22.3: Parabolas

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Learning Objectives

By the end of this section, you will be able to:

Graph vertical parabolas
Graph horizontal parabolas
Solve applications with parabolas

Be Prepared

Before you get started, take this readiness quiz.

Graph: $y=-3 x^{2}+12 x-12$.
If you missed this problem, review Example 9.47.
Solve by completing the square: $x^{2}-6 x+6=0$.
If you missed this problem, review Example 9.12.
Write in standard form: $y=3 x^{2}-6 x+5$.
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.

This figure shows a double cone. The bottom nappe is intersected by a plane in such a way that the intersection forms a parabola. — Figure 11.2.1

Definition $\PageIndex{1}$: Parabola, Focus, and Directrix

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.

This figure shows a parabola opening upwards. Below the parabola is a horizontal line labeled directrix. A vertical dashed line through the center of the parabola is labeled axis of symmetry. The point where the axis intersects the parabola is labeled vertex. A point on the axis, within the parabola is labeled focus. A line perpendicular to the directrix connects the directrix to a point on the parabola and another line connects this point to the focus. Both these lines are of the same length. — Figure 11.2.2

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

Table 11.2.1
	General form $y=a x^{2}+b x+c$	Standard Form $y=a(x-h)^{2}+k$
Orientation	$a>0$ up; $a<0$ down	$a>0$ up; $a<0$ down
Axis of Symmetry	$x=-\dfrac{b}{2 a}$	$x=h$
Vertex	Substitute $x=-\dfrac{b}{2 a}$ and solve for $y .$	$(h, k)$
$y$-intercept	Let $x=0$	Let $x=0$
$x$-intercepts	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 $y$-axis is merely an example.

This figure shows two parabolas with axis x equals h and vertex h, k. The one on the left opens up and A is greater than 0. The one on the right opens down. Here A is less than 0. — Figure 11.2.3

To graph a parabola from these forms, we used the following steps.

Graphing Vertical Parabolas

How to Graph Vertical Parabolas $y=a x^{2}+b x+c$ or $f(x)=a(x-h)^{2}+k$ 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 $y$-intercept. Find the point symmetric to the $y$-intercept across the axis of symmetry.
Step 5. Find the $x$-intercepts.
Step 6. Graph the parabola.

The next example reviews the method of graphing a parabola from the general form of its equation.

Example $\PageIndex{1}$

Graph $y=-x^{2}+6 x-8$ by using properties.

Solution:

Table 11.2.2
	$ \begin{align} \color{red}{y} &\color{red}{=} a x^{2}+b x+c \\[4pt] \color{black}{y} &=-x^{2}+6 x-8 \end{align}$
Since $a$ is $-1$, the parabola opens downward.
$.$
To find the axis of symmetry, find $x=-\dfrac{b}{2 a}$.	$ \begin{align} x &=-\dfrac{b}{2 a}\\[4pt] x &=-\dfrac{6}{2(-1)} \\[4pt] x &= 3 \end{align}$
	The axis of symmetry is $x=3$.
	$.$
The vertex is on the line $x=3$.	$y=-x^{2}+6 x-8$
Let $x=3$.	$.$
	$\begin{align} y &=-9+18-8 \\[4pt] y &=1 \end{align}$
	The vertex is $(3,1)$.
	$.$
The $y$-intercept occurs when $x=0$.	$y=-x^{2}+6 x-8$
Substitute $x=0$.	$y=-\color{red}{0}^{\color{black}{2}}+6 \cdot \color{red}{0} \color{black}{-} 8$
Simplify.	$y=-8$
	The $y$-intercept is $(0,-8)$.
The point $(0,−8)$ is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is $(6,−8)$.	Point symmetric to the $y$-intercept is $(6,−8)$.
	$.$
The $x$-intercept occurs when $y=0$.	$y=-x^{2}+6 x-8$
Let $y=0$.	$\color{red}{0} \color{black}{=}-x^{2}+6 x-8$
Factor the GCF.	$0=-\left(x^{2}-6 x+8\right)$
Factor the trinomial.	$0=-(x-4)(x-2)$
Solve for $x$.	$x=4, \quad x=2$
	The $x$-intercepts are $(4,0),(2,0)$.
Graph the parabola.	$.$

Exercise $\PageIndex{1}$

Graph $y=-x^{2}+5 x-6$ by using properties.

