# How do you find the axis of symmetry and vertex point of the function:  y=x^2+6x-7?

Oct 23, 2015

First you change this equation to complete square form;
y = ${x}^{2}$ +6x -7

$\left({x}^{2} + 2 \cdot 3 x + {3}^{2} - 7 - {3}^{2}\right)$

$\left[{\left({x}^{2} + 3\right)}^{2} - \left(7 + 9\right)\right]$

${\left({x}^{2} + 3\right)}^{2} - 16$ is in the form of $a {\left(x - h\right)}^{2} + k$

where $a = 1$ , $h = - 3$ and $k = - 16$

The vertex is given by $\left(h , k\right)$

So the vertex is ( -3,-16

and the line of symmetry is given by $\left(x - h\right) = 0$
So

x+3 = 0 is the axis of symmetry