# What is the vertex form of y= x^2 + 4x - 1 ?

Jan 17, 2016

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

#### Explanation:

The way I got this answer is by completing the square. The first step though, when looking at this equation, is to see if we can factor it. The way to check is to look at the coefficient for ${x}^{2}$, which is 1, and the constant, in this case -1. If we multiply those together, we get $- 1 {x}^{2}$ . Now we look at the middle term, $4 x$. We need to find any numbers that multiply to equal $- 1 {x}^{2}$ and add to $4 x$. There aren't any, which means it is not factorable.

After we have checked its factorability, lets to try to complete the square for ${x}^{2} + 4 x - 1$. The way completing the square works is by finding the numbers that will make the equation factorable and then rewriting the equation to fit them in.

The first step is to set $y$ equal to zero.

After that, we need to get the Xs by themselves, so we add 1 on both sides, like so:
$0 = {x}^{2} + 4 x - 1$
$\textcolor{red}{+ 1}$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .}$$\textcolor{red}{+ 1}$

Now the equation is $1 = {x}^{2} + 4 x$. We need to find a value that will make ${x}^{2} + 4 x$ factorable. I do this by taking $4 x$ and dividing $4$ by $2$. This equals $2$, which I would then square to equal $4$. This is a trick, taking the middle value, dividing it by two, then squaring the answer, which works for any quadratic as long as the coefficient of the ${x}^{2}$ is 1, as it is here. Now, if we rewrite the equation it looks like this:
$1 = {x}^{2} + 4 x$
$\textcolor{red}{+ 4}$$\textcolor{w h i t e}{\ldots \ldots \ldots \ldots . .} \textcolor{red}{+ 4}$

note we have to add 4 to both sides to keep the equation equal.

Now the equation is $5 = {x}^{2} + 4 x + 4$, which can be rewritten as
$5 = {\left(x + 2\right)}^{2}$. We can check this by expanding ${\left(x + 2\right)}^{2}$ to $\left(x + 2\right) \cdot \left(x + 2\right)$, which is ${x}^{2} + 2 x + 2 x + 4$, and can be simplified to ${x}^{2} + 4 x + 4$.

Now all that is left is to subtract 5 on both sides and set the equation equal to $y$ again.

So ${x}^{2} + 4 x - 1$ is ${\left(x + 2\right)}^{2} - 5$, which can be double checked by graphing ${x}^{2} + 4 x - 1$ and finding the vertex or lowest point. The coordinate pair is (-2,-5). It might seem wrong that the 2 in ${\left(x + 2\right)}^{2}$ is positive while the vertex has 2 as a negative, but the format for vertex form is a(x – h)^2 + k. Its ${\left(x - \left(- 2\right)\right)}^{2}$ which becomes ${\left(x - + 2\right)}^{2}$ when simplified.

Hope this helped!