# How do you write the Vertex form equation of the parabola y=x^2 + 8x - 7?

Jul 12, 2017

$y = {\left(x + 4\right)}^{2} - 23$

#### Explanation:

Vertex form looks like this:

$y = a {\left(x - h\right)}^{2} + k$

Where $\left(h , k\right)$ is the vertex of the parabola and $a$ is the same as the $a$ coefficient in standard form $y = a {x}^{2} + b x + c$. In this case, we can see that $a = 1$. So, to get something of the form:

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

We need to complete the square.

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So we have

$y = {x}^{2} + 8 x - 7$

Remember that to find $c$ in order to complete the square, we use this trick:

c = b^2/(4a

We can derive this from the quadratic formula, but that's a problem for another time. Anyway, in this case we have $b = 8$ and $a = 1$, so

$c = {8}^{2} / \left(4 \left(1\right)\right) = \frac{64}{4} = 16$

So what we need to do is add and subtract $16$ from the right side of the equation.

$y = {x}^{2} + 8 x + 16 - 7 - 16$

Notice that the first three terms are a perfect square.

$y = \left({x}^{2} + 2 \left(4\right) x + {4}^{2}\right) - 23$

$y = {\left(x + 4\right)}^{2} - 23$

This is the vertex form of our parabola.