# Question #f1c3c

Apr 4, 2017

Vertex is at $\left(- 4 , - 21\right)$, y-intercept is at $\left(0 , - 5\right)$

#### Explanation:

$y = {x}^{2} + 8 x - 5 \mathmr{and} y = {x}^{2} + 8 x + 16 - 16 - 5 \mathmr{and} {\left(x + 4\right)}^{2} - 21$
Comparing with standard equation $y = a {\left(x - h\right)}^{2} + k$
we get vertex as $h = - 4 , k = - 21 \mathmr{and} \left(- 4 , - 21\right)$

Putting $x = 0$ in the equation we get y intercept as $y = - 5$
Vertex is at $\left(- 4 , - 21\right)$, y-intercept is at $\left(0 , - 5\right)$ graph{x^2+8x-5 [-80, 80, -40, 40]}[Ans]