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)#