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

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Answer 1 · 2015-11-21T02:45:28

This equation is in vertex form ...

Vertex Form
$y=a(x-h)^2+k$

Where, the vertex $=(h,k)$

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

y-intercept $=-2.5(0-4)^2-5=-45$

x-intercepts: solve for x

$0=-2.5(x-4)^2-5$

$(x-4)^2=5/(-2.5)=-2$

$(x-4)=+-sqrt-2$

$x=4+-isqrt2$ (no real solutions)

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

hope that helped

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