Question 818708
I'll do problem "b":
{{{f(x)=(x-3)(x^2-36)(x+2)}}}
This will be easier to do if we finish factoring. The middle factor is a difference of squares. So we can use that pattern to factor it:
{{{f(x)=(x-3)(x+6)(x-6)(x+2)}}}<br>
For the x-intercepts we make the y equal to zero and solve for x:
{{{0=(x-3)(x+6)(x-6)(x+2)}}}
Since we have a product equal to zero, one of the factors must be zero. Perhaps you can see what x values these would be. (If not, then set each factor equal to zero (x-3 = 0 or x+6 = 0 or ...) and solve.) They are:
x = 3 or x = -6 or x = 6 or x = -2
So the x-intercepts are: (3, 0), (-6, 0), (6, 0) and (-2, 0)<br>
For the y-intercept, make the x zero:
{{{f(x)=(0-3)(0+6)(0-6)(0+2)}}}
which simplifies ...
{{{f(x)=(-3)(6)(-6)(2)}}}
{{{f(x)=216}}}
So the y-intercept is (0, 216)<br>
The graph might be difficult to sketch with just these intercepts. You probably want to try to build a table of values using different x-vales so you have additional points. FWIW, here's a rough sketch (with different scales on the x and y axes so you can see the most interesting parts):
{{{graph(600, 600, -8, 8, -300, 300, (x-3)(x^2-36)(x+2))}}}
The problem in part a is already fully factored. So it should be easier than part b.