Graph Quadratic Functions Using Properties

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

Recognize the Graph of a Quadratic Function

Previously we very briefly looked at the function $f(x)=x^2$ , 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 $f(x)=ax^2+bx+c$ if $a\ne0$ . We call this kind of function a quadratic function.

Quadratic Function. A quadratic function, where $a$ , $b$ , and $c$ are real numbers and $a\ne0$ , is a function of the form

f(x)=ax^2+bx+c.

We graphed the quadratic function $f(x)=x^2$ by plotting points.

$x$	$f(x)=x^2$	$(x,f(x))$
$-3$	$9$	$(-3,9)$
$-2$	$4$	$(-2,4)$
$-1$	$1$	$(-1,1)$
$0$	$0$	$(0,0)$
$1$	$1$	$(1,1)$
$2$	$4$	$(2,4)$
$3$	$9$	$(3,9)$

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. Graph $f(x)=x^2-1$ .

Solution.

We will graph the function by plotting points. Choose integer values for $x$ , substitute them into the equation and simplify to find $f(x)$ . Record the values of the ordered pairs in the chart.

$x$	$f(x)=x^2-1$
$0$	$-1$
$1$	$0$
$-1$	$0$
$2$	$3$
$-2$	$3$

Plot the points, and then connect them with a smooth curve. The result will be the graph of the function $f(x)=x^2-1$ .

Graph $f(x)=-x^2$ .

Graph $f(x)=x^2+1$ .

All graphs of quadratic functions of the form $f(x)=ax^2+bx+c$ are parabolas that open upward or downward.

Notice that the only difference in the two functions is the negative sign before the quadratic term ( $x^2$ in the equation of the graph). 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 $f(x)=ax^2+bx+c$ , if

$a>0$ , the parabola opens upward.
$a<0$ , the parabola opens downward.

Example. Determine whether each parabola opens upward or downward:

(a) $f(x)=-3x^2+2x-4$
(b) $f(x)=6x^2+7x-9$ .

Solution.

(a) In $f(x)=-3x^2+2x-4$ , $a=-3$ . Since $a$ is negative, the parabola will open downward.

(b) In $f(x)=6x^2+7x-9$ , $a=6$ . Since $a$ is positive, the parabola will open upward.

Determine whether the graph of $f(x)=2x^2+5x-2$ is a parabola that opens upward or downward.

Determine whether the graph of $f(x)=-3x^2-4x+7$ is a parabola that opens upward or downward.

Find the Axis of Symmetry and Vertex of a Parabola

Look again at the preceding two parabolas. 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.

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 $f(x)=ax^2+bx+c$ is

x=-\tfrac{b}{2a}.

For $f(x)=x^2+4x+3$ ,

x=-\tfrac{4}{2(1)}=-2.

For $f(x)=-x^2+4x+3$ ,

x=-\tfrac{4}{2(-1)}=2.

Notice that these are the equations of the dashed 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 $-\tfrac{b}{2a}$ . To find the $y$ -coordinate of the vertex we substitute the value of the $x$ -coordinate into the quadratic function.

For $f(x)=x^2+4x+3$ , the axis of symmetry is $x=-2$ , and

f(-2)=(-2)^2+4(-2)+3=-1.

The vertex is $(-2,-1)$ .

For $f(x)=-x^2+4x+3$ , the axis of symmetry is $x=2$ , and

f(2)=-(2)^2+4(2)+3=7.

The vertex is $(2,7)$ .

Axis of Symmetry and Vertex of a Parabola. The graph of the function $f(x)=ax^2+bx+c$ is a parabola where:

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

Example. For the graph of $f(x)=3x^2-6x+2$ find:

(a) the axis of symmetry
(b) the vertex.

Solution.

(a) The axis of symmetry is the vertical line $x=-\tfrac{b}{2a}$ .

x=-\tfrac{-6}{2\cdot3}=1.

The axis of symmetry is the line $x=1$ .

(b) The vertex is a point on the line of symmetry, so its $x$ -coordinate will be $x=1$ . Find $f(1)$ .

f(1)=3(1)^2-6(1)+2=-1.

The vertex is $(1,-1)$ .

For the graph of $f(x)=2x^2-8x+1$ , find the axis of symmetry.

For the graph of $f(x)=2x^2-8x+1$ , find the vertex as an ordered pair.

For the graph of $f(x)=2x^2-4x-3$ , find the axis of symmetry.

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.

For both $f(x)=x^2+4x+3$ and $f(x)=-x^2+4x+3$ ,

f(0)=3,

so the $y$ -intercept is $(0,3)$ .