Answer: $This graph shows a parabola opening downward, with x intercepts (2, 0) and (3, 0) and y intercept (0, negative 6).$
Figure 11.2.24

Exercise $\PageIndex{2}$

Graph $y=-x^{2}+8 x-12$ by using properties.

Answer: $This graph shows a parabola opening downward, with vertex (4, 4) and x intercepts (2, 0) and (6, 0).$
Figure 11.2.25

The next example reviews the method of graphing a parabola from the standard form of its equation, $y=a(x-h)^{2}+k$.

Example $\PageIndex{2}$

Write $y=3 x^{2}-6 x+5$ in standard form and then use properties of standard form to graph the equation.

Solution:

Table 11.2.3
Rewrite the function in $y=a(x-h)^{2}+k$ form by completing the square.	$\begin{align} y &=3 x^{2}-6 x+5 \\[4pt] y &=3\left(x^{2}-2 x\right)+5 \\[4pt] y &=3\left(x^{2}-2 x+1\right) + 5-3 \\[4pt] y &=3(x-1)^{2}+2 \end{align}$
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 $(h,k)$.	The vertex is $(1,2)$.
Find the $y$-intercept by substituting $x=0$,	$ \begin{align} y &=3(x-1)^{2}+2 \\[4pt] y &=3 \cdot 0^{2}-6 \cdot 0+5 \\[4pt] y &=0 \end{align} $
	$y$-intercept $(0,5)$
Find the point symmetric to $(0,5)$ across the axis of symmetry.	$(2,5)$
Find the $x$-intercepts.	$\begin{aligned} y &=3(x-1)^{2}+2 \\[4pt] 0 &=3(x-1)^{2}+2 \\[4pt] -2 &=3(x-1)^{2} \\[4pt] -\dfrac{2}{3} &=(x-1)^{2} \\[4pt] \pm \sqrt{-\dfrac{2}{3}} &=x-1 \end{aligned}$
	The square root of a negative number tells us the solutions are complex numbers. So there are no $x$-intercepts.
Graph the parabola.	$.$

Exercise $\PageIndex{3}$

Write $y=2 x^{2}+4 x+5$ in standard form and
use properties of standard form to graph the equation.

Answer

$y=2(x+1)^{2}+3$

This graph shows a parabola opening upwards, with vertex (negative 1, 3) and y intercept (0, 5). It has the point minus (2, 5) on it. — Figure 11.2.28

Exercise $\PageIndex{4}$

Write $y=-2 x^{2}+8 x-7$ in standard form and
use properties of standard form to graph the equation.

Answer

$y=-2(x-2)^{2}+1$

This graph shows a parabola opening downwards, with vertex (2, 1) and axis of symmetry x equals 2. Its y intercept is (0, negative 7). — Figure 11.2.29

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

Table 11.2.4
	General form $x=a y^{2}+b y+c$	Standard form $x=a(y-k)^{2}+h$
Orientation	$a>0$ right; $a<0$ left	$a>0$ right; $a<0$ left
Axis of Symmetry	$y=-\dfrac{b}{2 a}$	$y=k$
Vertex	Substitute $y=-\dfrac{b}{2 a}$ and solve for $x .$	$(h, k)$
$x$-intercepts	Let $x=0$	Let $x=0$
$y$-intercept	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 $y$-axis is merely an example.

This figure shows two parabolas with axis of symmetry y equals k,) and vertex (h, k. The one on the left is labeled a greater than 0 and opens to the right. The other parabola opens to the left. — Figure 11.2.30

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.

Howto: Graph Horizontal Parabolas $y=a x^{2}+b x+c$ or $f(x)=a(x-h)^{2}+k$ 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 $x$-intercept. Find the point symmetric to the $x$-intercept across the axis of symmetry.
Step 5: Find the $y$-intercepts.
Step 6: Graph the parabola.

Example $\PageIndex{3}$

Graph $x=2 y^{2}$ by using properties.

Solution:

Table 11.2.5
	$.$
Since $a=2$, the parabola opens to the right.
$.$
To find the axis of symmetry, find $y=-\dfrac{b}{2 a}$	$y=-\dfrac{b}{2 a}$
	$y=-\dfrac{0}{2(2)}$
	$y=0$
	The axis of symmetry is $y=0$.
The vertex is on the line $y=0$.	$x=2 y^{2}$
Let $y=0$.	$.$
	$x=0$
	The vertex is $(0,0)$.

