What is the vertex form of #y= (3x+9)(x-2) #?

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Answer 1 · 2017-03-12T03:37:48

#y=3(x+0.5)^2 -18.75#

First let's expand the equation:

#(3x+9)(x−2)# #=# #3x^2 -6x+9x-18#

which simplifies to:
#3x^2 +3x-18#

Let's find our vertex using #x=-b/(2a)# where a and b are of #ax^2 +bx+c#

We find the x value of our vertex to be #-0.5#
(#-3/(2(3))#)

Plug it into our equation and find y to be #-18.75#
#3(-0.5)^2+3(-0.5)-18#

so our vertex is at #(-0.5, -18.75)#

We can also check this with a graph:
graph{(3x^2+3x-18) [-10.3, 15.15, -22.4, -9.68]}

Now that we have our vertex, we can plug it into the vertex form!

#f(x)=a(x-h)^2+k#

where #h# is our x value of the vertex, and #k# is the y value of the vertex.

so #h=-0.5# and #k=-18.75#

In the end we find:

#y=3(x+0.5)^2 -18.75#