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 $0=ax^2+bx+c$ for $x$ . Solving quadratic equations like this is exactly what we have done earlier in this chapter!

For $f(x)=x^2+4x+3$ ,

\begin{array}{rcl} 0&=&x^2+4x+3 \\[4pt] 0&=&(x+1)(x+3) \\[4pt] x+1=0 && x+3=0 \\[4pt] x=-1 && x=-3 \end{array}

The $x$ -intercepts are $(-1,0)$ and $(-3,0)$ .

For $f(x)=-x^2+4x+3$ , the quadratic does not factor, so we use the Quadratic Formula.

\begin{array}{rcl} x&=&\tfrac{-4\pm\sqrt{4^2-4(-1)(3)}}{2(-1)} \\[10pt] &=&\tfrac{-4\pm\sqrt{28}}{-2} \\[10pt] &=&\tfrac{-4\pm2\sqrt7}{-2} \\[10pt] &=&2\pm\sqrt7 \end{array}

The $x$ -intercepts are $(2+\sqrt7,0)$ and $(2-\sqrt7,0)$ . We will use the decimal approximations of the $x$ -intercepts so that we can locate these points on the graph:

(2+\sqrt7,0)\approx(4.6,0),\qquad (2-\sqrt7,0)\approx(-0.6,0).

Do these results agree with our graphs?

Find the Intercepts of a Parabola. To find the intercepts of a parabola whose function is $f(x)=ax^2+bx+c$ :

$y$ -intercept	$x$ -intercepts
Let $x=0$ and solve for $f(x)$ .	Let $f(x)=0$ and solve for $x$ .

Example. Find the intercepts of the parabola whose function is $f(x)=x^2-2x-8$ .

Solution.

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

f(0)=0^2-2(0)-8=-8.

When $x=0$ , then $f(0)=-8$ . The $y$ -intercept is the point $(0,-8)$ .

To find the $x$ -intercepts, let $f(x)=0$ and solve for $x$ .

\begin{array}{rcl} 0&=&x^2-2x-8 \\[4pt] 0&=&(x-4)(x+2) \\[4pt] 0=x-4 && 0=x+2 \\[4pt] x=4 && x=-2 \end{array}

The $x$ -intercepts are the points $(4,0)$ and $(-2,0)$ .

Find the intercepts of the parabola $f(x)=x^2+2x-8$ . Enter the y-intercept first, then the two x-intercepts with the smaller x-value first, separated by commas.

Find the intercepts of the parabola $f(x)=x^2-4x-12$ . Enter the y-intercept first, then the two x-intercepts with the smaller x-value first, separated by commas.

In this chapter, we have been solving quadratic equations of the form $ax^2+bx+c=0$ . We solved for $x$ and the results were the solutions to the equation.

We are now looking at quadratic functions of the form $f(x)=ax^2+bx+c$ . The graphs of these functions are parabolas. The $x$ -intercepts of the parabolas occur where $f(x)=0$ .

For example, the quadratic equation

x^2-2x-15=0

has solutions $x=5$ and $x=-3$ . The quadratic function $f(x)=x^2-2x-15$ has $x$ -intercepts $(5,0)$ and $(-3,0)$ . 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 $ax^2+bx+c=0$ . Now we can use the discriminant to tell us how many $x$ -intercepts there are on the graph.

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

Example. Find the intercepts of the parabola for the function $f(x)=5x^2+x+4$ .

Solution.

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

f(0)=5(0)^2+0+4=4.

The $y$ -intercept is the point $(0,4)$ .

To find the $x$ -intercepts, let $f(x)=0$ and solve for $x$ . Find the value of the discriminant to predict the number of solutions, which is also the number of $x$ -intercepts.

b^2-4ac=1^2-4(5)(4)=-79.

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 $f(x)=3x^2+4x+4$ . Enter the y-intercept.

How many x-intercepts does $f(x)=3x^2+4x+4$ have?

Find the intercepts of $f(x)=x^2-4x-5$ . Enter the y-intercept first, then the two x-intercepts with the smaller x-value first, separated by commas.

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. How to Graph a Quadratic Function Using Properties. Graph $f(x)=x^2-6x+8$ by using its properties.

Solution.

Step 1. Determine whether the parabola opens upward or downward. Since $a=1$ is positive, the parabola opens upward.

Step 2. Find the axis of symmetry.

x=-\tfrac{-6}{2(1)}=3.

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

Step 3. Find the vertex. The vertex is on the axis of symmetry. Substitute $x=3$ into the function.

f(3)=3^2-6(3)+8=-1.

The vertex is $(3,-1)$ .

Step 4. Find the $y$ -intercept. Find the point symmetric to the $y$ -intercept across the axis of symmetry.

f(0)=0^2-6(0)+8=8.

