# How do I find the vertex, axis of symmetry, y-intercept, x-intercept, domain and range of y=-2x^2-8x+3?

Sep 29, 2015

Vertex(-2,11). Axis of symmetry x= -2
Y intercept is 3, x intercepts are $- 2 + \frac{\sqrt{22}}{2}$ and $- 2 - \frac{\sqrt{22}}{2}$

Domain $\left\{x \in \mathbb{R}\right\}$, Range $\left\{y \in \mathbb{R} , y \le 11\right\}$

#### Explanation:

Complete the square for x as follows:

$y = - 2 \left({x}^{2} + 4 x\right) + 3$

$y = - 2 \left({x}^{2} + 4 x + 4 - 4\right) + 3$

$y = - 2 \left({x}^{2} + 4 x + 4\right) + 8 + 3$

$y = - 2 {\left(x + 2\right)}^{2} + 11$

This gives vertex (-2, 11) axis of symmetry x=-2,
For y intercept put x=0 in the given equation to get y=3. For x intercept put y= 0 to get $- 2 {x}^{2} - 8 x + 3 = 0$. Solve for x, using

quadratic formula x= $\frac{8 \pm \sqrt{64 + 24}}{-} 4$

$- 2 - \frac{\sqrt{22}}{2} , - 2 + \frac{\sqrt{22}}{2}$.

The parabola opens downwards because of negative coefficient of ${x}^{2}$, has vertex at (-2,11), hence its domain would all of $\mathbb{R}$ and range would be {y in RR, y<=11}