# How do find the vertex and axis of symmetry, and intercepts for a quadratic equation y = -x^2-2x+6?

Jun 6, 2015

$y = - {x}^{2} - 2 x + 6$

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

$= - \left({\left(x + 1\right)}^{2} - 7\right)$

This has vertex where ${\left(x + 1\right)}^{2} = 0$, that is where $x = - 1$
and $y = - \left(- 7\right) = 7$. That is $\left(- 1 , 7\right)$

The axis of symmetry is the vertical line with equation $x = - 1$

The intercept with the $y$ axis is where $x = 0$ and $y = 6$, that is at $\left(0 , 6\right)$.

The intercepts with the $x$ axis are the roots of

$- \left({\left(x + 1\right)}^{2} - 7\right) = 0$

Hence ${\left(x + 1\right)}^{2} = 7$

So

$x + 1 = \pm \sqrt{7}$

$x = - 1 \pm \sqrt{7}$

So the intercepts are:

$\left(- 1 + \sqrt{7} , 0\right)$ and $\left(- 1 - \sqrt{7} , 0\right)$