The $y$ -intercept is $(0,8)$ . The $y$ -intercept is 3 units left of the axis of symmetry, $x=3$ . A point 3 units to the right of the axis of symmetry has $x=6$ . The point symmetric to the $y$ -intercept is $(6,8)$ .

Step 5. Find the $x$ -intercepts. Find additional points if needed.

\begin{array}{rcl} 0&=&x^2-6x+8 \\[4pt] 0&=&(x-2)(x-4) \\[4pt] x&=&2\quad\text{or}\quad x=4 \end{array}

The $x$ -intercepts are $(2,0)$ and $(4,0)$ .

Step 6. Graph the parabola. We graph the vertex, intercepts, and the point symmetric to the $y$ -intercept. We connect these five points to sketch the parabola.

Graph $f(x)=x^2+2x-8$ by using its properties.

Which graph shows $f(x)=x^2-8x+12$ graphed 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.

Example. Graph $f(x)=-x^2+6x-9$ by using its properties.

Solution.

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

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

x=-\tfrac{6}{2(-1)}=3.

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

f(3)=-(3)^2+6(3)-9=0.

The vertex is $(3,0)$ .

The $y$ -intercept occurs when $x=0$ .

f(0)=-(0)^2+6(0)-9=-9.

The $y$ -intercept is $(0,-9)$ . The point $(0,-9)$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $(6,-9)$ .

The $x$ -intercept occurs when $f(x)=0$ .

\begin{array}{rcl} 0&=&-x^2+6x-9 \\[4pt] 0&=&-(x^2-6x+9) \\[4pt] 0&=&-(x-3)^2 \\[4pt] x&=&3 \end{array}

Connect the points to graph the parabola.

Graph $f(x)=-3x^2+12x-12$ by using its properties.

Graph $f(x)=4x^2+24x+36$ by using its properties.

For the graph of $f(x)=-x^2+6x-9$ , 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 $0=-x^2+6x-9$ 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 $f(x)=x^2+4x+5$ ?

Example. Graph $f(x)=x^2+4x+5$ by using its properties.

Solution.

Since $a=1$ , the parabola opens upward.

x=-\tfrac{4}{2(1)}=-2.

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

f(-2)=(-2)^2+4(-2)+5=1.

The vertex is $(-2,1)$ .

The $y$ -intercept occurs when $x=0$ .

f(0)=5.

The $y$ -intercept is $(0,5)$ . The point $(-4,5)$ is two units to the left of the line of symmetry. The point two units to the right of the line of symmetry is $(0,5)$ . The point symmetric to the $y$ -intercept is $(-4,5)$ .

The $x$ -intercept occurs when $f(x)=0$ . Test the discriminant:

b^2-4ac=4^2-4(1)(5)=-4.

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 $f(x)=x^2-2x+3$ by using its properties.

Graph $f(x)=-3x^2-6x-4$ by using its properties.

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.

Example. Graph $f(x)=2x^2-4x-3$ by using its properties.

Solution.

Since $a=2$ , the parabola opens upward.

x=-\tfrac{-4}{2(2)}=1.

The equation of the axis of symmetry is $x=1$ .

f(1)=2(1)^2-4(1)-3=-5.

The vertex is $(1,-5)$ .

The $y$ -intercept occurs when $x=0$ .

f(0)=-3.

The $y$ -intercept is $(0,-3)$ . The point $(0,-3)$ is one unit to the left of the line of symmetry. The point one unit to the right of the line of symmetry is $(2,-3)$ .

The $x$ -intercept occurs when $y=0$ . Find $f(x)=0$ and use the Quadratic Formula.

\begin{array}{rcl} x&=&\tfrac{-(-4)\pm\sqrt{(-4)^2-4(2)(-3)}}{2(2)} \\[10pt] &=&\tfrac{4\pm\sqrt{40}}{4} \\[10pt] &=&\tfrac{2\pm\sqrt{10}}{2} \end{array}

The approximate values are $x\approx2.6$ and $x\approx-0.6$ . The approximate values of the $x$ -intercepts are $(2.6,0)$ and $(-0.6,0)$ . Graph the parabola using the points found.

Graph $f(x)=5x^2+10x+3$ by using its properties.

Graph $f(x)=-3x^2-6x+5$ 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.

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.

Example. Find the minimum or maximum value of the quadratic function $f(x)=x^2+2x-8$ .

Solution.

Since $a$ is positive, the parabola opens upward. The quadratic equation has a minimum.

x=-\tfrac{2}{2(1)}=-1.