Since the vertex is $(0,0)$, both the $x$- and $y$-intercepts are the point $(0,0)$. To graph the parabola we need more points. In this case it is easiest to choose values of $y$.

In the equation x equals 2 y squared, when y is 1, x is 2 and when y is 2, x is 8. The points are (2, 1) and (8, 2). — Figure 11.2.38

We also plot the points symmetric to $(2,1)$ and $(8,2)$ across the $y$-axis, the points $(2,−1),(8,−2)$.

Graph the parabola.

This graph shows right opening parabola with vertex (0, 0). Four points are marked on it: point (2, 1), point (2, negative 1), point (8, 2) and point (8 minus 2). — Figure 11.2.39

Exercise $\PageIndex{5}$

Graph $x=y^{2}$ by using properties.

Answer: $This graph shows right opening parabola with vertex at origin. Two points on it are (4, 2) and (4, negative 2).$
Figure 11.2.40

Exercise $\PageIndex{6}$

Graph $x=-y^{2}$ by using properties.

Answer: $This graph shows left opening parabola with vertex at origin. Two points on it are (negative 4, 2) and (negative 4, negative 2).$
Figure 11.2.41

In the next example, the vertex is not the origin.

Example $\PageIndex{4}$

Graph $x=-y^{2}+2 y+8$ by using properties.

Solution:

Table 11.2.6
	$.$
Since $a=-1$, the parabola opens to the left.
$.$
To find the axis of symmetry, find $y=-\dfrac{b}{2 a}$	$y=-\dfrac{b}{2 a}$
	$y=-\dfrac{2}{2(-1)}$
	$y=1$
	The axis of symmetry is $y=1$.
The vertex is on the line $y=1$.	$x=-y^{2}+2 y+8$
Let $y=1$.	$.$
	$x=9$
	The vertex is $(9,1)$.
The $x$-intercept occurs when $y=0$.	$x=-y^{2}+2 y+8$
	$.$
	$x=8$
	The $x$-intercept is $(8,0)$.
The point $(8,0)$ is one unit below the line of symmetry. The symmetric point one unit above the line of symmetry is $(8,2)$	Symmetric point is $(8,2)$.
The $y$-intercept occurs when $x=0$.	$x=-y^{2}+2 y+8$
Substitute $x=0$.	$0=-y^{2}+2 y+8$
Solve.	$y^{2}-2 y-8=0$
	$(y-4)(y+2)=0$
	$y=4, \quad y=-2$
	The $y$ -intercepts are $(0,4)$ and $(0,-2)$.
Connect the points to graph the parabola.	$.$

Exercise $\PageIndex{7}$

Graph $x=-y^{2}-4 y+12$ by using properties.

Answer: $This graph shows left opening parabola with vertex (16, negative 2) and x intercept (12, 0).$
Figure 11.2.58

Exercise $\PageIndex{8}$

Graph $x=-y^{2}+2 y-3$ by using properties.

Answer: $This graph shows left opening parabola with vertex (negative 2, 1) and x intercept minus (3, 0).$
Figure 11.2.59

In Table 11.2.4, 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 $x=a(y-k)^{2}+h$. We will use this procedure in the next example.

Example $\PageIndex{5}$

Graph $x=2(y-2)^{2}+1$ using properties.

Solution:

Table 11.2.7
	$.$
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 $y=k$.	The axis of symmetry is $y=2$.
The vertex is $(h,k)$.	The vertex is $(1,2)$.
Find the $x$-intercept by substituting $y=0$.	$x=2(y-2)^{2}+1$ $x=2(0-2)^{2}+1$ $x=9$
	The $x$-intercept is $(9,0)$.
Find the point symmetric to $(9,0)$ across the axis of symmetry.	$(9,4)$
Find the $y$-intercepts. Let $x=0$.	$\begin{aligned} x &=2(y-2)^{2}+1 \\ 0 &=2(y-2)^{2}+1 \\-1 &=2(y-2)^{2} \end{aligned}$
	A square cannot be negative, so there is no real solution. So there are no $y$-intercepts.
Graph the parabola.	$.$

Exercise $\PageIndex{9}$

Graph $x=3(y-1)^{2}+2$ using properties.

Answer: $This graph shows a parabola opening right with vertex (2, 1) and x intercept (5, 0).$
Figure 11.2.63

Exercise $\PageIndex{10}$

Graph $x=2(y-3)^{2}+2$ using properties.

