# What is the vertex form of y=-4x^2 -x-3?

Oct 17, 2017

$y = - 4 {\left(x + \frac{1}{8}\right)}^{2} - \frac{47}{16}$

#### Explanation:

Begin by grouping the terms involving $x$ together.

$y = \left(- 4 {x}^{2} - x\right) - 3$

Factor out $- 4$ from the $x$ terms.

$y = - 4 \left({x}^{2} + \frac{1}{4} x\right) - 3$

Complete the square. Using the formula ${\left(\frac{b}{2}\right)}^{2}$ we get ${\left(\frac{- \frac{1}{4}}{2}\right)}^{2} = {\left(- \frac{1}{8}\right)}^{2} = \frac{1}{64}$.

We now know that to complete the square by adding $\frac{1}{64}$ within the parentheses. Because we are adding $\frac{1}{64}$, we must also subtract the amount by which it changed the problem.

$y = - 4 \left({x}^{2} + \frac{1}{4} x + \frac{1}{6464} /\right) - 3 + \frac{1}{16}$

Since the $\frac{1}{16}$ is within the parentheses, it is multiplied by $- 4$, meaning overall, it changes the problem by $- \frac{1}{16}$. To undo this change, we add $\frac{1}{16}$ outside the parentheses.

Now that we completed the square, the terms involving $x$ can be factored like so:

$y = - 4 {\left(x + \frac{1}{8}\right)}^{2} - \frac{47}{16}$

The equation is now written in vertex form.