Learning Objectives
 Graph vertical parabolas
 Graph horizontal parabolas
 Solve applications with parabolas
Before you get started, take this readiness quiz.
Graph: $y=\mathrm{3}{x}^{2}+12x12.$
If you missed this problem, review Example 9.47.
Solve by completing the square: ${x}^{2}6x+6=0.$
If you missed this problem, review Example 9.12.
Write in standard form: $y=3{x}^{2}6x+5.$
If you missed this problem, review Example 9.59.
Graph Vertical Parabolas
The next conic section we will look at is a parabola. We define a parabola as all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.
Parabola
A parabola is all points in a plane that are the same distance from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix of the parabola.
Previously, we learned to graph vertical parabolas from the general form or the standard form using properties. Those methods will also work here. We will summarize the properties here.
Vertical Parabolas  

General form $y=a{x}^{2}+bx+c$ 
Standard form $y=a{\left(xh\right)}^{2}+k$ 

Orientation  $a>0$ up; $a<0$ down  $a>0$ up; $a<0$ down 
Axis of symmetry  $x=\frac{b}{2a}$  $x=h$ 
Vertex  Substitute $x=\frac{b}{2a}$ and solve for y. 
$\left(h,k\right)$ 
yintercept  Let $x=0$  Let $x=0$ 
xintercepts  Let $y=0$  Let $y=0$ 
The graphs show what the parabolas look like when they open up or down. Their position in relation to the x or yaxis is merely an example.
To graph a parabola from these forms, we used the following steps.
How To
Graph vertical parabolas $(y=a{x}^{2}+bx+c\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}f\left(x\right)=a{\left(xh\right)}^{2}+k)$ using properties.
 Step 1. Determine whether the parabola opens upward or downward.
 Step 2. Find the axis of symmetry.
 Step 3. Find the vertex.
 Step 4. Find the yintercept. Find the point symmetric to the yintercept across the axis of symmetry.
 Step 5. Find the xintercepts.
 Step 6. Graph the parabola.
The next example reviews the method of graphing a parabola from the general form of its equation.
Example 11.12
Graph $y=\text{\u2212}{x}^{2}+6x8$ by using properties.
Since a is $\mathrm{1},$ the parabola opens downward.  
To find the axis of symmetry, find $x=\frac{b}{2a}.$  
The axis of symmetry is $x=3.$  
The vertex is on the line $x=3.$  
Let $x=3.$  
The vertex is $\left(3,1\right).$  
The yintercept occurs when $x=0.$  
Substitute $x=0.$  
Simplify.  
The yintercept is $\left(0,\mathrm{8}\right).$  
The point $\left(0,\mathrm{8}\right)$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $\left(6,\mathrm{8}\right).$ 
Point symmetric to the yintercept is $\left(6,\mathrm{8}\right).$ 
The xintercept occurs when $y=0.$  
Let $y=0.$  
Factor the GCF.  
Factor the trinomial.  
Solve for x.  
The xintercepts are $\left(4,0\right),\left(2,0\right).$  
Graph the parabola. 
Graph $y=\text{\u2212}{x}^{2}+5x6$ by using properties.
Graph $y=\text{\u2212}{x}^{2}+8x12$ by using properties.
The next example reviews the method of graphing a parabola from the standard form of its equation, $y=a{\left(xh\right)}^{2}+k.$
Example 11.13
Write$y=3{x}^{2}6x+5$ in standard form and then use properties of standard form to graph the equation.
Rewrite the function in $y=a{\left(xh\right)}^{2}+k$ form by completing the square. 
$y=3{x}^{2}6x+5$ 
$y=3\left({x}^{2}2x\phantom{\rule{1.6em}{0ex}}\right)+5$  
$y=3\left({x}^{2}2x+1\right)+53$  
$y=3{\left(x1\right)}^{2}+2$  
Identify the constants a, h, k.  $a=3$, $h=1$, $k=2$ 
Since $a=2,$ the parabola opens upward.  
The axis of symmetry is $x=h.$  The axis of symmetry is $x=1.$ 
The vertex is $\left(h,k\right).$  The vertex is $\left(1,2\right).$ 
Find the yintercept by substituting $x=0.$  $y=3{\left(x1\right)}^{2}+2$ $y=3\xb7{0}^{2}6\xb70+5$ 
$y=5$  
yintercept $\left(0,5\right)$  
Find the point symmetric to $\left(0,5\right)$ across the axis of symmetry.  $\left(2,5\right)$ 
Find the xintercepts.  $\begin{array}{ccc}\hfill y& =\hfill & 3{\left(x1\right)}^{2}+2\hfill \\ \hfill 0& =\hfill & 3{\left(x1\right)}^{2}+2\hfill \\ \hfill 2& =\hfill & 3{\left(x1\right)}^{2}\hfill \\ \hfill \frac{2}{3}& =\hfill & {\left(x1\right)}^{2}\hfill \\ \hfill \pm \sqrt{\frac{2}{3}}& =\hfill & x1\hfill \end{array}$ 
The square root of a negative number tells us the solutions are complex numbers. So there are no xintercepts. 

