# What is the vertex form of y=(5x-5)(x+20)?

Feb 1, 2016

vertex form: $y = 5 {\left(x + \frac{19}{2}\right)}^{2} - \frac{2205}{4}$

#### Explanation:

1. Expand.
Rewrite the equation in standard form.

$y = \left(5 x - 5\right) \left(x + 20\right)$

$y = 5 {x}^{2} + 100 x - 5 x - 100$

$y = 5 {x}^{2} + 95 x - 100$

2. Factor 5 from the first two terms.

$y = 5 \left({x}^{2} + 19 x\right) - 100$

3. Turn the bracketed terms into a perfect square trinomial.
When a perfect square trinomial is in the form $a {x}^{2} + b x + c$, the $c$ value is ${\left(\frac{b}{2}\right)}^{2}$. So you have to divide $19$ by $2$ and square the value.

$y = 5 \left({x}^{2} + 19 x + {\left(\frac{19}{2}\right)}^{2}\right) - 100$

$y = 5 \left({x}^{2} + 19 x + \frac{361}{4}\right) - 100$

4. Subtract 361/4 from the bracketed terms.
You can't just add $\frac{361}{4}$ to the equation, so you have to subtract it from the $\frac{361}{4}$ you just added.

y=5(x^2+19x+361/4 color(red)(-361/4))-100

5. Multiply -361/4 by 5.
You then need to remove the $- \frac{361}{4}$ from the brackets, so you multiply it by your $a$ value, $\textcolor{b l u e}{5}$.

$y = \textcolor{b l u e}{5} \left({x}^{2} + 19 x + \frac{361}{4}\right) - 100 \left[\textcolor{red}{\left(- \frac{361}{4}\right)} \cdot \textcolor{b l u e}{\left(5\right)}\right]$

6. Simplify.

$y = 5 \left({x}^{2} + 19 x + \frac{361}{4}\right) - 100 - \frac{1805}{4}$

$y = 5 \left({x}^{2} + 19 x + \frac{361}{4}\right) - \frac{2205}{4}$

7. Factor the perfect square trinomial.
The last step is to factor the perfect square trinomial. This will tell you the coordinates of the vertex.

$\textcolor{g r e e n}{y = 5 {\left(x + \frac{19}{2}\right)}^{2} - \frac{2205}{4}}$