Graphing Quadratic Equations in Two Variables

By the end of this section, you will be able to: recognize the graph of a quadratic equation in two variables, find the axis of symmetry and vertex of a parabola, find the intercepts of a parabola, graph quadratic equations in two variables, and solve maximum and minimum applications.

Recognize the graph of a quadratic equation in two variables

We have graphed equations of the form $Ax+By=C$ . We called equations like this linear equations because their graphs are straight lines.

Now, we will graph equations of the form $y=ax^2+bx+c$ . We call this kind of equation a quadratic equation in two variables.

Quadratic equation in two variables. A quadratic equation in two variables, where $a$ , $b$ , and $c$ are real numbers and $a\ne0$ , is an equation of the form

y=ax^2+bx+c

Just like we started graphing linear equations by plotting points, we will do the same for quadratic equations.

Let’s look first at graphing the quadratic equation $y=x^2$ . We will choose integer values of $x$ between $-2$ and $2$ and find their $y$ values.

$x$	$y=x^2$
$0$	$0$
$1$	$1$
$-1$	$1$
$2$	$4$
$-2$	$4$

Notice when we let $x=1$ and $x=-1$ , we got the same value for $y$ .

\begin{array}{rcl} y&=&x^2 \\[4pt] y&=&1^2 \\[4pt] y&=&1 \end{array} \qquad\text{and}\qquad \begin{array}{rcl} y&=&x^2 \\[4pt] y&=&(-1)^2 \\[4pt] y&=&1 \end{array}

The same thing happened when we let $x=2$ and $x=-2$ .

Now, we will plot the points to show the graph of $y=x^2$ .

The graph is not a line. This figure is called a parabola. Every quadratic equation has a graph that looks like this.

Example 10.43. Graph $y=x^2-1$ .

We will graph the equation by plotting points. Choose integer values for $x$ , substitute them into the equation and solve for $y$ .

$x$	$y=x^2-1$
$0$	$-1$
$1$	$0$
$-1$	$0$
$2$	$3$
$-2$	$3$

Record the values of the ordered pairs in the chart. Plot the points, and then connect them with a smooth curve.

The result is the graph of $y=x^2-1$ .

Graph $y=-x^2$ . Which description identifies its graph?

How do the equations $y=x^2$ and $y=x^2-1$ differ? What is the difference between their graphs? How are their graphs the same?

All parabolas of the form $y=ax^2+bx+c$ open upwards or downwards. Compare the two graphs below.

Notice that the only difference in the two equations is the negative sign before the $x^2$ in the equation of the second graph. When the $x^2$ term is positive, the parabola opens upward, and when the $x^2$ term is negative, the parabola opens downward.

Parabola orientation. For the quadratic equation $y=ax^2+bx+c$ , if:

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

Example 10.44. Determine whether each parabola opens upward or downward: (a) $y=-3x^2+2x-4$ ; (b) $y=6x^2+7x-9$ .

(a) Find the value of $a$ . Since $a=-3$ is negative, the parabola will open downward.

(b) Find the value of $a$ . Since $a=6$ is positive, the parabola will open upward.

Determine whether the parabola $y=-2x^2-2x-3$ opens upward or downward.

Find the axis of symmetry and vertex of a parabola

Look again at the two parabolas $y=x^2+4x+3$ and $y=-x^2+4x+3$ . Do you see that we could fold each parabola in half and that 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.

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 $y=ax^2+bx+c$ 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 $x=-\tfrac{b}{2a}$ .

\begin{array}{rcl}x&=&-\tfrac{4}{2\cdot1}\\[4pt]x&=&-2\end{array} \qquad\text{and}\qquad \begin{array}{rcl}x&=&-\tfrac{4}{2(-1)}\\[4pt]x&=&2\end{array}

The point on the parabola that is on the axis of symmetry is the lowest or highest point on the parabola, depending on whether the parabola opens upwards or downwards. 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 equation.

For $y=x^2+4x+3$ , the axis is $x=-2$ and $y=(-2)^2+4(-2)+3=-1$ , so the vertex is $(-2,-1)$ . For $y=-x^2+4x+3$ , the axis is $x=2$ and $y=-(2)^2+4(2)+3=7$ , so the vertex is $(2,7)$ .

Axis of symmetry and vertex of a parabola. For a parabola with equation $y=ax^2+bx+c$ :

The axis of symmetry is the line $x=-\tfrac{b}{2a}$ .
The vertex is on the axis of symmetry, so its $x$ -coordinate is $-\tfrac{b}{2a}$ .

