Graph Quadratic Functions Using Properties

Lynn Marecek

Quadratic Equations and Functions

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

Before you get started, take this readiness quiz.

Graph the function $f\left(x\right)={x}^{2}$ by plotting points.

If you missed this problem, review (Figure).
Solve: $2{x}^{2}+3x-2=0.$

If you missed this problem, review (Figure).
Evaluate $-\frac{b}{2a}$ when a = 3 and b = −6.

If you missed this problem, review (Figure).

Recognize the Graph of a Quadratic Function

Previously we very briefly looked at the function $f\left(x\right)={x}^{2}$ , 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 $f\left(x\right)=a{x}^{2}+bx+c$ if $a\ne 0.$ We call this kind of function a quadratic function.

Quadratic Function

A quadratic function, where a, b, and c are real numbers and $a\ne 0,$ is a function of the form

$f\left(x\right)=a{x}^{2}+bx+c$

We graphed the quadratic function $f\left(x\right)={x}^{2}$ by plotting points.

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.

Graph $f\left(x\right)={x}^{2}-1.$

We will graph the function by plotting points.

Choose integer values for x,

substitute them into the equation

and simplify to find $f\left(x\right)$ .

Record the values of the ordered pairs in the chart.

Plot the points, and then connect

them with a smooth curve. The

result will be the graph of the

function $f\left(x\right)={x}^{2}-1$ .

Graph $f\left(x\right)=\text{−}{x}^{2}.$ .

This figure shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, 0).

Graph $f\left(x\right)={x}^{2}+1.$

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (0, −1).

All graphs of quadratic functions of the form f (x) = ax² + bx + c are parabolas that open upward or downward. See (Figure).

This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The general form for the equation of this graph is f of x equals a x squared plus b x plus c. The equation of this parabola is x squared plus 4 x plus 3. The leading coefficient, a, is greater than 0, so this parabola opens upward.The graph on the right shows an downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The general form for the equation of this graph is f of x equals a x squared plus b x plus c. The equation of this parabola is negative x squared plus 4 x plus 3. The leading coefficient, a, is less than 0, so this parabola opens downward.

Notice that the only difference in the two functions is the negative sign before the quadratic term (x² in the equation of the graph in (Figure)). 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 f (x) = ax² + bx + c, if

This images shows a bulleted list. The first bullet notes that, if a is greater than 0, then the parabola opens upward and shows an image of an upward-opening parabola. The second bullet notes that, if a is less than 0, then the parabola opens downward and shows an image of a downward-opening parabola.

Determine whether each parabola opens upward or downward:

ⓐ $f\left(x\right)=-3{x}^{2}+2x-4$ ⓑ $f\left(x\right)=6{x}^{2}+7x-9.$

ⓐ

Find the value of “a”.
	Since the “a” is negative, the parabola will open downward.

ⓑ

Find the value of “a”.
	Since the “a” is positive, the parabola will open upward.

Determine whether the graph of each function is a parabola that opens upward or downward:

ⓐ $f\left(x\right)=2{x}^{2}+5x-2$ ⓑ $f\left(x\right)=-3{x}^{2}-4x+7.$

ⓐ up; ⓑ down

Determine whether the graph of each function is a parabola that opens upward or downward:

ⓐ $f\left(x\right)=-2{x}^{2}-2x-3$ ⓑ $f\left(x\right)=5{x}^{2}-2x-1.$

ⓐ down; ⓑ up

Find the Axis of Symmetry and Vertex of a Parabola

Look again at (Figure). 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).

This image shows 2 graphs side-by-side. The graph on the left shows an upward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1) and passes through the points (negative 4, 3) and (0, 3). The equation of this parabola is x squared plus 4 x plus 3. The vertical line passes through the point (negative 2, 0) and has the equation x equals negative 2. The graph on the right shows an downward-opening parabola and a dashed vertical line graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7) and passes through the points (0, 3) and (4, 3). The equation of this parabola is negative x squared plus 4 x plus 3. The vertical line passes through the point (2, 0) and has the equation x equals 2.

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 f (x) = ax² + bx + c is $x=-\frac{b}{2a}.$

So to find the equation of symmetry of each of the parabolas we graphed above, we will substitute into the formula $x=-\frac{b}{2a}.$

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

x-coordinate is $-\frac{b}{2a}.$ To find the y-coordinate of the vertex we substitute the value of the x-coordinate into the quadratic function.

Axis of Symmetry and Vertex of a Parabola

The graph of the function f (x) = ax² + bx + c is a parabola where:

the axis of symmetry is the vertical line $x=-\frac{b}{2a}.$
the vertex is a point on the axis of symmetry, so its x-coordinate is $-\frac{b}{2a}.$
the y-coordinate of the vertex is found by substituting $x=-\frac{b}{2a}$ into the quadratic equation.

