How do you find the vertex and the intercepts for #y=-x^2+8x+16#?

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Answer 1 · 2016-04-26T06:01:35

Vertex#(4,32)#
Y-Intercept #(0, 16)#

Given -

#y=-x^2+8x+16#

Vertex

#x=(-b)/(2a)=(-8)/(2 xx (-1))=(-8)/(-2)=4#
At #x=4 ; y=-(4^2)+8(4)+16#
#y=-16+32+16=32#

Vertex#(4,32)#

Y-Intercept

At #x=0; y=-(0^2)+8(0)+16=16#

Y-Intercept #(0, 16)#