How do you find the vertex and the intercepts for #y = -8x^2 + 32x#?

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Answer 1 · 2017-06-01T06:36:01

Vertex is #(2,32)# and intercepts are at #(0,0)# and #(4,0)#.

Vertex form of quadratic function is #y=a(x-h)^2+k#, where #(h,k)# is the vertex

Hence #y=-8x^2+32x#

= #-8(x^2-4x)#

= #-8(x^2-4x+4)+32#

= #-8)x-2)^2+32#

Hence vertex is #(2,32)#

for intercepts on #x#-axis, put #y=0# i.e. #-8x^2+32x=0# or #-8x(x-4)=0# i.e. #x=0# or #x=4#

and for intercepts on #y#-axis, put #x=0# and we have #y=0#

Hence intercepts are at #(0,0)# and #(4,0)#
graph{-8x^2+32x [-10, 10, -50, 50]}