For the graph of $f\left(x\right)=3{x}^{2}-6x+2$ find:

ⓐ the axis of symmetry ⓑ the vertex.

ⓐ


The axis of symmetry is the vertical line $x=-\frac{b}{2a}$ .
Substitute the values of $a,b$ into the equation.
Simplify.
	The axis of symmetry is the line $x=1$ .

ⓑ


The vertex is a point on the line of symmetry, so its x-coordinate will be $x=1$ . Find $f\left(1\right)$ .
Simplify.
The result is the y-coordinate.
	The vertex is $\left(1,-1\right)$ .

For the graph of $f\left(x\right)=2{x}^{2}-8x+1$ find:

ⓐ the axis of symmetry ⓑ the vertex.

ⓐ $x=2;$ ⓑ $\left(2,-7\right)$

For the graph of $f\left(x\right)=2{x}^{2}-4x-3$ find:

ⓐ the axis of symmetry ⓑ the vertex.

ⓐ $x=1;$ ⓑ $\left(1,-5\right)$

Find the Intercepts of a Parabola

When we graphed linear equations, we often used the x– and y-intercepts to help us graph the lines. Finding the coordinates of the intercepts will help us to graph parabolas, too.

Remember, at the y-intercept the value of x is zero. So to find the y-intercept, we substitute x = 0 into the function.

Let’s find the y-intercepts of the two parabolas shown in (Figure).

An x-intercept results when the value of f (x) is zero. To find an x-intercept, we let f (x) = 0. In other words, we will need to solve the equation 0 = ax² + bx + c for x.

$\begin{array}{ccc}\hfill f\left(x\right)& =\hfill & a{x}^{2}+bx+c\hfill \\ \hfill 0& =\hfill & a{x}^{2}+bx+c\hfill \end{array}$

Solving quadratic equations like this is exactly what we have done earlier in this chapter!

We can now find the x-intercepts of the two parabolas we looked at. First we will find the x-intercepts of the parabola whose function is f (x) = x² + 4x + 3.


Let $f\left(x\right)=0$ .
Factor.
Use the Zero Product Property.
Solve.
	The x-intercepts are $\left(-1,0\right)$ and $\left(-3,0\right)$ .

Now we will find the x-intercepts of the parabola whose function is f (x) = −x² + 4x + 3.


Let $f\left(x\right)=0$ .
This quadratic does not factor, so we use the Quadratic Formula.
$a=-1,b=4,c=3$
Simplify.



	The x-intercepts are $\left(2+\sqrt{7},0\right)$ and $\left(2-\sqrt{7},0\right)$ .

We will use the decimal approximations of the x-intercepts, so that we can locate these points on the graph,

$\begin{array}{cccccc}\left(2+\sqrt{7},0\right)\approx \left(4.6,0\right)\hfill & & & & & \left(2-\sqrt{7},0\right)\approx \left(-0.6,0\right)\hfill \end{array}$

Do these results agree with our graphs? See (Figure).

This image shows 2 graphs side-by-side. The graph on the left shows the upward-opening parabola defined by the function f of x equals x squared plus 4 x plus 3 and a dashed vertical line, x equals negative 2, graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (negative 2, negative 1). The y-intercept is (0, 3) and the x-intercepts are (negative 1, 0) and (negative 3, 0). The graph on the right shows the downward-opening parabola defined by the function f of x equals negative x squared plus 4 x plus 3 and a dashed vertical line, x equals 2, graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 10 to 10. The parabola has a vertex at (2, 7). The y-intercept is (0, 3) and the x-intercepts are (2 plus square root 7, 0), approximately (4.6, 0) and (2 minus square root, 0), approximately (negative 0.6, 0).

Find the Intercepts of a Parabola

To find the intercepts of a parabola whose function is $f\left(x\right)=a{x}^{2}+bx+c:$

$\begin{array}{cccccc}\hfill \mathbit{\text{y}}\mathbf{\text{-intercept}}\hfill & & & & & \hfill \mathbit{\text{x}}\mathbf{\text{-intercepts}}\hfill \\ \hfill \text{Let}\phantom{\rule{0.2em}{0ex}}x=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}f\left(x\right).\hfill & & & & & \hfill \text{Let}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}x.\hfill \end{array}$

Find the intercepts of the parabola whose function is $f\left(x\right)={x}^{2}-2x-8.$

To find the y–intercept, let $x=0$ and solve for $f\left(x\right)$ .


	When $x=0$ , then $f\left(0\right)=-8$ . The y–intercept is the point $\left(0,-8\right)$ .
To find the x–intercept, let $f\left(x\right)=0$ and solve for $x$ .

