How do you graph #f(x)= 2/3(x-3)^2 - 6#?

1 Answer
Alan P.
Jul 22, 2015

Plot the vertex, the intercepts (and possibly a few other points) then connect the points with a smooth curve.

Explanation:

#f(x) = 2/3(x-3)^2 - 6# is in vertex form
so as one of our points we have the vertex #(3,-6)#

The #f(x)# (or #y#) intercept occurs where #x=0#
#color(white)("XXXX")##f(0) = 2/3(-3)^2 -6 =0#
giving us another point: #(0,0)#

The x-intercepts occur where #f(x)=0#
#color(white)("XXXX")##2/3(x-3)^2-6 = 0#

#color(white)("XXXX")##(x-3)^2 = 6*(3/2)#

#color(white)("XXXX")##x-3 = +-3#

#color(white)("XXXX")##x = 0 or x=6#
Unfortunately this only gives us 1 more point (we already had #(0,0)#)
#color(white)("XXXX")##(6,0)#

We could evaluate #f(x)# for a few more values of #x# (#x=+-6# and #x=+-9# would seem promising; but the 3 points we have will do for now:

graph{2/3(x-3)^2-6 [-4.39, 13.39, -6.404, 2.485]}

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