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

Jul 27, 2016

${\left(x + 2\right)}^{2} - 6$

#### Explanation:

First, find the coordinates of the vertex.
x-coordinate of vertex
$x = - \frac{b}{2 a} = - \frac{4}{2} = - 2$
y-coordinate of vertex
y(-2) = 4 - 8 - 2 = -6
Vertex (-2, -6)
Vertex form of y:
$y = {\left(x + 2\right)}^{2} - 6$

Jul 27, 2016

$y = {\left(x + 2\right)}^{2} - 6$

#### Explanation:

We begin with $y = {x}^{2} + 4 x - 2$. In order to find the vetex form of this equation we need to factor it. If you try it, $y = {x}^{2} + 4 x - 2$ is not dactorable, so now we can either complete the square or use the quadratic formula. I'm going to use the quadratic formula because it is fool-proof, but learning how to complete the square is valuable too.

The quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 \cdot a \cdot c}}{2 \cdot a}$, where $a , b , c$ come from $a {x}^{2} + b x + c$. In our case, $a = 1$, $b = 4$, and $c = - 2$.

That gives us $x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$, or $\frac{- 4 \pm \sqrt{16 - \left(- 8\right)}}{2}$, which simplifies further to $\frac{- 4 \pm \sqrt{24}}{2}$.

From here we expand $\sqrt{24}$ to $2 \sqrt{6}$, which makes the equation $\frac{- 4 \pm 2 \sqrt{6}}{2}$, or $- 2 \pm \sqrt{6}$.

So we went from $x = \frac{- 4 \pm \sqrt{{4}^{2} - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$ to $x = - 2 \pm \sqrt{6}$. Now we add $2$ on both sides, leaving us with $\pm \sqrt{6} = x + 2$. From here, we need to get rid of the square root, so we'll square both sides, which will give us $6 = {\left(x + 2\right)}^{2}$. Subtarct $6$, and have $0 = {\left(x + 2\right)}^{2} - 6$. Since we're looking for the eqaution when $y = 0$ (the $x$-axis), we can use $0$ and $y$ interchanagbly.

Thus, $0 = {\left(x + 2\right)}^{2} - 6$ is the same thing as $y = {\left(x + 2\right)}^{2} - 6$. Nice work, we ow have the equation in Vertex form!