Solve by factoring.


	When $f\left(x\right)=0$ , then $x=4\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}x=-2$ . The x–intercepts are the points $\left(4,0\right)$ and $\left(-2,0\right)$ .

Find the intercepts of the parabola whose function is $f\left(x\right)={x}^{2}+2x-8.$

y-intercept: $\left(0,-8\right)$ x-intercepts $\left(-4,0\right),\left(2,0\right)$

Find the intercepts of the parabola whose function is $f\left(x\right)={x}^{2}-4x-12.$

y-intercept: $\left(0,-12\right)$ x-intercepts $\left(-2,0\right),\left(6,0\right)$

In this chapter, we have been solving quadratic equations of the form ax² + bx + c = 0. We solved for x and the results were the solutions to the equation.

We are now looking at quadratic functions of the form f (x) = ax² + bx + c. The graphs of these functions are parabolas. The x–intercepts of the parabolas occur where f (x) = 0.

For example:

$\begin{array}{ccccccccc}\hfill \mathbf{\text{Quadratic equation}}\hfill & & & & & & & & \hfill \phantom{\rule{4em}{0ex}}\mathbf{\text{Quadratic function}}\hfill \\ \hfill \begin{array}{}\\ \hfill {x}^{2}-2x-15& =\hfill & 0\hfill \\ \hfill \left(x-5\right)\left(x+3\right)& =\hfill & 0\hfill \\ \hfill x-5=0\phantom{\rule{0.8em}{0ex}}x+3& =\hfill & 0\hfill \\ \hfill x=5\phantom{\rule{2.3em}{0ex}}x& =\hfill & -3\hfill \end{array}\hfill & & & & \hfill \begin{array}{c}\hfill \text{Let}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=0.\hfill \\ \\ \end{array}\hfill & & & & \hfill \phantom{\rule{4em}{0ex}}\begin{array}{ccc}\hfill f\left(x\right)& =\hfill & {x}^{2}-2x-15\hfill \\ \hfill 0& =\hfill & {x}^{2}-2x-15\hfill \\ \hfill 0& =\hfill & \left(x-5\right)\left(x+3\right)\hfill \\ \hfill x-5& =\hfill & 0\phantom{\rule{0.8em}{0ex}}x+3=0\hfill \\ \hfill x& =\hfill & 5\phantom{\rule{2.3em}{0ex}}x=-3\hfill \end{array}\hfill \\ & & & & & & & & \hfill \phantom{\rule{4em}{0ex}}\left(5,0\right)\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\left(-3,0\right)\hfill \\ & & & & & & & & \hfill \phantom{\rule{4em}{0ex}}x\text{-intercepts}\hfill \end{array}$

The solutions of the quadratic function are the x values of the x–intercepts.

Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show examples of parabolas for these three cases. Since the solutions of the functions give the x–intercepts of the graphs, the number of x–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 $a{x}^{2}+bx+c=0.$ Now we can use the discriminant to tell us how many x-intercepts there are on the graph.

Before you to find the values of the x-intercepts, you may want to evaluate the discriminant so you know how many solutions to expect.

Find the intercepts of the parabola for the function $f\left(x\right)=5{x}^{2}+x+4.$


To find the y-intercept, let $x=0$ and solve for $f\left(x\right)$ .

	When $x=0$ , then $f\left(0\right)=4$ . The y-intercept is the point $\left(0,4\right)$ .
To find the x-intercept, let $f\left(x\right)=0$ and solve for $x$ .

Find the value of the discriminant to predict the number of solutions which is also the number of x-intercepts.
$\begin{array}{c}\hfill {b}^{2}-4ac\hfill \\ \hfill {1}^{2}-4·5·4\hfill \\ \hfill 1-80\hfill \\ \hfill -79\hfill \end{array}$
	Since the value of the discriminant is negative, there is no real solution to the equation. There are no x-intercepts.

Find the intercepts of the parabola whose function is $f\left(x\right)=3{x}^{2}+4x+4.$

y-intercept: $\left(0,4\right)$ no x-intercept

Find the intercepts of the parabola whose function is $f\left(x\right)={x}^{2}-4x-5.$

y-intercept: $\left(0,-5\right)$ x-intercepts $\left(-1,0\right),\left(5,0\right)$

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.

How to Graph a Quadratic Function Using Properties

Graph f (x) = x² −6x + 8 by using its properties.

Step 1 is to determine whether the parabola opens upward or downward. Loot at the leading coefficient, a, in the equation. If f of x equals x squared minus 6 x plus 8, then a equals 1. Since a is positive, the parabola opens upward.

