What if you had a quadratic function like *x*-intercepts and vertex to help you graph it? After completing this Concept, you'll be able to use the intercept form of quadratic functions to solve problems like this one.

### Watch This

CK-12 Foundation: 1002S Graph Quadratic Functions in Intercept Form

### Guidance

Now it’s time to learn how to graph a parabola without having to use a table with a large number of points.

Let’s look at the graph of

There are several things we can notice:

- The parabola crosses the
x− axis at two points:x=2 andx=4 . These points are called thex− **intercepts**of the parabola. - The lowest point of the parabola occurs at (3, -1).
- This point is called the
**vertex**of the parabola. - The vertex is the lowest point in any parabola that turns upward, or the highest point in any parabola that turns downward.
- The vertex is
*exactly halfway between the two*x− This will always be the case, and you can find the vertex using that property.*intercepts.*

- This point is called the
- The parabola is
**symmetric.**If you draw a vertical line through the vertex, you see that the two halves of the parabola are mirror images of each other. This vertical line is called the**line of symmetry.**

We said that the general form of a quadratic function is **intercept form:**

This form is very useful because it makes it easy for us to find the

Since the equation is already factored, we use the zero-product property to set each factor equal to zero and solve the individual linear equations:

So the

Once we find the

#### Example A

*Find the x−intercepts and the vertex of the following quadratic functions:*

a)

b)

**Solution**

a)

Write the quadratic function in intercept form by factoring the right hand side of the equation. Remember, to factor we need two numbers whose product is 15 and whose sum is –8. These numbers are –5 and –3.

The function in intercept form is

We find the

We have:

So **the** **intercepts are (5, 0) and (3, 0).**

The vertex is halfway between the two

We find the

So **the vertex is (4, -1).**

b)

Re-write the function in intercept form.

Factor the common term of 3 first:

Then factor completely:

Set

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

For the vertex,

\begin{align*}x = \frac{-4 + 2}{2} = -1\end{align*} and \begin{align*}y = 3(-1)^2 + 6( -1) - 24 = 3 - 6 -24 = -27\end{align*}

**The vertex is: (-1, -27)**

Knowing the vertex and \begin{align*}x-\end{align*}intercepts is a useful first step toward being able to graph quadratic functions more easily. Knowing the vertex tells us where the middle of the parabola is. When making a table of values, we can make sure to pick the vertex as a point in the table. Then we choose just a few smaller and larger values of \begin{align*}x\end{align*}. In this way, we get an accurate graph of the quadratic function without having to have too many points in our table.

#### Example B

*Find the* \begin{align*}x-\end{align*}*intercepts and vertex. Use these points to create a table of values and graph each function.*

a) \begin{align*}y = x^2 - 4\end{align*}

b) \begin{align*} y = -x^2 + 14x - 48\end{align*}

**Solution**

a) \begin{align*}y = x^2 - 4\end{align*}

Let’s find the \begin{align*}x-\end{align*}intercepts and the vertex:

Factor the right-hand side of the function to put the equation in intercept form:

\begin{align*}y = (x - 2)(x + 2)\end{align*}

Set \begin{align*}y = 0\end{align*} and solve:

\begin{align*}&0 = (x - 2)(x + 2)\\ & x - 2 = 0 & & & & x + 2 = 0\\ && & \text{or}\\ & x = 2 & & & & x = -2\end{align*}

The \begin{align*}x-\end{align*}intercepts are (2, 0) and (-2, 0).

Find the vertex:

\begin{align*}x = \frac{2 -2}{2} = 0 \quad y = (0)^2 - 4 = -4\end{align*}

The vertex is (0, -4).

Make a table of values using the vertex as the middle point. Pick a few values of \begin{align*}x\end{align*} smaller and larger than \begin{align*}x = 0\end{align*}. Include the \begin{align*}x-\end{align*}intercepts in the table.

\begin{align*}x\end{align*} | \begin{align*}y = x^2 - 4\end{align*} | |
---|---|---|

\begin{align*}-3\end{align*} | \begin{align*}y = (-3)^2 - 4 = 5\end{align*} | |

–2 | \begin{align*}y = (-2)^2 - 4 = 0\end{align*} | \begin{align*}x-\end{align*}intercept |

