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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