Step 2 is to find the axis of symmetry. The axis of symmetry is the line x equals negative b divided by the product 2 a. For the function f of x equals x squared minust 6 x plus 8, the axis of symmetry is negative b divided by the product 2 a. x equals the opposite of negative 6 divided by the product 2 times 1. X equals 3.

In step 3, find the vertex. The vertex is on the axis of symmetry. Substitute x equals 3 into the function. F of x equals x squared minus 6 x plus 8. F of 3 equals 3 squared minus 6 times 3 plus 8. F of 3 equals negative 1. The vertex is the point (3, negative 1).

Step 4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry. We first find f of 0 to find the y-intercept. F of x equals x squared plus 6 x plus 8, so f of 0 equals 0 squared plus 6 times 0 plus 8. F of 0 equals 8. The y-intercept is the point (0, 8). We use the axis of symmetry to find a point symmetric to the y-intercept. The y-intercept is 3 units left of the axis of symmetry, x equals 3. A point 3 units to the right of the axis of symmetry has x-value 6. The point symmetric to the y-intercept is the point (6, 8).

Step 5 is to find the x-intercepts. Find additional points if needed. We solve f of x equals 0. We can solve this quadratic equation by factoring. To find the x-intercepts, set f of x equal to 0. F of x equals x squared minus 6 x plus 8. 0 equals x squared minus 6 x plus 8. 0 equals the product of x minus 2 and x minus 4. So x equals 2 or x equals 4. The x-intercepts are the points (2, 0) and (4, 0).

The final step, step 6, is to graph the parabola. We graph the vertex, intercepts, and the point symmetric to the y-intercept. We connect these 5 points to sketch the parabola. An image shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 10. The y-axis of the plane runs from negative 3 to 7. The parabola has a vertex at (3, negative 1). Other points plotted include the x-intercepts, (2, 0) and (4, 0), the y-intercept, (0, 8), and the point (6, 8) that is symmetric to the y-intercept across the axis of symmetry.

Graph f (x) = x² + 2x − 8 by using its properties.

Graph f (x) = x² − 8x + 12 by using its properties.

We list the steps to take in order to graph a quadratic function here.

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 y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
Find the x-intercepts. Find additional points if needed.
Graph the parabola.

We were able to find the x-intercepts in the last example by factoring. We find the x-intercepts in the next example by factoring, too.

Graph f (x) = x² + 6x − 9 by using its properties.


Since a is $-1$ , the parabola opens downward.

To find the equation of the axis of symmetry, use $x=-\frac{b}{2a}$ .


	The axis of symmetry is $x=3$ . The vertex is on the line $x=3$ .

Find $f\left(3\right)$ .



	The vertex is $\left(3,0\right).$

The y-intercept occurs when $x=0$ . Find $f\left(0\right)$ .
Substitute $x=0$ .
Simplify.
	The y-intercept is $\left(0,-9\right).$
The point $\left(0,-9\right)$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $\left(6,-9\right)$ .
	Point symmetric to the y-intercept is $\left(6,-9\right)$
The x-intercept occurs when $f\left(x\right)=0$ .
Find $f\left(x\right)=0$ .
Factor the GCF.
Factor the trinomial.
Solve for x.
Connect the points to graph the parabola.

Graph f (x) = 3x² + 12x − 12 by using its properties.

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 10 to 10. The y-axis of the plane runs from negative 15 to 10. The parabola has a vertex at (2, 0). The y-intercept (0, negative 12) is plotted as well as the axis of symmetry, x equals 2.

Graph f (x) = 4x² + 24x + 36 by using its properties.

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 30 to 20. The y-axis of the plane runs from negative 10 to 40. The parabola has a vertex at (negative 3, 0). The y-intercept (0, 36) is plotted as well as the axis of symmetry, x equals negative 3.

For the graph of f (x) = −x² + 6x − 9, the vertex and the x-intercept were the same point. Remember how the discriminant determines the number of solutions of a quadratic equation? The discriminant of the equation 0 = −x² + 6x − 9 is 0, so there is only one solution. That means there is only one x-intercept, and it is the vertex of the parabola.

How many x-intercepts would you expect to see on the graph of f (x) = x² + 4x + 5?

Graph f (x) = x² + 4x + 5 by using its properties.


Since a is 1, the parabola opens upward.

To find the axis of symmetry, find $x=-\frac{b}{2a}$ .


	The equation of the axis of symmetry is $x=-2$ .

The vertex is on the line $x=-2.$
Find $f\left(x\right)$ when $x=-2.$



	The vertex is $\left(-2,1\right)$ .

