2}+2x-8[/latex].
Solution
Step 1: Find the direction of the parabola.
Since [latex]a[/latex] is positive, the parabola opens upward.
The quadratic equation has a minimum.
Step 2: Find the equation of the axis of symmetry.
[latex]\begin{eqnarray*}x&=&-\frac b{2a}\\x&=&-\frac2{2\times1}\\x&=&-1\end{eqnarray*}[/latex]
The equation of the axis of symmetry is [latex]x=-1[/latex].
The vertex is on the line [latex]x=-1[/latex].
[latex]f(x)=x^2+2x-8[/latex]
Step 3: Find [latex]f\left(-1\right)[/latex].
[latex]\begin{eqnarray*}f(-1)&=&{({\color{red}{-}}{\color{red}{1}})}^2+2({\color{red}{-}}{\color{red}{1}})-8\\f(-1)&=&1-2-8\\f(-1)&=&-9\end{eqnarray*}[/latex]
The vertex is [latex]\left(-1,-9\right)[/latex].
Since the parabola has a minimum, the [latex]y[/latex]-coordinate of the vertex is the minimum [latex]y[/latex]-value of the quadratic equation.
The minimum value of the quadratic is [latex]-9[/latex] and it occurs when [latex]x=-1[/latex].
Step 4: Show the graph to verify the result.
Try It
22) Find the maximum or minimum value of the quadratic function [latex]f\left(x\right)={x}^{2}-8x+12[/latex].
Solution
The minimum value of the quadratic function is [latex]−4[/latex] and it occurs when [latex]x = 4[/latex].
Try It
23) Find the maximum or minimum value of the quadratic function [latex]f\left(x\right)=-4{x}^{2}+16x-11[/latex].
Solution
The maximum value of the quadratic function is [latex]5[/latex] and it occurs when [latex]x = 2[/latex].
We have used the formula
to calculate the height in feet, [latex]h[/latex], of an object shot upwards into the air with initial velocity, [latex]v_0[/latex], after [latex]t[/latex] seconds .
This formula is a quadratic function, so its graph is a parabola. By solving for the coordinates of the vertex [latex](t,h)[/latex], we can find how long it will take the object to reach its maximum height. Then we can calculate the maximum height.
Example 4.6.11
The quadratic equation [latex]h(t)=−16t^2+176t+4[/latex] models the height of a volleyball hit straight upwards with velocity [latex]176[/latex] feet per second from a height of 4 feet.
a. How many seconds will it take the volleyball to reach its maximum height?
b. Find the maximum height of the volleyball.
Solution
Since [latex]a[/latex] is negative, the parabola opens downward.
The quadratic function has a maximum.
a.
Step 1: Find the equation of the axis of symmetry.
[latex]\begin{eqnarray*}t&=&-\frac b{2a}\\t&=&-\frac{176}{2\left(-16\right)}\\t&=&5.5\end{eqnarray*}[/latex]
Step 2: Find the axis of symmetry.
The equation of the axis of symmetry is [latex]t=5.5[/latex].
Step 3: Find the vertex.
The vertex is on the line [latex]t=5.5[/latex].
Step 4: Where is the maximum?
The maximum occurs when [latex]t=5.5[/latex] seconds.
b.
Step 1: Find [latex]h(5.5)[/latex].
[latex]\begin{eqnarray*}h(t)&=&-16t^2+176t+4\\h(t)&=&-16{({\color{red}{5}}{\color{red}{.}}{\color{red}{5}})}^2+176({\color{red}{5}}{\color{red}{.}}{\color{red}{5}})+4\\\end{eqnarray*}[/latex]
Step 2: Use a calculator to simplify.
[latex]h(t)=488[/latex]
Step 3: State the vertex.
The vertex is [latex](5.5,488)[/latex].
Step 4: Where is the maximum?
Since the parabola has a maximum, the [latex]h[/latex]-coordinate of the vertex is the maximum value of the quadratic function.
