What is the vertex form of $y= 2x^2 - 9x – 18$ ?

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Answer 1 · 2017-07-05T13:00:48

$y= 2(x-9/4x)^2 -28 1/8$

$a(x+b)^2 +c$

This is vertex form, giving the vertex as $(-b, c)$ which is:

$(2 1/4, -28 1/8)$

Write it in the form $a(x+b)^2 +c$

$y= 2[x^2color(blue)(-9/2)x -9]" "larr$ factor out $2$ to to get $1x^2$

Complete the square by adding and subtracting $color(blue)((b/2)^2)$

$color(blue)(((-9/2)div2)^2 = (-9/4)^2 = 81/16)$

$y= 2[x^2color(blue)(-9/2)x color(blue)( + 81/16-81/16)-9]$

Group to create a perfect square.

$y= 2[color(red)((x^2-9/2x + 81/16)) +(-81/16-9)]$

$y= 2[color(red)((x-9/4x)^2) +(-5 1/16-9)]" "larr$ distribute the $2$

$y= 2(x-9/4x)^2 +2(-14 1/16)$

$y= 2(x-9/4x)^2 -28 1/8$

This is now vertex form, giving the vertex at $(2 1/4, -28 1/8)$

What is the vertex form of y= 2x^2 - 9x – 18 y= 2x^2 - 9x – 18 ?