# How do you write f(x)= -2x^2+20x-49 in vertex form?

Feb 28, 2017

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

#### Explanation:

Standard Form equation: $y = a {x}^{2} + b x + c$
Vertex Form equation: $y = a {\left(x - h\right)}^{2} + k$

Where $a$ is equal to the $a$ value of the standard form equation and $\left(h , k\right)$ is equal to the vertex of the equation.

In order to convert it, let's first fill in what we know. The $a$ value of the given equation is -2.

So, we have:

$y =$ -2 $\left(x - {h}^{2}\right)$+$k$

In order to find the vertex, you must use the equation $- \frac{b}{2 a}$

Looking at the standard form equation, $b = 20$ and $a = - 2$
So plugging in, you get $- \frac{20}{2 \cdot - 2}$

Once solved, you're x value of your vertex is 5. Now, you plug 5 into your original standard form equation.

$y = - 2 {\left(5\right)}^{2} + 20 \left(5\right) - 49$

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