Graph the parabola. 
ⓐ Write $y=2{x}^{2}+4x+5$ in standard form and ⓑ use properties of standard form to graph the equation.
ⓐ Write $y=\mathrm{2}{x}^{2}+8x7$ in standard form and ⓑ use properties of standard form to graph the equation.
Graph Horizontal Parabolas
Our work so far has only dealt with parabolas that open up or down. We are now going to look at horizontal parabolas. These parabolas open either to the left or to the right. If we interchange the x and y in our previous equations for parabolas, we get the equations for the parabolas that open to the left or to the right.
Horizontal Parabolas  

General form $x=a{y}^{2}+by+c$ 
Standard form $x=a{\left(yk\right)}^{2}+h$ 

Orientation  $a>0$ right; $a<0$ left  $a>0$ right; $a<0$ left 
Axis of symmetry  $y=\frac{b}{2a}$  $y=k$ 
Vertex  Substitute $y=\frac{b}{2a}$ and solve for x. 
$\left(h,k\right)$ 
yintercepts  Let $x=0$  Let $x=0$ 
xintercept  Let $y=0$  Let $y=0$ 
The graphs show what the parabolas look like when they to the left or to the right. Their position in relation to the x or yaxis is merely an example.
Looking at these parabolas, do their graphs represent a function? Since both graphs would fail the vertical line test, they do not represent a function.
To graph a parabola that opens to the left or to the right is basically the same as what we did for parabolas that open up or down, with the reversal of the x and y variables.
How To
Graph horizontal parabolas $(x=a{y}^{2}+by+c\phantom{\rule{0.5em}{0ex}}\text{or}\phantom{\rule{0.5em}{0ex}}x=a{\left(yk\right)}^{2}+h)$ using properties.
 Step 1. Determine whether the parabola opens to the left or to the right.
 Step 2. Find the axis of symmetry.
 Step 3. Find the vertex.
 Step 4. Find the xintercept. Find the point symmetric to the xintercept across the axis of symmetry.
 Step 5. Find the yintercepts.
 Step 6. Graph the parabola.
Example 11.14
Graph $x=2{y}^{2}$ by using properties.
Since $a=2,$ the parabola opens to the right.  
To find the axis of symmetry, find $y=\frac{b}{2a}.$  
The axis of symmetry is $y=0.$  
The vertex is on the line$y=0.$  
Let $y=0.$  
The vertex is $\left(0,0\right).$ 
Since the vertex is $\left(0,0\right),$ both the x and yintercepts are the point $\left(0,0\right).$ To graph the parabola we need more points. In this case it is easiest to choose values of y.
We also plot the points symmetric to $\left(2,1\right)$ and $\left(8,2\right)$ across the yaxis, the points $\left(2,\mathrm{1}\right),$$\left(8,\mathrm{2}\right).$
Graph the parabola.
Graph $x={y}^{2}$ by using properties.
Graph $x=\text{\u2212}{y}^{2}$ by using properties.
In the next example, the vertex is not the origin.
Example 11.15
Graph $x=\text{\u2212}{y}^{2}+2y+8$ by using properties.
Since $a=\mathrm{1},$ the parabola opens to the left.  
To find the axis of symmetry, find $y=\frac{b}{2a}.$  
The axis of symmetry is $y=1.$  
The vertex is on the line$y=1.$  
Let $y=1.$  
The vertex is $\left(9,1\right).$  
The xintercept occurs when $y=0.$  
The xintercept is $\left(8,0\right).$  
The point $\left(8,0\right)$ is one unit below the line of symmetry. The symmetric point one unit above the line of symmetry is $\left(8,2\right)$ 
Symmetric point is $\left(8,2\right).$ 
The yintercept occurs when $x=0.$  
Substitute $x=0.$  
Solve.  
The yintercepts are $\left(0,4\right)$ and $\left(0,\mathrm{2}\right).$  
Connect the points to graph the parabola. 
Graph $x=\text{\u2212}{y}^{2}4y+12$ by using properties.
Graph $x=\text{\u2212}{y}^{2}+2y3$ by using properties.
In Table 11.1, we see the relationship between the equation in standard form and the properties of the parabola. The How To box lists the steps for graphing a parabola in the standard form $x=a{\left(yk\right)}^{2}+h.$ We will use this procedure in the next example.
Example 11.16
Graph $x=2{\left(y2\right)}^{2}+1$ using properties.
Identify the constants a, h, k.  $a=2,$$h=1,$$k=2$ 
Since $a=2,$ the parabola opens to the right.  
The axis of symmetry is $y=k.$  $\phantom{\rule{7.5em}{0ex}}$The axis of symmetry is $y=2.$ 
The vertex is $\left(h,k\right).$  $\phantom{\rule{7.5em}{0ex}}$The vertex is $\left(1,2\right).$ 
Find the xintercept by substituting $y=0.$  $\begin{array}{ccc}\hfill x& =\hfill & 2{\left(y2\right)}^{2}+1\hfill \\ \hfill x& =\hfill & 2{\left(02\right)}^{2}+1\hfill \\ \hfill x& =\hfill & 9\hfill \end{array}$ 
$\phantom{\rule{7.5em}{0ex}}$The xintercept is $\left(9,0\right).$  
Find the point symmetric to $\left(9,0\right)$ across the axis of symmetry. 
$\phantom{\rule{7.5em}{0ex}}\left(9,4\right)$ 
Find the yintercepts. Let $x=0.$  $\begin{array}{ccc}\hfill x& =\hfill & 2{\left(y2\right)}^{2}+1\hfill \\ \hfill 0& =\hfill & 2{\left(y2\right)}^{2}+1\hfill \\ \hfill \mathrm{1}& =\hfill & 2{\left(y2\right)}^{2}\hfill \end{array}$ 
A square cannot be negative, so there is no real solution. So there are no yintercepts. 