–1 | \begin{align*}y = (-1)^2 - 4 = -3\end{align*} | |

0 | \begin{align*}y = (0)^2 - 4 = -4\end{align*} | vertex |

1 | \begin{align*}y = (1)^2 - 4 = -3\end{align*} | |

2 | \begin{align*}y = (2)^2 - 4 = 0\end{align*} | \begin{align*}x-\end{align*}intercept |

3 | \begin{align*}y = (3)^2 - 4 = 5\end{align*} |

Then plot the graph:

b) \begin{align*}y = -x^2 + 14x - 48\end{align*}

Let’s find the \begin{align*}x-\end{align*}intercepts and the vertex:

Factor the right-hand-side of the function to put the equation in intercept form:

\begin{align*}y = -(x^2 - 14x + 48) = -(x - 6)(x - 8)\end{align*}

Set \begin{align*}y = 0\end{align*} and solve:

\begin{align*}&0 = -(x - 6)(x - 8)\\ & x - 6 = 0 & & & & x - 8 = 0\\ && & \text{or} \\ & x = 6 & & & & x = 8\end{align*}

The \begin{align*}x-\end{align*}intercepts are (6, 0) and (8, 0).

Find the vertex:

\begin{align*}x = \frac{6 + 8}{2} = 7 \quad y = -(7)^2 + 14(7) - 48 = 1\end{align*}

The vertex is (7, 1).

Make a table of values using the vertex as the middle point. Pick a few values of \begin{align*}x\end{align*} smaller and larger than \begin{align*}x = 7\end{align*}. Include the \begin{align*}x-\end{align*}intercepts in the table.

\begin{align*}x\end{align*} | \begin{align*}y = -x^2 + 14x - 48\end{align*} |
---|---|

4 | \begin{align*}y = -(4)^2 + 14(4) - 48 = -8\end{align*} |

5 | \begin{align*}y = -(5)^2 + 14(5) - 48 = -3\end{align*} |

6 | \begin{align*}y = -(6)^2 + 14(6) - 48 = 0\end{align*} |

7 | \begin{align*}y = -(7)^2 + 14(7) - 48 = 1\end{align*} |

8 | \begin{align*}y = -(8)^2 + 14(8) - 48 = 0\end{align*} |

9 | \begin{align*}y = -(9)^2 + 14(9) - 48 = -3\end{align*} |

10 | \begin{align*}y = -(10)^2 + 14(10) - 48 = -8\end{align*} |

Then plot the graph:

**Applications of Quadratic Functions to Real-World Problems**

As we mentioned in a previous concept, parabolic curves are common in real-world applications. Here we will look at a few graphs that represent some examples of real-life application of quadratic functions.

#### Example C

*Andrew has 100 feet of fence to enclose a rectangular tomato patch. What should the dimensions of the rectangle be in order for the rectangle to have the greatest possible area?*

**Solution**

Drawing a picture will help us find an equation to describe this situation:

If the length of the rectangle is \begin{align*}x\end{align*}, then the width is \begin{align*}50 - x\end{align*}. (The length and the width add up to 50, not 100, because two lengths and two widths together add up to 100.)

If we let \begin{align*}y\end{align*} be the area of the triangle, then we know that the area is length \begin{align*}\times\end{align*} width, so \begin{align*}y = x (50 - x) = 50x - x^2\end{align*}.

Here’s the graph of that function, so we can see how the area of the rectangle depends on the length of the rectangle:

We can see from the graph that the highest value of the area occurs when the length of the rectangle is 25. The area of the rectangle for this side length equals 625. (Notice that the width is also 25, which makes the shape a square with side length 25.)

This is an example of an *optimization problem.* These problems show up often in the real world, and if you ever study calculus, you’ll learn how to solve them without graphs.

Watch this video for help with the Examples above.

CK-12 Foundation: 1002 Graph Quadratic Functions in Intercept Form

### Vocabulary

- A parabola can be divided in half by a vertical line. Because of this, parabolas have
. The vertical line dividing the parabola into two equal portions is called the line of*symmetry*.*symmetry*

- The point where the parabola crosses the \begin{align*}x-\end{align*}axis is called the \begin{align*}x -\end{align*}
**intercepts**of the parabola.