The equation of the axis of symmetry is $x=-1$ . The vertex is on the line $x=-1$ .

f(-1)=(-1)^2+2(-1)-8=-9.

The vertex is $(-1,-9)$ . 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 $x=-1$ .

Find the maximum or minimum value of $f(x)=x^2-8x+12$ .

Find the maximum or minimum value of $f(x)=-4x^2+16x-11$ .

We have used the formula

h(t)=-16t^2+v_0t+h_0

to calculate the height in feet, $h$ , of an object shot upwards into the air with initial velocity, $v_0$ , 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.

Example. The quadratic function $h(t)=-16t^2+176t+4$ models the height of a volleyball hit straight upwards with velocity 176 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 $a$ is negative, the parabola opens downward. The quadratic function has a maximum.

(a) Find the equation of the axis of symmetry.

t=-\tfrac{176}{2(-16)}=5.5.

The equation of the axis of symmetry is $t=5.5$ . The maximum occurs when $t=5.5$ seconds.

(b) Find $h(5.5)$ .

h(5.5)=-16(5.5)^2+176(5.5)+4=488.

The vertex is $(5.5,488)$ . 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.

The quadratic function $h(t)=-16t^2+128t+32$ gives the height of a stone thrown upward from a height of 32 feet at 128 ft/sec. How long will it take the stone to reach its maximum height? Round to the nearest tenth.

For $h(t)=-16t^2+128t+32$ , what is the maximum height? Round to the nearest tenth.

A toy rocket’s path is $h(t)=-16t^2+208t$ . When will the rocket reach its maximum height? Round to the nearest tenth.

Key terms

A quadratic function is a function of the form $f(x)=ax^2+bx+c$ , where $a\ne0$ . Its graph is a parabola. The vertical line through the vertex is the axis of symmetry. The vertex is the lowest point of a parabola that opens upward or the highest point of a parabola that opens downward.

Practice

Recognize the Graph of a Quadratic Function

Determine whether the graph of $f(x)=-2x^2-6x-7$ is a parabola that opens upward or downward.

Determine whether the graph of $f(x)=6x^2+2x+3$ is a parabola that opens upward or downward.

Determine whether the graph of $f(x)=-3x^2+5x-1$ is a parabola that opens upward or downward.

Determine whether the graph of $f(x)=2x^2-4x+5$ is a parabola that opens upward or downward.

Find the Axis of Symmetry and Vertex of a Parabola

For the graph of $f(x)=x^2+8x-1$ , find the axis of symmetry.

For the graph of $f(x)=x^2+8x-1$ , find the vertex as an ordered pair.

For the graph of $f(x)=-x^2+2x+5$ , find the axis of symmetry.

For the graph of $f(x)=-x^2+2x+5$ , find the vertex as an ordered pair.

Find the Intercepts of a Parabola

Find the y-intercept of the parabola $f(x)=x^2+7x+6$ .

Find the x-intercepts of the parabola $f(x)=x^2+7x+6$ . Enter the intercept with the smaller x-value first.

Find the y-intercept of the parabola $f(x)=-x^2+8x-19$ .

How many x-intercepts does the parabola $f(x)=-x^2+8x-19$ have?

Graph Quadratic Functions Using Properties

Graph $f(x)=x^2+6x+5$ by using its properties.

Graph $f(x)=-x^2+2x-7$ by using its properties.

Solve Maximum and Minimum Applications

The quadratic function $h(t)=-16t^2+168t+45$ gives the height of an arrow shot upward from a platform 45 feet high at 168 ft/sec. How long will it take the arrow to reach its maximum height? Round to the nearest tenth.

For $h(t)=-16t^2+168t+45$ , what is the maximum height? Round to the nearest tenth.

A retailer selling pairs of boots for $x$ dollars each can sell $70-x$ pairs a week. The revenue function is $R(x)=-x^2+70x$ . Find the selling price that gives the maximum weekly revenue.

For $R(x)=-x^2+70x$ , what is the maximum weekly revenue?

Adapted from [Intermediate Algebra 2e, Section 9.6](https://openstax.org/books/intermediate-algebra-2e/pages/9-6-graph-quadratic-functions-using-properties) by Lynn Marecek and Andrea Honeycutt Mathis, © OpenStax, licensed under [CC BY-NC-SA 4.0](https://creativecommons.org/licenses/by-nc-sa/4.0/). Access the original for free at OpenStax. Changes: omitted the readiness quiz, media links, self-check, and review apparatus; converted Try It exercises to interactive checks, recreated instructional graphs for the web, and adapted selected end-of-section exercises into an interactive Practice block.

Solve Applications of Quadratic Equations Graph Quadratic Functions Using Transformations