Graph the parabola. 
Graph $x=3{\left(y1\right)}^{2}+2$ using properties.
Graph $x=2{\left(y3\right)}^{2}+2$ using properties.
In the next example, we notice the a is negative and so the parabola opens to the left.
Example 11.17
Graph $x=\mathrm{4}{\left(y+1\right)}^{2}+4$ using properties.
Identify the constants a, h, k.  $a=\mathrm{4},$$h=4,$$k=\mathrm{1}$ 
Since $a=\mathrm{4},$ the parabola opens to the left.  
The axis of symmetry is $y=k.$  $\phantom{\rule{8em}{0ex}}$The axis of symmetry is $y=\mathrm{1}.$ 
The vertex is $\left(h,k\right).$  $\phantom{\rule{8em}{0ex}}$The vertex is $\left(4,\mathrm{1}\right).$ 
Find the xintercept by substituting $y=0.$  $\phantom{\rule{1.7em}{0ex}}\begin{array}{ccc}\hfill x& =\hfill & \mathrm{4}{\left(y+1\right)}^{2}+4\hfill \\ \hfill x& =\hfill & \mathrm{4}{\left(0+1\right)}^{2}+4\hfill \\ \hfill x& =\hfill & 0\hfill \end{array}$ 
$\phantom{\rule{8em}{0ex}}$The xintercept is $\left(0,0\right).$  
Find the point symmetric to $\left(0,0\right)$ across the axis of symmetry. 
$\phantom{\rule{8em}{0ex}}\left(0,\mathrm{2}\right)$ 
Find the yintercepts.  $\phantom{\rule{8em}{0ex}}x=\mathrm{4}{\left(y+1\right)}^{2}+4$ 
Let $x=0.$  $\begin{array}{ccc}\hfill 0& =\hfill & \mathrm{4}{\left(y+1\right)}^{2}+4\hfill \\ \hfill \mathrm{4}& =\hfill & \mathrm{4}{\left(y+1\right)}^{2}\hfill \\ \hfill 1& =\hfill & {\left(y+1\right)}^{2}\hfill \\ \hfill y+1& =\hfill & \pm 1\hfill \end{array}$ 
$y=\mathrm{1}+1\phantom{\rule{1.5em}{0ex}}y=\mathrm{1}1$  
$y=0\phantom{\rule{4em}{0ex}}y=\mathrm{2}$  
The yintercepts are $\left(0,0\right)$ and $\left(0,\mathrm{2}\right).$  
Graph the parabola. 
Graph $x=\mathrm{4}{\left(y+2\right)}^{2}+4$ using properties.
Graph $x=\mathrm{2}{\left(y+3\right)}^{2}+2$ using properties.
The next example requires that we first put the equation in standard form and then use the properties.
Example 11.18
Write $x=2{y}^{2}+12y+17$ in standard form and then use the properties of the standard form to graph the equation.
Rewrite the function in $x=a{\left(yk\right)}^{2}+h$ form by completing the square. 