The maximum value of the quadratic is [latex]488[/latex] feet and it occurs when [latex]t=5.5[/latex] seconds.
Step 5: State the findings.
After [latex]5.5[/latex] seconds, the volleyball will reach its maximum height of [latex]488[/latex] feet.
Try It
24) Solve, rounding answers to the nearest tenth.
The quadratic function [latex]h(t)=−16t^2+128t+32[/latex] is used to find the height of a stone thrown upward from a height of [latex]32[/latex] feet at a rate of [latex]128[/latex] ft/sec.
How long will it take for the stone to reach its maximum height? What is the maximum height?
Solution
It will take [latex]4[/latex] seconds for the stone to reach its maximum height of [latex]288[/latex] feet.
Try It
25) A path of a toy rocket thrown upward from the ground at a rate of [latex]208[/latex] ft/sec is modelled by the quadratic function of [latex]h(t)=−16t^2+208t[/latex].
When will the rocket reach its maximum height? What will be the maximum height?
Solution
It will [latex]6.5[/latex] seconds for the rocket to reach its maximum height of [latex]676[/latex] feet.
Access these online resources for additional instruction and practice with graphing quadratic functions using properties.
Key Concepts
- Parabola Orientation
- For the graph of the quadratic function [latex]f\left(x\right)=a{x}^{2}+bx+c[/latex], if
- [latex]a>0[/latex], the parabola opens upward.
- [latex]a<0[/latex], the parabola opens downward.
- For the graph of the quadratic function [latex]f\left(x\right)=a{x}^{2}+bx+c[/latex], if
- Axis of Symmetry and Vertex of a Parabola The graph of the function [latex]f\left(x\right)=a{x}^{2}+bx+c[/latex] is a parabola where:
- the axis of symmetry is the vertical line [latex]x=-\frac{b}{2a}[/latex].
- the vertex is a point on the axis of symmetry, so its [latex]x[/latex]-coordinate is [latex]-\frac{b}{2a}[/latex].
- the [latex]y[/latex]-coordinate of the vertex is found by substituting [latex]x=-\frac{b}{2a}[/latex] into the quadratic equation.
- Find the Intercepts of a Parabola
- To find the intercepts of a parabola whose function is [latex]f\left(x\right)=a{x}^{2}+bx+c:[/latex]
[latex]\begin{array}{cccccc}y\textbf{-intercept}&&&&&x\textbf{-intercepts}\\\text{Let}\;x=0\;\text{and solve for}\;f\left(x\right)&&&&&\text{Let}\;f\left(x\right)=0\;\text{and solve for}\;x\end{array}[/latex]
- To find the intercepts of a parabola whose function is [latex]f\left(x\right)=a{x}^{2}+bx+c:[/latex]
- How to graph a quadratic function using properties.
- Determine whether the parabola opens upward or downward.
- Find the equation of the axis of symmetry.
- Find the vertex.
- Find the [latex]y[/latex]-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
- Find the [latex]x[/latex]-intercepts. Find additional points if needed.
- Graph the parabola.
- Minimum or Maximum Values of a Quadratic Equation
- The [latex]y[/latex]-coordinate of the vertex of the graph of a quadratic equation is the
- minimum value of the quadratic equation if the parabola opens upward.
- maximum value of the quadratic equation if the parabola opens downward.
Self Check
a. After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
b. After looking at the checklist, do you think you are well-prepared for the next section? Why or why not?
Glossary
- Quadratic Function
- A quadratic function, where [latex]a[/latex], [latex]b[/latex], and [latex]c[/latex] are real numbers and [latex]a\ne{0}[/latex], is a function of the form [latex]f\left(x\right)=a{x}^{2}+bx+c[/latex].
A quadratic function, where a, b, and c are real numbers and [latex]a\ne 0,[/latex] is a function of the form [latex]f\left(x\right)=a{x}^{2}+bx+c.[/latex]
