# What is the vertex form of y=5x^2 - 10x - 75 ?

##### 1 Answer
Dec 19, 2015

$y = 5 {\left(x - 1\right)}^{2} - 80$, meaning the vertex is at the point $\left(x , y\right) = \left(1 , - 80\right)$.

#### Explanation:

First, factor out the coefficient of ${x}^{2}$, which is 5, out of the first two terms:

$y = 5 {x}^{2} - 10 x - 75 = 5 \left({x}^{2} - 2 x\right) - 75$.

Next, complete the square on the expression inside the parentheses. Take the coefficient of $x$, which is $- 2$, divide it by 2 and square it to get $1$. Add this number inside the parentheses and compensate for this change by subtracting $5 \cdot 1 = 5$ outside of the parentheses as follows:

$y = 5 \left({x}^{2} - 2 x + 1\right) - 75 - 5$.

This trick makes the expression inside the parentheses a perfect square to get the final answer:

$y = 5 {\left(x - 1\right)}^{2} - 80$.

The graph of this function is a parabola opening upward with a minimum at the vertex $\left(x , y\right) = \left(1 , - 80\right)$.