Question 134056
{{{(x+3) (x-2)^2=0}}} doesn't actually have a graph other than the three values (-3, 2, and 2) on the number line that satisfy the equation.  You can, however, graph {{{f(x)=(x+3)(x-2)^2}}}, and I suspect that is what you really meant.  (I sincerely hope your instructor didn't pose the question that way.  If s/he did, go get your money back)


Zeros, or x-intercepts at -3, 2, and 2  (we should have an extrema at 2)


Expand:
{{{f(x)=x^3-x^2-8x+12}}}


Extreme points where f'(x) = 0, so:


f'(x) = {{{3x^2-2x-8}}}.  Set the first derivative equal to zero:


{{{3x^2-2x-8=0}}}
{{{(3x+4)(x-2)=0}}}, hence critical points are at ({{{-4/3}}},{{{f(-4/3)}}}) and ({{{2}}},{{{f(2)}}}).


We know that {{{f(2)=0}}}, so we have a critical point at (2,0).


{{{f(-4/3)=((-4/3)+3)((-4/3)-2)^2}}}
{{{f(-4/3)=(5/3)(-10/3)^2}}}
{{{f(-4/3)=(5/3)(100/9)}}}
{{{f(-4/3)=(500/27)}}}


And we have an extreme point at {{{-4/3}}},{{{500/27}}})


f"(x)={{{6x-2}}}
f"(2)={{{8-2=6>0}}} => (2,f(2)) is a local minimum
f"(-4/3)={{{6(-4/3)-2=-8-2=-10<0}}} => ({{{-4/3}}},{{{f(-4/3)}}}) is a local maximum


The y-intercept is found by evaluating f(0):
{{{f(0)=0^3-0^2-80+12=12}}} hence the y-intercept is (0,12)



Using this information, we can make a rough sketch of the graph:


{{{drawing(600,600,-20,20,-20,20,
grid(1),
graph(600,600,-20,20,-20,20,x^3-x^2-8x+12),
circle(2,0,.3),
circle(-4/3,500/27,.3),
circle(-3,0,.3),
circle(0,12,.3)
)}}}