Answer: $This graph shows a parabola opening right with vertex (2, 3) and symmetric points (4, 2) and (4, 4).$
Figure 11.2.64

In the next example, we notice the a is negative and so the parabola opens to the left.

Example $\PageIndex{6}$

Graph $x=-4(y+1)^{2}+4$ using properties.

Solution:

Table 11.2.8
	$.$
Identify the constants $a, h, k$.	$a=-4, h=4, k=-1$
Since $a=-4$, the parabola opens to the left.
$.$
The axis of symmetry is $y=k$.	The axis of symmetry is $y=-1$.
The vertex is $(h,k)$.	The vertex is $(4,-1)$.
Find the $x$-intercept by substituting $y=0$.	$x=-4(y+1)^{2}+4$ $x=-4(0+1)^{2}+4$ $x=0$
	The $x$-intercept is $(0,0)$.
Find the point symmetric to $(0,0)$ across the axis of symmetry.	$(0,-2)$
Find the $y$-intercepts.	$x=-4(y+1)^{2}+4$
Let $x=0$.	$\begin{aligned} 0 &=-4(y+1)^{2}+4 \\-4 &=-4(y+1)^{2} \\ 1 &=(y+1)^{2} \\ y+1 &=\pm 1 \end{aligned}$
	$y=-1+1 \quad y=-1-1$
	$y=0 \quad\quad y=-2$
	The $y$-intercepts are $(0,0)$ and $(0,-2)$.
Graph the parabola.	$.$

Exercise $\PageIndex{11}$

Graph $x=-4(y+2)^{2}+4$ using properties.

Answer: $This figure shows a parabola opening to the left with vertex (4, negative 2) and y intercepts (0, negative 1) and (0, negative 3).$
Figure 11.2.68

Exercise $\PageIndex{12}$

Graph $x=-2(y+3)^{2}+2$ using properties.

Answer: $This figure shows a parabola opening to the left with vertex (2, negative 3) and y intercepts (0, negative 2) and (0, negative 4).$
Figure 11.2.69

The next example requires that we first put the equation in standard form and then use the properties.

Example $\PageIndex{7}$

Write $x=2 y^{2}+12 y+17$ in standard form and then use the properties of the standard form to graph the equation.

Solution:

Table 11.2.9
	$x=2 y^{2}+12 y+17$
Rewrite the function in $x=a(y-k)^{2}+h$ form by completing the square.	$x=2\left(y^{2}+6 y\right)+17$
	$.$
	$x=2(y+3)^{2}-1$
	$.$
Identify the constants $a, h, k$.	$a=2, h=-1, k=-3$
Since $a=2$, the parabola opens to the right.
$.$
The axis of symmetry is $y=k$.	The axis of symmetry is $y=-3$.
The vertex is $(h,k)$.	The vertex is $(-1,-3)$.
Find the $x$-intercept by substituting $y=0$.	$x=2(y+3)^{2}-1$ $x=2(0+3)^{2}-1$ $x=17$
	The $x$-intercept is $(17,0)$.
Find the point symmetric to $(17,0)$ across the axis of symmetry.	$(17,-6)$
Find the $y$-intercepts. Let $x=0$.	$\begin{aligned} x &=2(y+3)^{2}-1 \\ 0 &=2(y+3)^{2}-1 \\ 1 &=2(y+3)^{2} \\ \dfrac{1}{2} &=(y+3)^{2} \\ y+3 &=\pm \sqrt{\dfrac{1}{2}} \\ y &=-3 \pm \dfrac{\sqrt{2}}{2} \end{aligned}$
	$y=-3+\dfrac{\sqrt{2}}{2} \quad y=-3-\dfrac{\sqrt{2}}{2}$
	$y \approx-2.3 \quad y \approx-3.7$
	The $y$-intercepts are $\left(0,-3+\dfrac{\sqrt{2}}{2}\right),\left(0,-3-\dfrac{\sqrt{2}}{2}\right)$.
Graph the parabola.	$.$

Exercise $\PageIndex{13}$

Write $x=3 y^{2}+6 y+7$ in standard form and
Use properties of the standard form to graph the equation.

