# How do you find the vertex and intercepts for y = -2.5(x - 4)^2 - 5?

Nov 21, 2015

This equation is in vertex form ...

#### Explanation:

Vertex Form
$y = a {\left(x - h\right)}^{2} + k$

Where, the vertex $= \left(h , k\right)$

For this problem, vertex $= \left(4 , - 5\right)$

y-intercept $= - 2.5 {\left(0 - 4\right)}^{2} - 5 = - 45$

x-intercepts: solve for x

$0 = - 2.5 {\left(x - 4\right)}^{2} - 5$

${\left(x - 4\right)}^{2} = \frac{5}{- 2.5} = - 2$

$\left(x - 4\right) = \pm \sqrt{-} 2$

$x = 4 \pm i \sqrt{2}$ (no real solutions)

Since the solution for the x-intercepts are imaginary numbers , there are no x-intercepts !

hope that helped