Identify the constants a, h, k.  $a=2,\phantom{\rule{1em}{0ex}}h=\mathrm{1},\phantom{\rule{1em}{0ex}}k=\mathrm{3}$ 
Since $a=2,$ the parabola opens to the right. 

The axis of symmetry is $y=k.$  $\phantom{\rule{10.4em}{0ex}}$The axis of symmetry is $y=\mathrm{3}.$ 
The vertex is $\left(h,k\right).$  $\phantom{\rule{10.4em}{0ex}}$The vertex is $\left(\mathrm{1},\mathrm{3}\right).$ 
Find the xintercept by substituting $y=0.$ 
$\phantom{\rule{1.8em}{0ex}}\begin{array}{ccc}\hfill x& =\hfill & 2{\left(y+3\right)}^{2}1\hfill \\ \hfill x& =\hfill & 2{\left(0+3\right)}^{2}1\hfill \\ \hfill x& =\hfill & 17\hfill \end{array}$ 
$\phantom{\rule{10.4em}{0ex}}$The xintercept is $\left(17,0\right).$  
Find the point symmetric to $\left(17,0\right)$ across the axis of symmetry. 
$\phantom{\rule{10.4em}{0ex}}\left(17,\mathrm{6}\right)$ 
Find the yintercepts. Let $x=0.$ 
$\begin{array}{ccc}\hfill x& =\hfill & 2{\left(y+3\right)}^{2}1\hfill \\ \hfill 0& =\hfill & 2{\left(y+3\right)}^{2}1\hfill \\ \hfill 1& =\hfill & 2{\left(y+3\right)}^{2}\hfill \\ \hfill \frac{1}{2}& =\hfill & {\left(y+3\right)}^{2}\hfill \\ \hfill y+3& =\hfill & \pm \sqrt{\frac{1}{2}}\hfill \\ \hfill y& =\hfill & \mathrm{3}\pm \frac{\sqrt{2}}{2}\hfill \end{array}$ 
$y=\mathrm{3}+\frac{\sqrt{2}}{2}\phantom{\rule{1em}{0ex}}y=\mathrm{3}\frac{\sqrt{2}}{2}$  
$y\approx 2.3\phantom{\rule{2em}{0ex}}y\approx 3.7$  
The yintercepts are $\left(0,\mathrm{3}+\frac{\sqrt{2}}{2}\right),\left(0,\mathrm{3}\frac{\sqrt{2}}{2}\right).$  
Graph the parabola. 
ⓐ Write $x=3{y}^{2}+6y+7$ in standard form and ⓑ use properties of the standard form to graph the equation.
ⓐ Write $x=\mathrm{4}{y}^{2}16y12$ in standard form and ⓑ use properties of the standard form to graph the equation.
Solve Applications with Parabolas
Many architectural designs incorporate parabolas. It is not uncommon for bridges to be constructed using parabolas as we will see in the next example.
Example 11.19
Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.
We will first set up a coordinate system and draw the parabola. The graph will give us the information we need to write the equation of the graph in the standard form$y=a{\left(xh\right)}^{2}+k.$
Let the lower left side of the bridge be the origin of the coordinate grid at the point $\left(0,0\right).$ Since the base is 20 feet wide the point $\left(20,0\right)$ represents the lower right side. The bridge is 10 feet high at the highest point. The highest point is the vertex of the parabola so the ycoordinate of the vertex will be 10. Since the bridge is symmetric, the vertex must fall halfway between the left most point, $\left(0,0\right),$ and the rightmost point $\left(20,0\right).$ From this we know that the xcoordinate of the vertex will also be 10. 