Answer

$x=3(y+1)^{2}+4$

This graph shows a parabola opening to the right with vertex (4, negative 1) and x intercept (7, 0). — Figure 11.2.77

Exercise $\PageIndex{14}$

Write $x=-4 y^{2}-16 y-12$ in standard form and
Use properties of the standard form to graph the equation.

Answer

$x=-4(y+2)^{2}+4$

This graph shows a parabola opening to the left with vertex (4, negative 2) and x intercept minus (12, 0). — Figure 11.2.78

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 $\PageIndex{8}$

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 10 feet high and 20 feet wide at the base. — Figure 11.2.79

Solution:

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 $y=a(x-h)^{2}+k$.

Table 11.2.10
Let the lower left side of the bridge be the origin of the coordinate grid at the point $(0,0)$. Since the base is $20$ feet wide the point $(20,0)$ 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 $y$-coordinate of the vertex will be $10$. Since the bridge is symmetric, the vertex must fall halfway between the left most point, $(0,0)$, and the rightmost point $(20,0)$. From this we know that the $x$-coordinate of the vertex will also be $10$.	$.$
Identify the vertex, $(h,k)$.	$(h, k)=(10,10)$
	$h=10, \quad k=10$
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.	$\begin{aligned} y &=a(x-h)^{2}+k \\ y &=a(x-10)^{2}+10 \\(x, y) &=(0,0) \end{aligned}$
Substitute the values of the other point into the equation.	$y=a(x-10)^{2}+10$ $0=a(0-10)^{2}+10$
Solve for $a$.	$\begin{aligned} 0 &=a(0-10)^{2}+10 \\-10 &=a(-10)^{2} \\-10 &=100 a \\ \dfrac{-10}{100} &=a \\ a &=-\dfrac{1}{10} \end{aligned}$
	$y=a(x-10)^{2}+10$
Substitute the value for $a$ into the equation.	$y=-\dfrac{1}{10}(x-10)^{2}+10$

Exercise $\PageIndex{15}$

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 20 feet high and 40 feet wide at the base. — Figure 11.2.81

Answer: $y=-\dfrac{1}{20}(x-20)^{2}+20$

Exercise $\PageIndex{16}$

Find the equation of the parabolic arch formed in the foundation of the bridge shown. Write the equation in standard form.

This figure shows a parabolic arch formed in the foundation of a bridge. It is 5 feet high and 10 feet wide at the base. — Figure 11.2.82

Answer: $y=-\dfrac{1}{5} x^{2}+2 x y=-\dfrac{1}{5}(x-5)^{2}+5$

Access these online resources for additional instructions and practice with quadratic functions and parabolas.

Quadratic Functions
Introduction to Conics and Graphing Horizontal Parabolas

Key Concepts

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.

Vertical Parabolas

Table 11.2.1
	General form $y=a x^{2}+b x+c$	Standard Form $y=a(x-h)^{2}+k$
Orientation	$a>0$ up; $a<0$ down	$a>0$ up; $a<0$ down
Axis of Symmetry	$x=-\dfrac{b}{2 a}$	$x=h$
Vertex	Substitute $x=-\dfrac{b}{2 a}$ and solve for $y .$	$(h, k)$
$y$-intercept	Let $x=0$	Let $x=0$
$x$-intercepts	Let $y=0$	Let $y=0$

How to graph vertical parabolas $y=a x^{2}+b x+c$ or $f(x)=a(x-h)^{2}+k)$ using properties.

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.

Horizontal Parabolas

Table 11.2.4
	General form $x=a y^{2}+b y+c$	Standard form $x=a(y-k)^{2}+h$
Orientation	$a>0$ right; $a<0$ left	$a>0$ right; $a<0$ left
Axis of Symmetry	$y=-\dfrac{b}{2 a}$	$y=k$
Vertex	Substitute $y=-\dfrac{b}{2 a}$ and solve for $x .$	$(h, k)$
$x$-intercepts	Let $x=0$	Let $x=0$
$y$-intercept	Let $y=0$	Let $y=0$

Graphing Horizontal Parabolas

How to graph horizontal parabolas $x=a y^{2}+b y+c$ or $x=a(y-k)^{2}+h$ using properties.

Determine whether the parabola opens to the left or to the right.
Find the axis of symmetry.
Find the vertex.
Find the $x$-intercept. Find the point symmetric to the $x$-intercept across the axis of symmetry.
Find the $y$-intercepts.
Graph the parabola.

Glossary

parabola: A parabola is all points in a plane that are the same distance from a fixed point and a fixed line.