The y-intercept occurs when $x=0$ .
Find $f\left(0\right).$
Simplify.
	The y-intercept is $\left(0,5\right)$ .
The point $\left(-4,5\right)$ is two units to the left of the line of symmetry. The point two units to the right of the line of symmetry is $\left(0,5\right)$ .
	Point symmetric to the y-intercept is $\left(-4,5\right)$ .
The x-intercept occurs when $f\left(x\right)=0$ .
Find $f\left(x\right)=0$ .
Test the discriminant.




Since the value of the discriminant is negative, there is no real solution and so no x-intercept.
Connect the points to graph the parabola. You may want to choose two more points for greater accuracy.

Graph f (x) = x² − 2x + 3 by using its properties.

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 2 to 4. The y-axis of the plane runs from negative 1 to 5. The parabola has a vertex at (1, 2). The y-intercept (0, 3) is plotted as is the line of symmetry, x equals 1.

Graph f (x) = −3x² − 6x − 4 by using its properties.

This figure shows a downward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 2. The y-axis of the plane runs from negative 5 to 1. The parabola has a vertex at (negative 1, negative 2). The y-intercept (0, negative 4) is plotted as is the line of symmetry, x equals negative 1.

Finding the y-intercept by finding f (0) is easy, isn’t it? Sometimes we need to use the Quadratic Formula to find the x-intercepts.

Graph f (x) = 2x² − 4x − 3 by using its properties.


Since a is 2, the parabola opens upward.
To find the equation of the axis of symmetry, use $x=-\frac{b}{2a}$ .


	The equation of the axis of symmetry is $x=1.$
The vertex is on the line $x=1$ .
Find $f\left(1\right)$ .


	The vertex is $\left(1,-5\right).$
The y-intercept occurs when $x=0$ .
Find $f\left(0\right)$ .
Simplify.
	The y-intercept is $\left(0,-3\right).$
The point $\left(0,-3\right)$ is one unit to the left of the line of symmetry.	Point symmetric to the y-intercept is $\left(2,-3\right)$
The point one unit to the right of the line of symmetry is $\left(2,-3\right)$ .
The x-intercept occurs when $y=0$ .
Find $f\left(x\right)=0$ .
Use the Quadratic Formula.
Substitute in the values of $a,b,$ and $c.$
Simplify.
Simplify inside the radical.
Simplify the radical.
Factor the GCF.
Remove common factors.
Write as two equations.
Approximate the values.
	The approximate values of the x-intercepts are $\left(2.5,0\right)$ and $\left(-0.6,0\right).$
Graph the parabola using the points found.

Graph f (x) = 5x² + 10x + 3 by using its properties.

This figure shows an upward-opening parabola graphed on the x y-coordinate plane. The x-axis of the plane runs from negative 4 to 4. The y-axis of the plane runs from negative 4 to 4. The axis of symmetry, x equals negative 1, is graphed as a dashed line. The parabola has a vertex at (negative 1, negative 2). The y-intercept of the parabola is the point (0, 3). The x-intercepts of the parabola are approximately (negative 1.6, 0) and (negative 0.4, 0).

Graph f (x) = −3x² − 6x + 5 by using its properties.

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 y-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 (Figure).

This figure shows 2 graphs side-by-side. The left graph shows a downward opening parabola plotted in the x y-plane. An arrow points to the vertex with the label maximum. The right graph shows an upward opening parabola plotted in the x y-plane. An arrow points to the vertex with the label minimum.

Minimum or Maximum Values of a Quadratic Function

The y-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.

Find the minimum or maximum value of the quadratic function $f\left(x\right)={x}^{2}+2x-8.$


Since a is positive, the parabola opens upward. The quadratic equation has a minimum.
Find the equation of the axis of symmetry.


	The equation of the axis of symmetry is $x=-1$ .
The vertex is on the line $x=-1$ .
Find $f\left(-1\right)$ .


	The vertex is $\left(-1,-9\right)$ .
Since the parabola has a minimum, the y-coordinate of the vertex is the minimum y-value of the quadratic equation. The minimum value of the quadratic is $-9$ and it occurs when $x=-1$ .

Show the graph to verify the result.

Find the maximum or minimum value of the quadratic function $f\left(x\right)={x}^{2}-8x+12.$

The minimum value of the quadratic function is −4 and it occurs when x = 4.

Find the maximum or minimum value of the quadratic function $f\left(x\right)=-4{x}^{2}+16x-11.$

The maximum value of the quadratic function is 5 and it occurs when x = 2.

We have used the formula

$h\left(t\right)=-16{t}^{2}+{v}_{0}t+{h}_{0}$

to calculate the height in feet, h , of an object shot upwards into the air with initial velocity, v₀, after t seconds .

This formula is a quadratic function, so its graph is a parabola. By solving for the coordinates of the vertex (t, h), we can find how long it will take the object to reach its maximum height. Then we can calculate the maximum height.