To find the $y$ -coordinate of the vertex, substitute $x=-\tfrac{b}{2a}$ into the quadratic equation.

Example 10.45. For the parabola $y=3x^2-6x+2$ find: (a) the axis of symmetry and (b) the vertex.

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

\begin{array}{lrcl} \text{Substitute the values of }a,b\text{ into the equation.}&x&=&-\tfrac{-6}{2\cdot3}\\[10pt] \text{Simplify.}&x&=&1 \end{array}

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

(b) The vertex is on the line of symmetry, so its $x$ -coordinate will be $x=1$ . Substitute $x=1$ into the equation and solve for $y$ .

\begin{array}{lrcl} &y&=&3x^2-6x+2\\[4pt] \text{Substitute.}&y&=&3(1)^2-6(1)+2\\[4pt] \text{Simplify.}&y&=&3\cdot1-6+2\\[4pt] \text{This is the }y\text{-coordinate.}&y&=&-1 \end{array}

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

For the parabola $y=2x^2-8x+1$ , find the $x$ -coordinate of 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 equation. Both $y=x^2+4x+3$ and $y=-x^2+4x+3$ have $y$ -intercept $(0,3)$ .

At an x-intercept, the value of $y$ is zero. To find an $x$ -intercept, we substitute $y=0$ into the equation. In other words, we will need to solve $0=ax^2+bx+c$ for $x$ .

But solving quadratic equations like this is exactly what we have done earlier in this chapter.

First, we will find the $x$ -intercepts of $y=x^2+4x+3$ .

\begin{array}{lrcl} \text{Let }y=0.&0&=&x^2+4x+3\\[4pt] \text{Factor.}&0&=&(x+1)(x+3)\\[4pt] \text{Use the zero product property.}&&&x+1=0,\ x+3=0\\[4pt] \text{Solve.}&&&x=-1,\ x=-3 \end{array}

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

Now, we will find the $x$ -intercepts of $y=-x^2+4x+3$ .

\begin{array}{lrcl} \text{Let }y=0.&0&=&-x^2+4x+3\\[4pt] \text{Use the Quadratic Formula.}&x&=&\tfrac{-4\pm\sqrt{4^2-4(-1)(3)}}{2(-1)}\\[10pt] \text{Simplify.}&x&=&2\pm\sqrt7 \end{array}

The $x$ -intercepts are $(2+\sqrt7,0)$ and $(2-\sqrt7,0)$ , approximately $(4.6,0)$ and $(-0.6,0)$ .

Find the intercepts of a parabola. For $y=ax^2+bx+c$ :

y-intercept: Let $x=0$ and solve for $y$ .
x-intercepts: Let $y=0$ and solve for $x$ .

Example 10.46. Find the intercepts of $y=x^2-2x-8$ .

To find the $y$ -intercept, let $x=0$ and solve for $y$ : $y=0^2-2\cdot0-8=-8$ . The $y$ -intercept is $(0,-8)$ .

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

\begin{array}{lrcl} &0&=&x^2-2x-8\\[4pt] \text{Solve by factoring.}&0&=&(x-4)(x+2)\\[4pt] &&&0=x-4\quad 0=x+2\\[4pt] &&&4=x\quad -2=x \end{array}

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

Find the $x$ -intercepts of $y=x^2+2x-8$ . Enter the $x$ -values separated by commas from least to greatest.

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 equations in two variables of the form $y=ax^2+bx+c$ . The graphs of these equations are parabolas. The $x$ -intercepts of the parabolas occur where $y=0$ .

The solutions of the quadratic equation are the $x$ values of the $x$ -intercepts.

Earlier, we saw that quadratic equations have 2, 1, or 0 solutions. The graphs below show parabolas for these three cases. Since the solutions of the equations 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 equation 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.

Discriminant	Solutions	Graph
$b^2-4ac>0$	Two solutions	Two $x$ -intercepts
$b^2-4ac=0$	One solution	One $x$ -intercept
$b^2-4ac<0$	No real solution	No $x$ -intercept

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

Example 10.47. Find the intercepts of $y=5x^2+x+4$ .

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

y=5\cdot0^2+0+4=4.

When $x=0$ , then $y=4$ . The $y$ -intercept is the point $(0,4)$ .

To find the $x$ -intercept, let $y=0$ and solve for $x$ . Find the value of the discriminant to predict the number of solutions and so $x$ -intercepts.

b^2-4ac=1^2-4\cdot5\cdot4=1-80=-79.

Since the value of the discriminant is negative, there is no real solution to the equation. There are no $x$ -intercepts.

Example 10.48. Find the intercepts of $y=4x^2-12x+9$ .

