# 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 \begin{align*}-5p^2 + 400p - 8000\end{align*}*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, \begin{align*}y=ax^2+bx+c\end{align*}**intercept form** of a quadratic equation is \begin{align*}y=a(x-p)(x-q)\end{align*}

#### Example A

Change \begin{align*}y=2x^2+9x+10\end{align*} to intercept form and find the vertex. Graph the parabola.

**Solution:** First, let’s change this equation into intercept form by factoring. \begin{align*}ac = 20\end{align*} and the factors of 20 that add up to 9 are 4 and 5. Expand the \begin{align*}x-\end{align*}term.

\begin{align*}y &=2x^2+9x+10\\ y &=2x^2+4x+5x+10\\ y &=2x(x+2)+5(x+2)\\ y &=(2x+5)(x+2)\end{align*}

Notice, this does not exactly look like the definition. The factors cannot have a number in front of \begin{align*}x\end{align*}. Pull out the 2 from the first factor to get \begin{align*}y=2\left(x+\frac{5}{2}\right)(x+2)\end{align*}. Now, find the vertex. Recall that all parabolas are symmetrical. This means that the axis of symmetry is *halfway* between the \begin{align*}x-\end{align*}intercepts or their average.

\begin{align*}\text{axis of symmetry} = \frac{p+q}{2}= \frac{-\frac{5}{2}-2}{2}= -\frac{9}{2} \div 2 = -\frac{9}{2}\cdot\frac{1}{2}=-\frac{9}{4}\end{align*}

This is also the \begin{align*}x-\end{align*}coordinate of the vertex. To find the \begin{align*}y-\end{align*}coordinate, plug the \begin{align*}x-\end{align*}value into either form of the quadratic equation. We will use Intercept form.

\begin{align*}y &=2\left(-\frac{9}{4}+\frac{5}{2}\right)\left(-\frac{9}{4}+2\right)\\ y &=2\cdot\frac{1}{4}\cdot-\frac{1}{4}\\ y &=-\frac{1}{8}\end{align*}

The vertex is \begin{align*}\left(-2\frac{1}{4}, -\frac{1}{8}\right)\end{align*}. Plot the \begin{align*}x-\end{align*}intercepts and the vertex to graph.

The last form is vertex form. **Vertex form** is written \begin{align*}y=a(x-h)^2+k\end{align*}, where \begin{align*}(h, k)\end{align*} is the vertex and \begin{align*}a\end{align*} is the same is in the other two forms. Notice that \begin{align*}h\end{align*} is negative in the equation, but positive when written in coordinates of the vertex.

#### Example B

Find the vertex of \begin{align*}y=\frac{1}{2}(x-1)^2+3\end{align*} and graph the parabola.

**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 \begin{align*}x = -3\end{align*}, then \begin{align*}y=\frac{1}{2}(-3-1)^2+3=11\end{align*}. -3 is 4 units away from 1 (the \begin{align*}x-\end{align*}coordinate of the vertex). 4 units on the *other* side of 1 is 5. Therefore, the \begin{align*}y-\end{align*}coordinate will be 11. Plot (1, 3), (-3, 11), and (5, 11) to graph the parabola.

#### Example C

Change \begin{align*}y=x^2-10x+16\end{align*} into vertex form.

**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.

\begin{align*}y &=x^2-10x+16\\ y-16+{\color{red}25} &=x^2-10x+{\color{red}25} \quad \text{Move 16 to the other side and add} \ \left(\frac{b}{2}\right)^2 \text{to both sides}.\\ y+9 &=(x-5)^2 \qquad \quad \ \ \text{Simplify left side and factor the right side}\\ y &=(x-5)^2-9 \qquad \text{Subtract 9 from both sides to get} \ y \ \text{by itself}.\end{align*}

To solve an equation in vertex form, set \begin{align*}y = 0\end{align*} and solve for \begin{align*}x\end{align*}.

\begin{align*}(x-5)^2-9 &=0\\ (x-5)^2 &=9\\ x-5 &=\pm 3\\ x &=5 \pm 3 \ or \ 8 \ and \ 2\end{align*}

**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.

\begin{align*}-5p^2 + 400p - 8000 = -5(p^2 - 80p + 1600)\\ -5(p - 40)(p - 40)\end{align*}

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.

\begin{align*}-5(40)^2 + 400(40) - 8000 = -8000 + 16000 - 8000 = 0\end{align*}

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 \begin{align*}y=2(x-7)(x+2)\end{align*} and change it to standard form.

2. Find the vertex of \begin{align*}y = -\frac{1}{2}(x+4)^2-5\end{align*} and change it to standard form.

3. Change \begin{align*}y=x^2+18x+45\end{align*} to intercept form and graph.

4. Change \begin{align*}y=x^2-6x-7\end{align*} to vertex form and graph.

#### 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 \begin{align*}a\end{align*}.

\begin{align*}y &=2(x-7)(x+2)\\ y &=2(x^2-5x-14)\\ y &=2x^2-10x-28\end{align*}

2. The vertex is (-4, -5). To change the equation into standard form, FOIL \begin{align*}(x+4)^2\end{align*}, distribute \begin{align*}a\end{align*}, and then subtract 5.

\begin{align*}y &=-\frac{1}{2}(x+4)(x+4)-5\\ y &=-\frac{1}{2}(x^2+8x+16)-5\\ y &=-\frac{1}{2}x^2-4x-21\end{align*}

3. To change \begin{align*}y=x^2+18x+45\end{align*} into intercept form, factor the equation. The factors of 45 that add up to 18 are 15 and 3. Intercept form would be \begin{align*}y = (x+15)(x+3)\end{align*}. The intercepts are (-15, 0) and (-3, 0). The \begin{align*}x-\end{align*}coordinate of the vertex is halfway between -15 and -3, or -9. The \begin{align*}y-\end{align*}coordinate of the vertex is \begin{align*}y=(-9)^2+18(-9)+45=-36\end{align*}. Here is the graph:

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.