The quadratic equation h(t) = −16t² + 176t + 4 models the height of a volleyball hit straight upwards with velocity 176 feet per second from a height of 4 feet.

ⓐ How many seconds will it take the volleyball to reach its maximum height? ⓑ Find the maximum height of the volleyball.

$\begin{array}{cccc}& & & \phantom{\rule{4em}{0ex}}h\left(t\right)=-16{t}^{2}+176t+4\hfill \\ \begin{array}{c}\text{Since}\phantom{\rule{0.2em}{0ex}}a\phantom{\rule{0.2em}{0ex}}\text{is negative, the parabola opens}\hfill \\ \text{downward.}\hfill \\ \text{The quadratic function has a maximum.}\hfill \end{array}\hfill & & & \end{array}$

ⓐ

$\begin{array}{cccc}\text{Find the equation of the axis of symmetry.}\hfill & & & \phantom{\rule{4em}{0ex}}t=-\frac{b}{2a}\hfill \\ & & & \phantom{\rule{4em}{0ex}}t=-\frac{176}{2\left(-16\right)}\hfill \\ & & & \phantom{\rule{4em}{0ex}}t=5.5\hfill \\ & & & \phantom{\rule{4em}{0ex}}\begin{array}{c}\text{The equation of the axis of symmetry is}\hfill \\ t=5.5.\hfill \end{array}\hfill \\ \text{The vertex is on the line}\phantom{\rule{0.2em}{0ex}}t=5.5.\hfill & & & \phantom{\rule{4em}{0ex}}\begin{array}{c}\text{The maximum occurs when}\phantom{\rule{0.2em}{0ex}}t=5.5\hfill \\ \text{seconds.}\hfill \end{array}\hfill \end{array}$

ⓑ

$\begin{array}{cccccc}\text{Find}\phantom{\rule{0.2em}{0ex}}h\left(5.5\right).\hfill & & & & & \phantom{\rule{9em}{0ex}}h\left(t\right)=-16{t}^{2}+176t+4\hfill \\ & & & & & \phantom{\rule{9em}{0ex}}h\left(t\right)=-16{\left(5.5\right)}^{2}+176\left(5.5\right)+4\hfill \\ \text{Use a calculator to simplify.}\hfill & & & & & \phantom{\rule{9em}{0ex}}h\left(t\right)=488\hfill \\ & & & & & \phantom{\rule{9em}{0ex}}\text{The vertex is}\phantom{\rule{0.2em}{0ex}}\left(5.5,488\right).\hfill \end{array}$

Since the parabola has a maximum, the h-coordinate of the vertex is the maximum value of the quadratic function.

The maximum value of the quadratic is 488 feet and it occurs when t = 5.5 seconds.

After 5.5 seconds, the volleyball will reach its maximum height of 488 feet.

Solve, rounding answers to the nearest tenth.

The quadratic function h(t) = −16t² + 128t + 32 is used to find the height of a stone thrown upward from a height of 32 feet at a rate of 128 ft/sec. How long will it take for the stone to reach its maximum height? What is the maximum height?

It will take 4 seconds for the stone to reach its maximum height of 288 feet.

A path of a toy rocket thrown upward from the ground at a rate of 208 ft/sec is modeled by the quadratic function of h(t) = −16t² + 208t. When will the rocket reach its maximum height? What will be the maximum height?

It will 6.5 seconds for the rocket to reach its maximum height of 676 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 if
  - a > 0, the parabola opens upward.
  - a < 0, the parabola opens downward.
Axis of Symmetry and Vertex of a Parabola The graph of the function is a parabola where:
- the axis of symmetry is the vertical line $x=-\frac{b}{2a}.$
- the vertex is a point on the axis of symmetry, so its x-coordinate is $-\frac{b}{2a}.$
- the y-coordinate of the vertex is found by substituting $x=-\frac{b}{2a}$ into the quadratic equation.
Find the Intercepts of a Parabola
- To find the intercepts of a parabola whose function is $f\left(x\right)=a{x}^{2}+bx+c:$
  
  $\begin{array}{cccccc}\hfill \mathbit{\text{y}}\mathbf{\text{-intercept}}\hfill & & & & & \hfill \mathbit{\text{x}}\mathbf{\text{-intercepts}}\hfill \\ \hfill \text{Let}\phantom{\rule{0.2em}{0ex}}x=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}f\left(x\right).\hfill & & & & & \hfill \text{Let}\phantom{\rule{0.2em}{0ex}}f\left(x\right)=0\phantom{\rule{0.2em}{0ex}}\text{and solve for}\phantom{\rule{0.2em}{0ex}}x.\hfill \end{array}$
How to graph a quadratic function using properties.
1. Determine whether the parabola opens upward or downward.
2. Find the equation of the axis of symmetry.
3. Find the vertex.
4. Find the y-intercept. Find the point symmetric to the y-intercept across the axis of symmetry.
5. Find the x-intercepts. Find additional points if needed.
6. Graph the parabola.
Minimum or Maximum Values of a Quadratic Equation
- The y-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.