	General form \(y=a x^{2}+b x+c\)	Standard Form \(y=a(x-h)^{2}+k\)
Orientation	\(a>0\) up; \(a<0\) down	\(a>0\) up; \(a<0\) down
Axis of Symmetry	\(x=-\dfrac{b}{2 a}\)	\(x=h\)
Vertex	Substitute \(x=-\dfrac{b}{2 a}\) and solve for \(y .\)	\((h, k)\)
\(y\)-intercept	Let \(x=0\)	Let \(x=0\)
\(x\)-intercepts	Let \(y=0\)	Let \(y=0\)

	\( \begin{align} \color{red}{y} &\color{red}{=} a x^{2}+b x+c \\[4pt] \color{black}{y} &=-x^{2}+6 x-8 \end{align}\)
Since \(a\) is \(-1\), the parabola opens downward.
$.$
To find the axis of symmetry, find \(x=-\dfrac{b}{2 a}\).	\( \begin{align} x &=-\dfrac{b}{2 a}\\[4pt] x &=-\dfrac{6}{2(-1)} \\[4pt] x &= 3 \end{align}\)
	The axis of symmetry is \(x=3\).
	$.$
The vertex is on the line \(x=3\).	\(y=-x^{2}+6 x-8\)
Let \(x=3\).	$.$
	\(\begin{align} y &=-9+18-8 \\[4pt] y &=1 \end{align}\)
	The vertex is \((3,1)\).
	$.$
The \(y\)-intercept occurs when \(x=0\).	\(y=-x^{2}+6 x-8\)
Substitute \(x=0\).	\(y=-\color{red}{0}^{\color{black}{2}}+6 \cdot \color{red}{0} \color{black}{-} 8\)
Simplify.	\(y=-8\)
	The \(y\)-intercept is \((0,-8)\).
The point \((0,−8)\) is three units to the left of the line of symmetry. The point three units to the right of the line of symmetry is \((6,−8)\).	Point symmetric to the \(y\)-intercept is \((6,−8)\).
	$.$
The \(x\)-intercept occurs when \(y=0\).	\(y=-x^{2}+6 x-8\)
Let \(y=0\).	\(\color{red}{0} \color{black}{=}-x^{2}+6 x-8\)
Factor the GCF.	\(0=-\left(x^{2}-6 x+8\right)\)
Factor the trinomial.	\(0=-(x-4)(x-2)\)
Solve for \(x\).	\(x=4, \quad x=2\)
	The \(x\)-intercepts are \((4,0),(2,0)\).
Graph the parabola.	$.$

Rewrite the function in \(y=a(x-h)^{2}+k\) form by completing the square.	\(\begin{align} y &=3 x^{2}-6 x+5 \\[4pt] y &=3\left(x^{2}-2 x\right)+5 \\[4pt] y &=3\left(x^{2}-2 x+1\right) + 5-3 \\[4pt] y &=3(x-1)^{2}+2 \end{align}\)
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 \((h,k)\).	The vertex is \((1,2)\).
Find the \(y\)-intercept by substituting \(x=0\),	\( \begin{align} y &=3(x-1)^{2}+2 \\[4pt] y &=3 \cdot 0^{2}-6 \cdot 0+5 \\[4pt] y &=0 \end{align} \)
	\(y\)-intercept \((0,5)\)
Find the point symmetric to \((0,5)\) across the axis of symmetry.	\((2,5)\)
Find the \(x\)-intercepts.	\(\begin{aligned} y &=3(x-1)^{2}+2 \\[4pt] 0 &=3(x-1)^{2}+2 \\[4pt] -2 &=3(x-1)^{2} \\[4pt] -\dfrac{2}{3} &=(x-1)^{2} \\[4pt] \pm \sqrt{-\dfrac{2}{3}} &=x-1 \end{aligned}\)
	The square root of a negative number tells us the solutions are complex numbers. So there are no \(x\)-intercepts.
Graph the parabola.	$.$

	General form \(x=a y^{2}+b y+c\)	Standard form \(x=a(y-k)^{2}+h\)
Orientation	\(a>0\) right; \(a<0\) left	\(a>0\) right; \(a<0\) left
Axis of Symmetry	\(y=-\dfrac{b}{2 a}\)	\(y=k\)
Vertex	Substitute \(y=-\dfrac{b}{2 a}\) and solve for \(x .\)	\((h, k)\)
\(x\)-intercepts	Let \(x=0\)	Let \(x=0\)
\(y\)-intercept	Let \(y=0\)	Let \(y=0\)

