# How do you write f(x)=2x^2+4x-5 in vertex form?

Aug 2, 2016

$2 {\left(x + 1\right)}^{2} - 7$

#### Explanation:

We begin with $y = 2 {x}^{2} + 4 x - 5$. We need to rewrite this into vertex form. The easiest way to do that is to complete the square. This is a very useful way to factor, and I don't want to spend too much time explaining it right now, but I'll cover it as I go.

In order to begin completing the square we need to factor the equation until the leading coefficient (meaning $a$ from $a {x}^{2} + b x + c$) is $1$. Sometimes we get lucky and the leading coefficient is already $1$, but in this case we'll need to factor. So, we factor out the $2$, which leaves us with $2 \left({x}^{2} + 2 x - \frac{5}{2}\right)$. Now, I realize that fraction might look scary, but don't worry about it. Fractions are no big deal, and I advise you not to convert it into decimal form. I don't want to lose any value based on rounding.

So, our next step is to take the middle component ($b$, or in this case $2$) and divide it in half. That gives us $1$. We take that value and square it. ${1}^{2}$ is just $1$. We add that to our equation, BUT we need to subtract it. IF we add $1$ and then subtract $1$, we haven't changed the value of the equation, which is important. We could add $523 , 650 , 320$, as long as we also subtract it.

We have $2 \left({x}^{2} + 2 x + 1 - 1 - \frac{5}{2}\right)$. Now, please notice how lovely ${x}^{2} + 2 x + 1$ looks.T hat is a perfect square, and we can combine it to this: ${\left(x + 1\right)}^{2}$. If you don't belive me, just multiply $\left(x + 1\right) \cdot \left(x + 1\right)$. Now, we still need to deal with $- 1 + \frac{5}{2}$. Let's combine them and see what we have to work with.

If we put it all together, we'll have $2 \left({\left(x + 1\right)}^{2} - \frac{7}{2}\right)$. I want to clean this up a bit, so let's mutiply that $2$ out. $- \frac{7}{2} \cdot 2$ leaves us with just $- 7$. We are left with $2 {\left(x + 1\right)}^{2} - 7$, which is in vertex form . Nice work, we're done!