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 $(-1/2 , 3 1/2)$

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