# 5.16: Vertex, Intercept, and Standard Form

**At Grade**Created by: CK-12

**Practice**Quadratic Functions and Equations

The profit on your school fundraiser is represented by the quadratic expression *p* is your price point. What price point will result in a maximum profit and what is that profit?

### Guidance

So far, we have only used the **standard form** of a quadratic equation, **intercept form** of a quadratic equation is

#### Example A

Change

**Solution:** First, let’s change this equation into intercept form by factoring.

Notice, this does not exactly look like the definition. The factors cannot have a number in front of *halfway* between the

This is also the

The vertex is

The last form is vertex form. **Vertex form** is written

#### Example B

Find the vertex of

**Solution:** The vertex is going to be (1, 3). To graph this parabola, use the symmetric properties of the function. Pick a value on the left side of the vertex. If *other* side of 1 is 5. Therefore, the

#### Example C

Change

**Solution:** To change an equation from standard form into vertex form, you must complete the square. Review the *Completing the Square* Lesson if needed. The major difference is that you will not need to solve this equation.

To solve an equation in vertex form, set

**Intro Problem Revisit** The vertex will give us the price point that will result in the maximum profit and that profit, so let’s change this equation into intercept form by factoring. First factor out –5.

From this we can see that the x-intercepts are 40 and 40. The average of 40 and 40 is 40 we plug 40 into our original equation.

Therefore, the price point that results in a maximum profit is $40 and that price point results in a profit of $0. You're not making any money, so you better rethink your fundraising approach!

### Guided Practice

1. Find the intercepts of

2. Find the vertex of

3. Change

4. Change

#### Answers

1. The intercepts are the opposite sign from the factors; (7, 0) and (-2, 0). To change the equation into standard form, FOIL the factors and distribute

2. The vertex is (-4, -5). To change the equation into standard form, FOIL

3. To change

4. To change \begin{align*}y=x^2-6x-7\end{align*} into vertex form, complete the square.

\begin{align*}y+7+9 &=x^2-6x+9\\ y+16 &=(x-3)^2 \\ y &=(x-3)^2-16\end{align*}

The vertex is (3, -16).

For vertex form, we could solve the equation by using square roots or we could factor the standard form. Either way, we will get that the \begin{align*}x-\end{align*}intercepts are (7, 0) and (-1, 0).

### Vocabulary

- Standard form
- \begin{align*}y=ax^2+bx+c\end{align*}

- Intercept form
- \begin{align*}y=a(x-p)(x-q)\end{align*}, where \begin{align*}p\end{align*} and \begin{align*}q\end{align*} are the \begin{align*}x-\end{align*}intercepts.

- Vertex form
- \begin{align*}y=a(x-h)^2+k\end{align*}, where \begin{align*}(h, k)\end{align*} is the vertex.

### Practice

- Fill in the table below. Either describe how to find each entry or use a formula.

Find the vertex and \begin{align*}x-\end{align*}intercepts of each function below. Then, graph the function. If a function does not have any \begin{align*}x-\end{align*}intercepts, use the symmetry property of parabolas to find points on the graph.

- \begin{align*}y=(x-4)^2-9\end{align*}
- \begin{align*}y=(x+6)(x-8)\end{align*}
- \begin{align*}y=x^2+2x-8\end{align*}
- \begin{align*}y=-(x-5)(x+7)\end{align*}
- \begin{align*}y=2(x+1)^2-3\end{align*}
- \begin{align*}y=3(x-2)^2+4\end{align*}
- \begin{align*}y=\frac{1}{3}(x-9)(x+3)\end{align*}
- \begin{align*}y=-(x+2)^2+7\end{align*}
- \begin{align*}y=4x^2-13x-12\end{align*}

Change the following equations to intercept form.

- \begin{align*}y=x^2-3x+2\end{align*}
- \begin{align*}y=-x^2-10x+24\end{align*}
- \begin{align*}y=4x^2+18x+8\end{align*}

Change the following equations to vertex form.

- \begin{align*}y=x^2+12x-28\end{align*}
- \begin{align*}y=-x^2-10x+24\end{align*}
- \begin{align*}y=2x^2-8x+15\end{align*}

Change the following equations to standard form.

- \begin{align*}y=(x-3)^2+8\end{align*}
- \begin{align*}y=2\left(x-\frac{3}{2}\right)(x-4)\end{align*}
- \begin{align*}y=-\frac{1}{2}(x+6)^2-11\end{align*}

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Intercept form

The intercept form of a quadratic function is , where and are the -intercepts of the function.standard form

The standard form of a quadratic function is .Vertex form

The vertex form of a quadratic function is , where is the vertex of the parabola.### Image Attributions

Here you'll explore the different forms of the quadratic equation.