- All parabolas have a
, the ordered pair that represents the bottom (or the top) of the curve. The line of symmetry always goes through the vertex.*vertex*- The vertex is the lowest point in any parabola that turns upward, or the highest point in any parabola that turns downward.
- The vertex is
\begin{align*}x-\end{align*}*exactly halfway between the two*This will always be the case, and you can find the vertex using that property.*intercepts.*

- If the \begin{align*}x-\end{align*}intercepts are at points \begin{align*}(m, 0)\end{align*} and \begin{align*}(n, 0)\end{align*}. The \begin{align*}x-\end{align*}value of the
**vertex**is halfway between the two \begin{align*}x-\end{align*}intercepts, so we can find it by taking the average of the two values: \begin{align*}\frac{m + n}{2}\end{align*}. Then we can find the \begin{align*}y-\end{align*}value by plugging the value of \begin{align*}x\end{align*} back into the equation of the function.

### Guided Practice

*Anne is playing golf. On the* \begin{align*}4^{th}\end{align*} *tee, she hits a slow shot down the level fairway. The ball follows a parabolic path described by the equation* \begin{align*}y = x - 0.04x^2\end{align*}, *where* \begin{align*}y\end{align*} *is the ball’s height in the air and* \begin{align*}x\end{align*} *is the horizontal distance it has traveled from the tee. The distances are measured in feet. How far from the tee does the ball hit the ground? At what distance from the tee does the ball attain its maximum height? What is the maximum height?*

**Solution**

Let’s graph the equation of the path of the ball:

\begin{align*}x(1 - 0.04x) = 0\end{align*} has solutions \begin{align*}x = 0\end{align*} and \begin{align*} x = 25\end{align*}.

From the graph, we see that the ball hits the ground **25 feet from the tee.** (The other \begin{align*}x-\end{align*}intercept, \begin{align*}x = 0,\end{align*} tells us that the ball was also on the ground when it was on the tee!)

We can also see that the ball reaches its maximum height of **about 6.25 feet** when it is **12.5 feet from the tee.**

### Practice

For 1-4, rewrite the following functions in intercept form. Find the \begin{align*}x-\end{align*}intercepts and the vertex.

- \begin{align*}y =x^2 - 2x - 8\end{align*}
- \begin{align*}y = -x^2 + 10x - 21\end{align*}
- \begin{align*}y =2x^2 + 6x + 4\end{align*}
- \begin{align*}y =3(x+5)(x - 2)\end{align*}

For 5-8, the vertex of which parabola is higher?

- \begin{align*}y = x^2 + 4\end{align*} or \begin{align*}y = x^2 +1\end{align*}
- \begin{align*}y= -2x^2\end{align*} or \begin{align*}y = -2x^2 - 2\end{align*}
- \begin{align*}y = 3x^2 - 3\end{align*} or \begin{align*}y = 3x^2 - 6\end{align*}
- \begin{align*}y = 5 - 2x^2\end{align*} or \begin{align*}y = 8 - 2x^2\end{align*}

For 9-14, graph the following functions by making a table of values. Use the vertex and \begin{align*}x-\end{align*}intercepts to help you pick values for the table.

- \begin{align*}y = 4x^2 - 4\end{align*}
- \begin{align*}y = -x^2 + x + 12\end{align*}
- \begin{align*}y = 2x^2 + 10x + 8\end{align*}
- \begin{align*}y = \frac{1}{2} x^2 - 2x\end{align*}
- \begin{align*}y = x - 2x^2\end{align*}
- \begin{align*}y = 4x^2 -8x + 4\end{align*}

- Nadia is throwing a ball to Peter. Peter does not catch the ball and it hits the ground. The graph shows the path of the ball as it flies through the air. The equation that describes the path of the ball is \begin{align*}y = 4 + 2x - 0.16x^2\end{align*}. Here \begin{align*}y\end{align*} is the height of the ball and \begin{align*}x\end{align*} is the horizontal distance from Nadia. Both distances are measured in feet.
- How far from Nadia does the ball hit the ground?
- At what distance \begin{align*}x\end{align*} from Nadia, does the ball attain its maximum height?
- What is the maximum height?

- Jasreel wants to enclose a vegetable patch with 120 feet of fencing. He wants to put the vegetable against an existing wall, so he only needs fence for three of the sides. The equation for the area is given by \begin{align*}A = 120x - x^2\end{align*}. From the graph, find what dimensions of the rectangle would give him the greatest area.