What is the vertex form of #y=-8x^2 +8x+32 #?

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Answer 1 · 2016-06-03T08:16:10

#y = -8[(x + (1x)/2)^2 + 3 1/2]#

This gives the vertex as #(-1/2 , 3 1/2)#

Vertex form is #y = a(x b)^2 + c# This is obtained by the process of completing the square.

Step 1. Divide the coefficient of #x^2# out as a common factor.

#y = -8[x^2 + x + 4]#

Step 2: Add in the missing square number to create the square of a binomial. Subtract it as well to keep the value of the right hand side the same.

#y = -8[x^2 + x + color(red) ((1/2))^2+ 4 -color(red) ((1/2))^2 ]#

Step 3: Write the first 3 terms in the bracket as #("binomial" )^2#

#y = -8[(x + (1x)/2)^2 + 3 1/2]#

This gives the vertex as #(-1/2 , 3 1/2)#