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

y=4\cdot0^2-12\cdot0+9=9.

When $x=0$ , then $y=9$ . The $y$ -intercept is the point $(0,9)$ .

To find the $x$ -intercept, let $y=0$ and solve for $x$ . Find the value of the discriminant to predict the number of solutions and so $x$ -intercepts.

b^2-4ac=(-12)^2-4\cdot4\cdot9=144-144=0.

Since the value of the discriminant is 0, there is only one real solution to the equation. Therefore, there is only one $x$ -intercept.

\begin{array}{lrcl} \text{Factor the perfect square trinomial.}&0&=&(2x-3)^2\\[4pt] \text{Use the Zero Product Property.}&0&=&2x-3\\[4pt] \text{Solve for }x.&3&=&2x\\[4pt] &x&=&\tfrac32 \end{array}

When $y=0$ , then $\tfrac32=x$ . The $x$ -intercept is the point $(\tfrac32,0)$ .

Graph quadratic equations in two variables

Now, we have all the pieces we need in order to graph a quadratic equation in two variables. We just need to put them together.

Example 10.49. How To Graph a Quadratic Equation in Two Variables. Graph $y=x^2-6x+8$ .

Step 1. Write the quadratic equation with $y$ on one side. This equation has $y$ on one side.

y=x^2-6x+8\qquad a=1,\ b=-6,\ c=8

Step 2. Determine whether the parabola opens upward or downward. Look at $a$ in the equation. Since $a$ is positive, the parabola opens upward.

Step 3. Find the axis of symmetry.

\begin{array}{rcl} x&=&-\tfrac{b}{2a}\\[10pt] x&=&-\tfrac{-6}{2\cdot1}\\[10pt] x&=&3 \end{array}

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

Step 4. Find the vertex. The vertex is on the axis of symmetry. Substitute $x=3$ into the equation and solve for $y$ .

\begin{array}{rcl} y&=&x^2-6x+8\\[4pt] y&=&(3)^2-6(3)+8\\[4pt] y&=&-1 \end{array}

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

Step 5. Find the $y$ -intercept. Find the point symmetric to the $y$ -intercept across the axis of symmetry. We substitute $x=0$ into the equation.

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

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

Step 6. Find the $x$ -intercepts. We substitute $y=0$ into the equation. We can solve this quadratic equation by factoring.

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

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

Step 7. 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.

Graph a quadratic equation in two variables.

Write the quadratic equation with $y$ on one side.
Determine whether the parabola opens upward or downward.
Find 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.
Graph the parabola.

Example 10.50. Graph $y=-x^2+6x-9$ .

The equation has $y$ on one side. Since $a$ is $-1$ , the parabola opens downward. To find the axis of symmetry, find $x=-\tfrac{b}{2a}$ .

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

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

y=-3^2+6\cdot3-9=-9+18-9=0.

The vertex is $(3,0)$ . The $y$ -intercept occurs when $x=0$ . Substitute $x=0$ . Simplify.

y=-0^2+6\cdot0-9=-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 $(6,-9)$ . Point symmetric to the $y$ -intercept is $(6,-9)$ .

\begin{array}{lrcl} \text{Substitute }y=0.&0&=&-x^2+6x-9\\[4pt] \text{Factor the GCF.}&0&=&-(x^2-6x+9)\\[4pt] \text{Factor the trinomial.}&0&=&-(x-3)^2\\[4pt] \text{Solve for }x.&x&=&3 \end{array}

Connect the points to graph the parabola.

For the graph of $y=-x^2+6x-9$ , 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 $y=x^2+4x+5$ ?

Example 10.51. Graph $y=x^2+4x+5$ .

The equation has $y$ on one side. Since $a$ is 1, the parabola opens upward. To find the axis of symmetry, find $x=-\tfrac{b}{2a}$ .

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

The axis of symmetry is $x=-2$ . The vertex is on the line $x=-2$ . Find $y$ when $x=-2$ .

y=(-2)^2+4(-2)+5=4-8+5=1.

The vertex is $(-2,1)$ . The $y$ -intercept occurs when $x=0$ . Substitute $x=0$ . Simplify.

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

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

The $x$ -intercept occurs when $y=0$ . Substitute $y=0$ . Test the discriminant.

b^2-4ac=4^2-4\cdot1\cdot5=16-20=-4.

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

Example 10.52. Graph $y=2x^2-4x-3$ .

The equation has $y$ on one side. Since $a$ is 2, the parabola opens upward. To find the axis of symmetry, find $x=-\tfrac{b}{2a}$ .

x=-\frac{-4}{2\cdot2}=1.

