# How do you find the vertex and the intercepts for y = -x^2 - 4x + 7?

##### 1 Answer
May 10, 2017

Vertex is at $\left(- 2 , 11\right)$, y intercept is at $\left(0 , 7\right)$, x-interceps are at $\left(- 5.32 , 0\right) \mathmr{and} \left(1.32 , 0\right)$

#### Explanation:

$y = - {x}^{2} - 4 x + 7 \mathmr{and} y = - \left({x}^{2} + 4 x\right) + 7 \mathmr{and} y = - \left({x}^{2} + 4 x + 4\right) + 4 + 7$ or
$y = - {\left(x + 2\right)}^{2} + 11$ comparing with standard equation $y = a {\left(x - h\right)}^{2} + k , \left(h , k\right)$ being vertex we find here vertex at h= -2 ,k=11) or (-2,11)

y-intercept can be found by putting $x = 0$ in the equation, $y = 7$
y intercept is at $\left(0 , 7\right)$

x-intercept can be found by putting $y = 0$ in the equation , $- {\left(x + 2\right)}^{2} + 11 = 0 \mathmr{and} {\left(x + 2\right)}^{2} = 11 \mathmr{and} \left(x + 2\right) = \pm \sqrt{11} \therefore x = - 2 + \sqrt{11} \approx 1.32 \left(2 \mathrm{dp}\right)$ or $x = - 2 - \sqrt{11} \approx - 5.32 \left(2 \mathrm{dp}\right)$
x-intercepts are at $\left(- 5.32 , 0\right) \mathmr{and} \left(1.32 , 0\right)$ [Ans]