Practice Makes Perfect

Recognize the Graph of a Quadratic Function

In the following exercises, graph the functions by plotting points.

$f\left(x\right)={x}^{2}+3$

$f\left(x\right)={x}^{2}-3$

$y=\text{−}{x}^{2}+1$

$f\left(x\right)=\text{−}{x}^{2}-1$

For each of the following exercises, determine if the parabola opens up or down.

ⓐ $f\left(x\right)=-2{x}^{2}-6x-7$

ⓑ $f\left(x\right)=6{x}^{2}+2x+3$

ⓐ down ⓑ up

ⓐ $f\left(x\right)=4{x}^{2}+x-4$

ⓑ $f\left(x\right)=-9{x}^{2}-24x-16$

ⓐ $f\left(x\right)=-3{x}^{2}+5x-1$

ⓑ $f\left(x\right)=2{x}^{2}-4x+5$

ⓐ down ⓑ up

ⓐ $f\left(x\right)={x}^{2}+3x-4$

ⓑ $f\left(x\right)=-4{x}^{2}-12x-9$

Find the Axis of Symmetry and Vertex of a Parabola

In the following functions, find ⓐ the equation of the axis of symmetry and ⓑ the vertex of its graph.

$f\left(x\right)={x}^{2}+8x-1$

ⓐ $x=-4$ ; ⓑ $\left(-4,-17\right)$

$f\left(x\right)={x}^{2}+10x+25$

$f\left(x\right)=\text{−}{x}^{2}+2x+5$

ⓐ $x=1$ ; ⓑ $\left(1,2\right)$

$f\left(x\right)=-2{x}^{2}-8x-3$

Find the Intercepts of a Parabola

In the following exercises, find the intercepts of the parabola whose function is given.

$f\left(x\right)={x}^{2}+7x+6$

y-intercept: $\left(0,6\right);$ x-intercept $\left(-1,0\right),\left(-6,0\right)$

$f\left(x\right)={x}^{2}+10x-11$

$f\left(x\right)={x}^{2}+8x+12$

y-intercept: $\left(0,12\right);$ x-intercept $\left(-2,0\right),\left(-6,0\right)$

$f\left(x\right)={x}^{2}+5x+6$

$f\left(x\right)=\text{−}{x}^{2}+8x-19$

y-intercept: $\left(0,-19\right);$ x-intercept: none

$f\left(x\right)=-3{x}^{2}+x-1$

$f\left(x\right)={x}^{2}+6x+13$

y-intercept: $\left(0,13\right);$ x-intercept: none

$f\left(x\right)={x}^{2}+8x+12$

$f\left(x\right)=4{x}^{2}-20x+25$

y-intercept: $\left(0,-16\right);$ x-intercept $\left(\frac{5}{2},0\right)$

$f\left(x\right)=\text{−}{x}^{2}-14x-49$

$f\left(x\right)=\text{−}{x}^{2}-6x-9$

y-intercept: $\left(0,9\right);$ x-intercept $\left(-3,0\right)$

$f\left(x\right)=4{x}^{2}+4x+1$

Graph Quadratic Functions Using Properties

In the following exercises, graph the function by using its properties.

$f\left(x\right)={x}^{2}+6x+5$

$f\left(x\right)={x}^{2}+4x-12$

$f\left(x\right)={x}^{2}+4x+3$

$f\left(x\right)={x}^{2}-6x+8$

$f\left(x\right)=9{x}^{2}+12x+4$

$f\left(x\right)=\text{−}{x}^{2}+8x-16$

$f\left(x\right)=\text{−}{x}^{2}+2x-7$

$f\left(x\right)=5{x}^{2}+2$

$f\left(x\right)=2{x}^{2}-4x+1$

$f\left(x\right)=3{x}^{2}-6x-1$

$f\left(x\right)=2{x}^{2}-4x+2$

$f\left(x\right)=-4{x}^{2}-6x-2$

$f\left(x\right)=\text{−}{x}^{2}-4x+2$

$f\left(x\right)={x}^{2}+6x+8$

$f\left(x\right)=5{x}^{2}-10x+8$

$f\left(x\right)=-16{x}^{2}+24x-9$

$f\left(x\right)=3{x}^{2}+18x+20$

$f\left(x\right)=-2{x}^{2}+8x-10$

Solve Maximum and Minimum Applications

In the following exercises, find the maximum or minimum value of each function.

