# What is the vertex form of y=4x^2-17x-16?

Feb 1, 2016

$y = 4 {\left(x - \frac{17}{8}\right)}^{2} - \frac{545}{16}$

#### Explanation:

We start with $4 {x}^{2} - 17 x - 16 = y$
$4 {x}^{2} - 17 x - 16$ cannot be factored, so we will have to complete the square. To do that, we first have to make the coefficient of ${x}^{2}$ $1$. That makes the equation now $4 \left({x}^{2} - \frac{17}{4} x - 4\right)$.

The way completing the square works is, because ${x}^{2} - \frac{17}{4} x$ isn't factorable, we find a value that makes it factorable. We do that by taking the middle value, $- \frac{17}{4} x$, dividing it by two and then squaring the answer. In this case it would look this: $\frac{- \frac{17}{4}}{2}$, which equals $- \frac{17}{8}$. If we square it, that becomes $\frac{289}{64}$.

We can rewrite the equation as $4 \left({x}^{2} - \frac{17}{4} x + \frac{289}{64} - 4\right)$, but we can't just stick a number into an equation and not add it on both sides. We could add $\frac{289}{64}$ to both sides, but I would prefer to just add $\frac{289}{64}$ and then immedietly subtract it.

Now, we can rewrite this equation as $4 \left({x}^{2} - \frac{17}{4} x + \frac{289}{64} - \frac{289}{64} - 4\right)$. Because ${x}^{2} - \frac{17}{4} x + \frac{289}{64}$ is factorable, I can rewrite it as ${\left(x - \frac{17}{8}\right)}^{2}$. Putting it together we have $4 {\left(x - \frac{17}{8}\right)}^{2} - \frac{289}{64} - 4$ or $4 {\left(x - \frac{17}{8}\right)}^{2} - \frac{545}{64}$. The last step is to multiply $- \frac{545}{64}$ by $4$.

The final form is $4 {\left(x - \frac{17}{8}\right)}^{2} - \frac{545}{16}$