What is the vertex form of #y= 4t^2-12t+8 #?

1 Answer
Aug 18, 2017

#y = 4(t-3/2)^2 -1#

Explanation:

Vertex form is given as #y = a(x+b)^2+c#,

where the vertex is at #(-b,c)#

Use the process of completing the square.

#y = 4t^2 -12t +8#

#y = 4(t^2 -color(blue)(3)t +2)" "larr# take out the factor of #4#

#y = 4(t^2 -3t color(blue)( +(3/2)^2 -(3/2)^2) +2)#

#[color(blue)(+(3/2)^2 -(3/2)^2=0)]" "larr +(b/2)^2 -(b/2)^2#

#y = 4(color(red)(t^2 -3t +(3/2)^2) color(forestgreen)( -(3/2)^2 +2))#

#y = 4(color(red)((t-3/2)^2) color(forestgreen)( -9/4 +2))#

#y = 4(color(red)((t-3/2)^2) color(forestgreen)( -1/4))#

Now distribute the #4# into the bracket.

#y = color(red)(4(t-3/2)^2) +color(forestgreen)(4( -1/4))#

#y = 4(t-3/2)^2 -1#