Identify the vertex, $\left(h,k\right).$  $\left(h,k\right)=\left(10,10\right)$ 
$h=10,\text{\hspace{1em}}k=10$  
Substitute the values into the standard form. The value of a is still unknown. To find the value of a use one of the other points on the parabola. 
$\phantom{\rule{1.3em}{0ex}}\begin{array}{ccc}\hfill y& =\hfill & a{\left(xh\right)}^{2}+k\hfill \\ \hfill y& =\hfill & a{\left(x10\right)}^{2}+10\hfill \\ \hfill \left(x,y\right)& =\hfill & \left(0,0\right)\hfill \end{array}$ 
Substitute the values of the other point into the equation. 
$\phantom{\rule{3.35em}{0ex}}\begin{array}{ccc}\hfill y& =\hfill & a{\left(x10\right)}^{2}+10\hfill \\ \hfill 0& =\hfill & a{\left(010\right)}^{2}+10\hfill \end{array}$ 
Solve for a.  $\phantom{\rule{2em}{0ex}}\begin{array}{ccc}\hfill 0& =\hfill & a{\left(010\right)}^{2}+10\hfill \\ \hfill \mathrm{10}& =\hfill & a{\left(\mathrm{10}\right)}^{2}\hfill \\ \hfill \mathrm{10}& =\hfill & 100a\hfill \\ \hfill \frac{\mathrm{10}}{100}& =\hfill & a\hfill \\ \hfill a& =\hfill & \frac{1}{10}\hfill \end{array}$ 
$y=a{\left(x10\right)}^{2}+10$  
Substitute the value for a into the equation. 
$\phantom{\rule{1.8em}{0ex}}y=\frac{1}{10}{\left(x10\right)}^{2}+10$ 
Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.
Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.
Media Access Additional Online Resources
Access these online resources for additional instructions and practice with quadratic functions and parabolas.
Section 11.2 Exercises
Practice Makes Perfect
Graph Vertical Parabolas
In the following exercises, graph each equation by using properties.
$y=\text{\u2212}{x}^{2}+8x15$
$y=8{x}^{2}10x+3$
In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.
$y=2{x}^{2}+4x+6$
$y=3{x}^{2}12x+7$
Graph Horizontal Parabolas
In the following exercises, graph each equation by using properties.
$x=3{y}^{2}$
$x=\mathrm{4}{y}^{2}$
$x=\text{\u2212}{y}^{2}4y+5$
$x={y}^{2}4y12$
$x={\left(y1\right)}^{2}+4$
$x=\text{\u2212}{\left(y4\right)}^{2}+3$
$x={\left(y+1\right)}^{2}+2$
$x=\text{\u2212}{\left(y+4\right)}^{2}+3$
$x=\mathrm{2}{\left(y1\right)}^{2}+2$
$x=2{\left(y+4\right)}^{2}2$
In the following exercises, ⓐ write the equation in standard form and ⓑ use properties of the standard form to graph the equation.
$x={y}^{2}+2y3$
$x=\mathrm{3}{y}^{2}6y5$
Mixed Practice
In the following exercises, match each graph to one of the following equations: ⓐ x^{2} + y^{2} = 64 ⓑ x^{2} + y^{2} = 49
ⓒ (x + 5)^{2} + (y + 2)^{2} = 4 ⓓ (x − 2)^{2} + (y − 3)^{2} = 9 ⓔ y = −x^{2} + 8x − 15 ⓕ y = 6x^{2} + 2x − 1
Solve Applications with Parabolas
Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.
Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.
Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.
Write the equation in standard form of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.
Writing Exercises
Is the parabola $y={x}^{2}$ a function? Is the parabola $x={y}^{2}$ a function? Explain why or why not.
Write the equation of a parabola that opens up or down in standard form and the equation of a parabola that opens left or right in standard form. Provide a sketch of the parabola for each one, label the vertex and axis of symmetry.
Explain in your own words, how you can tell from its equation whether a parabola opens up, down, left or right.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ After reviewing this checklist, what will you do to become confident for all objectives?