$f\left(x\right)=2{x}^{2}+x-1$

The minimum value is $-\frac{9}{8}$ when $x=-\frac{1}{4}.$

$y=-4{x}^{2}+12x-5$

$y={x}^{2}-6x+15$

The maximum value is 6 when x = 3.

$y=\text{−}{x}^{2}+4x-5$

$y=-9{x}^{2}+16$

The maximum value is 16 when x = 0.

$y=4{x}^{2}-49$

In the following exercises, solve. Round answers to the nearest tenth.

An arrow is shot vertically upward from a platform 45 feet high at a rate of 168 ft/sec. Use the quadratic function h(t) = −16t² + 168t + 45 find how long it will take the arrow to reach its maximum height, and then find the maximum height.

In 5.3 sec the arrow will reach maximum height of 486 ft.

A stone is thrown vertically upward from a platform that is 20 feet height at a rate of 160 ft/sec. Use the quadratic function h(t) = −16t² + 160t + 20 to find how long it will take the stone to reach its maximum height, and then find the maximum height.

A ball is thrown vertically upward from the ground with an initial velocity of 109 ft/sec. Use the quadratic function h(t) = −16t² + 109t + 0 to find how long it will take for the ball to reach its maximum height, and then find the maximum height.

In 3.4 seconds the ball will reach its maximum height of 185.6 feet.

A ball is thrown vertically upward from the ground with an initial velocity of 122 ft/sec. Use the quadratic function h(t) = −16t² + 122t + 0 to find how long it will take for the ball to reach its maximum height, and then find the maximum height.

A computer store owner estimates that by charging x dollars each for a certain computer, he can sell 40 − x computers each week. The quadratic function R(x) = −x² +40x is used to find the revenue, R, received when the selling price of a computer is x, Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

20 computers will give the maximum of ?400 in receipts.

A retailer who sells backpacks estimates that by selling them for x dollars each, he will be able to sell 100 − x backpacks a month. The quadratic function R(x) = −x² +100x is used to find the R, received when the selling price of a backpack is x. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue.

A retailer who sells fashion boots estimates that by selling them for x dollars each, he will be able to sell 70 − x boots a week. Use the quadratic function R(x) = −x² +70x to find the revenue received when the average selling price of a pair of fashion boots is x. Find the selling price that will give him the maximum revenue, and then find the amount of the maximum revenue per day.

He will be able to sell 35 pairs of boots at the maximum revenue of ?1,225.

A cell phone company estimates that by charging x dollars each for a certain cell phone, they can sell 8 − x cell phones per day. Use the quadratic function R(x) = −x² +8x to find the revenue received per day when the selling price of a cell phone is x. Find the selling price that will give them the maximum revenue per day, and then find the amount of the maximum revenue.

A rancher is going to fence three sides of a corral next to a river. He needs to maximize the corral area using 240 feet of fencing. The quadratic equation A(x) = x(240 − 2x) gives the area of the corral, A, for the length, x, of the corral along the river. Find the length of the corral along the river that will give the maximum area, and then find the maximum area of the corral.

The length of the side along the river of the corral is 120 feet and the maximum area is 7,200 square feet.

A veterinarian is enclosing a rectangular outdoor running area against his building for the dogs he cares for. He needs to maximize the area using 100 feet of fencing. The quadratic function A(x) = x(100 − 2x) gives the area, A, of the dog run for the length, x, of the building that will border the dog run. Find the length of the building that should border the dog run to give the maximum area, and then find the maximum area of the dog run.

A land owner is planning to build a fenced in rectangular patio behind his garage, using his garage as one of the “walls.” He wants to maximize the area using 80 feet of fencing. The quadratic function A(x) = x(80 − 2x) gives the area of the patio, where x is the width of one side. Find the maximum area of the patio.

The maximum area of the patio is 800 feet.

A family of three young children just moved into a house with a yard that is not fenced in. The previous owner gave them 300 feet of fencing to use to enclose part of their backyard. Use the quadratic function A(x) = x(300 − 2x) to determine the maximum area of the fenced in yard.

Writing Exercise

How do the graphs of the functions $f\left(x\right)={x}^{2}$ and $f\left(x\right)={x}^{2}-1$ differ? We graphed them at the start of this section. What is the difference between their graphs? How are their graphs the same?

Answers will vary.

Explain the process of finding the vertex of a parabola.

Explain how to find the intercepts of a parabola.

Answers will vary.

How can you use the discriminant when you are graphing a quadratic function?

Self Check

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ 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 a, b, and c are real numbers and $a\ne 0,$ is a function of the form $f\left(x\right)=a{x}^{2}+bx+c.$

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Learning Objectives

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

Key Concepts

Practice Makes Perfect

Writing Exercise

Self Check

Glossary

License

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