Question

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

Answer 1

`y = (x+4)^2-23`

Explanation:

Vertex form looks like this:

`y = a(x-h)^2+k`

Where `(h,k)` is the vertex of the parabola and `a` is the same as the `a` coefficient in standard form `y = ax^2+bx+c`. In this case, we can see that `a=1`. So, to get something of the form:

`y = (x-h)^2+k`

We need to complete the square.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

So we have

`y = x^2+8x-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/(4(1)) = 64/4 = 16`

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

`y = x^2+8x+16-7-16`

Notice that the first three terms are a perfect square.

`y = (x^2+2(4)x+4^2 )- 23`

`y = (x+4)^2-23`

This is the vertex form of our parabola.

Final Answer