	$.$
Since \(a=2\), the parabola opens to the right.
$.$
To find the axis of symmetry, find \(y=-\dfrac{b}{2 a}\)	\(y=-\dfrac{b}{2 a}\)
	\(y=-\dfrac{0}{2(2)}\)
	\(y=0\)
	The axis of symmetry is \(y=0\).
The vertex is on the line \(y=0\).	\(x=2 y^{2}\)
Let \(y=0\).	$.$
	\(x=0\)
	The vertex is \((0,0)\).

	$.$
Since \(a=-1\), the parabola opens to the left.
$.$
To find the axis of symmetry, find \(y=-\dfrac{b}{2 a}\)	\(y=-\dfrac{b}{2 a}\)
	\(y=-\dfrac{2}{2(-1)}\)
	\(y=1\)
	The axis of symmetry is \(y=1\).
The vertex is on the line \(y=1\).	\(x=-y^{2}+2 y+8\)
Let \(y=1\).	$.$
	\(x=9\)
	The vertex is \((9,1)\).
The \(x\)-intercept occurs when \(y=0\).	\(x=-y^{2}+2 y+8\)
	$.$
	\(x=8\)
	The \(x\)-intercept is \((8,0)\).
The point \((8,0)\) is one unit below the line of symmetry. The symmetric point one unit above the line of symmetry is \((8,2)\)	Symmetric point is \((8,2)\).
The \(y\)-intercept occurs when \(x=0\).	\(x=-y^{2}+2 y+8\)
Substitute \(x=0\).	\(0=-y^{2}+2 y+8\)
Solve.	\(y^{2}-2 y-8=0\)
	\((y-4)(y+2)=0\)
	\(y=4, \quad y=-2\)
	The \(y\) -intercepts are \((0,4)\) and \((0,-2)\).
Connect the points to graph the parabola.	$.$

	$.$
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 \(y=k\).	The axis of symmetry is \(y=2\).
The vertex is \((h,k)\).	The vertex is \((1,2)\).
Find the \(x\)-intercept by substituting \(y=0\).	\(x=2(y-2)^{2}+1\) \(x=2(0-2)^{2}+1\) \(x=9\)
	The \(x\)-intercept is \((9,0)\).
Find the point symmetric to \((9,0)\) across the axis of symmetry.	\((9,4)\)
Find the \(y\)-intercepts. Let \(x=0\).	\(\begin{aligned} x &=2(y-2)^{2}+1 \\ 0 &=2(y-2)^{2}+1 \\-1 &=2(y-2)^{2} \end{aligned}\)
	A square cannot be negative, so there is no real solution. So there are no \(y\)-intercepts.
Graph the parabola.	$.$

	$.$
Identify the constants \(a, h, k\).	\(a=-4, h=4, k=-1\)
Since \(a=-4\), the parabola opens to the left.
$.$
The axis of symmetry is \(y=k\).	The axis of symmetry is \(y=-1\).
The vertex is \((h,k)\).	The vertex is \((4,-1)\).
Find the \(x\)-intercept by substituting \(y=0\).	\(x=-4(y+1)^{2}+4\) \(x=-4(0+1)^{2}+4\) \(x=0\)
	The \(x\)-intercept is \((0,0)\).
Find the point symmetric to \((0,0)\) across the axis of symmetry.	\((0,-2)\)
Find the \(y\)-intercepts.	\(x=-4(y+1)^{2}+4\)
Let \(x=0\).	\(\begin{aligned} 0 &=-4(y+1)^{2}+4 \\-4 &=-4(y+1)^{2} \\ 1 &=(y+1)^{2} \\ y+1 &=\pm 1 \end{aligned}\)
	\(y=-1+1 \quad y=-1-1\)
	\(y=0 \quad\quad y=-2\)
	The \(y\)-intercepts are \((0,0)\) and \((0,-2)\).
Graph the parabola.	$.$