The axis of symmetry is $x=1$ . The vertex is on the line $x=1$ . Find $y$ when $x=1$ .

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

The vertex is $(1,-5)$ . The $y$ -intercept occurs when $x=0$ . Substitute $x=0$ . Simplify.

y=2\cdot0^2-4\cdot0-3=-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 $(2,-3)$ . Point symmetric to the $y$ -intercept is $(2,-3)$ .

The $x$ -intercept occurs when $y=0$ . Substitute $y=0$ . Use the Quadratic Formula.

\begin{array}{lrcl} \text{Use the Quadratic Formula.}&x&=&\tfrac{-(-4)\pm\sqrt{(-4)^2-4\cdot2\cdot(-3)}}{2\cdot2}\\[10pt] \text{Simplify.}&x&=&\tfrac{4\pm\sqrt{16+24}}4\\[10pt] \text{Simplify inside the radical.}&x&=&\tfrac{4\pm\sqrt{40}}4\\[10pt] \text{Simplify the radical.}&x&=&\tfrac{4\pm2\sqrt{10}}4\\[10pt] \text{Factor the GCF.}&x&=&\tfrac{2(2\pm\sqrt{10})}4\\[10pt] \text{Remove common factors.}&x&=&\tfrac{2\pm\sqrt{10}}2 \end{array}

Write as two equations: $x=\tfrac{2+\sqrt{10}}2$ , $x=\tfrac{2-\sqrt{10}}2$ . Approximate the values: $x\approx2.5$ , $x\approx-0.6$ . The approximate values of the $x$ -intercepts are $(2.5,0)$ and $(-0.6,0)$ . Graph the parabola using the points found.

Graph $y=5x^2+10x+3$ . Which description identifies its vertex and orientation?

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 equation. The $y$ -coordinate of the vertex is the minimum $y$ -value of a parabola that opens upward. It is the maximum $y$ -value of a parabola that opens downward.

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.

Example 10.53. Find the minimum value of $y=x^2+2x-8$ .

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

x=-\frac{2}{2\cdot1}=-1.

The axis of symmetry is $x=-1$ . The vertex is on the line $x=-1$ . Find $y$ when $x=-1$ .

y=(-1)^2+2(-1)-8=1-2-8=-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 $x=-1$ . Show the graph to verify the result.

Find the minimum value of $y=x^2-8x+12$ .

We have used the formula

h=-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 equation in the variable $t$ , so its graph is a parabola. By solving for the coordinates of the vertex, we can find how long it will take the object to reach its maximum height. Then, we can calculate the maximum height.

Example 10.54. The quadratic equation $h=-16t^2+v_0t+h_0$ 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.

Here $h=-16t^2+176t+4$ . Since $a$ is negative, the parabola opens downward. The quadratic equation has a maximum.

(a) Find the axis of symmetry.

t=-\frac{b}{2a}=-\frac{176}{2(-16)}=5.5.

The axis of symmetry is $t=5.5$ . The vertex is on the line $t=5.5$ . The maximum occurs when $t=5.5$ seconds.

(b) Find $h$ when $t=5.5$ .

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

Use a calculator to simplify: $h=488$ . The vertex is $(5.5,488)$ . Since the parabola has a maximum, the $h$ -coordinate of the vertex is the maximum $y$ -value of the quadratic equation. The maximum value of the quadratic is 488 feet and it occurs when $t=5.5$ seconds.

For $h=-16t^2+128t+32$ , how many seconds will it take the stone to reach its maximum height?

Key terms

quadratic equation in two variables — an equation of the form $y=ax^2+bx+c$ , where $a$ , $b$ , and $c$ are real numbers and $a\ne0$ . parabola — the graph of a quadratic equation in two variables. axis of symmetry — the vertical line passing through the middle of the parabola. vertex — the point on the parabola that is on the axis of symmetry; it is the lowest or highest point on the parabola. x-intercepts of a parabola — the points where $y=0$ . y-intercept of a parabola — the point where $x=0$ .

This section is adapted from Elementary Algebra 2e, Section 10.5: Graphing Quadratic Equations in Two Variables by Lynn Marecek and MaryAnne Anthony-Smith, © OpenStax, licensed under CC BY-NC-SA 4.0. Access the original for free at openstax.org. Changes: recreated the source parabola graphs as accessible inline graphics, recast source data tables as markdown tables, omitted the Be Prepared quiz, media links, and end-of-section exercises, and converted selected practice problems (“Try Its”) into interactive exercises with instant feedback.

Solve Applications Modeled by Quadratic Equations Knowledge Check: Chapters 6–10