# How do find the vertex and axis of symmetry for a quadratic equation y=-2x^2-8x+3?

Jun 15, 2015

See the explanation below.

#### Explanation:

General equation for the quadratic formula: $y = a {x}^{2} + b x + c$

The graph of a quadratic equation is a parabola.The axis of symmetry is the vertical line that separates the parabola into two equal halves.The point on the x-axis where the vertical axis is placed is determined by the formula $x = \frac{- b}{2 a}$

$y = a {x}^{2} + b x + c$

$y = - 2 {x}^{2} - 8 x + 3$

$a = - 2$ and $b = - 8$

$x = \frac{- \left(- 8\right)}{2 \left(- 2\right)} = \frac{8}{-} 4 = - 2$

$x = - 2$

Substitute $- 2$ for $x$ into the equation to find $y$.

$y = - 2 {\left(- 2\right)}^{2} - 8 \left(- 2\right) + 3$

$y = - 2 \left(4\right) + 16 + 3$

$y = - 8 + 16 + 3 = 11$

$y = 11$

The vertex of a parabola is the point where the parabola crosses its axis of symmetry. Vertex$=$$\left(x , y\right) = \left(- 2 , 11\right)$

graph{y=-2x^2-8x+3 [-16.42, 15.6, -3.2, 12.82]}