# How do find the vertex and axis of symmetry for a quadratic equation y=-x^2+2x-5?

Jun 25, 2015

vertex at $\left(1 , - 4\right)$
axis of symmetry: $x = 1$

#### Explanation:

By rewriting $y = - {x}^{2} + 2 x - 5$ into vertex form:
$\textcolor{w h i t e}{\text{XXXX}}$$y = m {\left(x - a\right)}^{2} + b$ (which has a vertex at $\left(a , b\right)$)
we can easily extract the vertex.

$y = - {x}^{2} + 2 x - 5$
$\textcolor{w h i t e}{\text{XXXX}}$extract $m$
$y = \left(- 1\right) \left({x}^{2} - 2 x\right) - 5$
$\textcolor{w h i t e}{\text{XXXX}}$complete the square
$y = \left(- 1\right) \left({x}^{2} - 2 x + 1\right) - 5 + 1$
$\textcolor{w h i t e}{\text{XXXX}}$simplify into vertex form
$y = \left(- 1\right) {\left(x - 1\right)}^{2} + \left(- 4\right)$

with a vertex at $\left(1 , - 4\right)$

Since this is a parabola in standard position (vertical axis)
the axis of symmetry runs though the x coordinate of the vertex.