	\(x=2 y^{2}+12 y+17\)
Rewrite the function in \(x=a(y-k)^{2}+h\) form by completing the square.	\(x=2\left(y^{2}+6 y\right)+17\)
	$.$
	\(x=2(y+3)^{2}-1\)
	$.$
Identify the constants \(a, h, k\).	\(a=2, h=-1, k=-3\)
Since \(a=2\), the parabola opens to the right.
$.$
The axis of symmetry is \(y=k\).	The axis of symmetry is \(y=-3\).
The vertex is \((h,k)\).	The vertex is \((-1,-3)\).
Find the \(x\)-intercept by substituting \(y=0\).	\(x=2(y+3)^{2}-1\) \(x=2(0+3)^{2}-1\) \(x=17\)
	The \(x\)-intercept is \((17,0)\).
Find the point symmetric to \((17,0)\) across the axis of symmetry.	\((17,-6)\)
Find the \(y\)-intercepts. Let \(x=0\).	\(\begin{aligned} x &=2(y+3)^{2}-1 \\ 0 &=2(y+3)^{2}-1 \\ 1 &=2(y+3)^{2} \\ \dfrac{1}{2} &=(y+3)^{2} \\ y+3 &=\pm \sqrt{\dfrac{1}{2}} \\ y &=-3 \pm \dfrac{\sqrt{2}}{2} \end{aligned}\)
	\(y=-3+\dfrac{\sqrt{2}}{2} \quad y=-3-\dfrac{\sqrt{2}}{2}\)
	\(y \approx-2.3 \quad y \approx-3.7\)
	The \(y\)-intercepts are \(\left(0,-3+\dfrac{\sqrt{2}}{2}\right),\left(0,-3-\dfrac{\sqrt{2}}{2}\right)\).
Graph the parabola.	$.$

Let the lower left side of the bridge be the origin of the coordinate grid at the point \((0,0)\). Since the base is \(20\) feet wide the point \((20,0)\) 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 \(y\)-coordinate of the vertex will be \(10\). Since the bridge is symmetric, the vertex must fall halfway between the left most point, \((0,0)\), and the rightmost point \((20,0)\). From this we know that the \(x\)-coordinate of the vertex will also be \(10\).	$.$
Identify the vertex, \((h,k)\).	\((h, k)=(10,10)\)
	\(h=10, \quad k=10\)
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.	\(\begin{aligned} y &=a(x-h)^{2}+k \\ y &=a(x-10)^{2}+10 \\(x, y) &=(0,0) \end{aligned}\)
Substitute the values of the other point into the equation.	\(y=a(x-10)^{2}+10\) \(0=a(0-10)^{2}+10\)
Solve for \(a\).	\(\begin{aligned} 0 &=a(0-10)^{2}+10 \\-10 &=a(-10)^{2} \\-10 &=100 a \\ \dfrac{-10}{100} &=a \\ a &=-\dfrac{1}{10} \end{aligned}\)
	\(y=a(x-10)^{2}+10\)
Substitute the value for \(a\) into the equation.	\(y=-\dfrac{1}{10}(x-10)^{2}+10\)

Learning Objectives

Be Prepared

Graph Vertical Parabolas

Definition \(\PageIndex{1}\): Parabola, Focus, and Directrix

Vertical Parabolas

Graphing Vertical Parabolas

Example \(\PageIndex{1}\)

Exercise \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)

Example \(\PageIndex{2}\)

Exercise \(\PageIndex{3}\)

Exercise \(\PageIndex{4}\)

Graph Horizontal Parabolas

Horizontal Parabolas

Howto: Graph Horizontal Parabolas \(y=a x^{2}+b x+c\) or \(f(x)=a(x-h)^{2}+k\) using Properties

Example \(\PageIndex{3}\)

Exercise \(\PageIndex{5}\)

Exercise \(\PageIndex{6}\)

Example \(\PageIndex{4}\)

Exercise \(\PageIndex{7}\)

Exercise \(\PageIndex{8}\)

Example \(\PageIndex{5}\)

Exercise \(\PageIndex{9}\)

Exercise \(\PageIndex{10}\)

Example \(\PageIndex{6}\)

Exercise \(\PageIndex{11}\)

Exercise \(\PageIndex{12}\)

Example \(\PageIndex{7}\)

Exercise \(\PageIndex{13}\)

Exercise \(\PageIndex{14}\)

Solve Applications with Parabolas

Example \(\PageIndex{8}\)

Exercise \(\PageIndex{15}\)

Exercise \(\PageIndex{16}\)

Key Concepts

Vertical Parabolas

Horizontal Parabolas

Graphing Horizontal Parabolas

Glossary