SOLUTION: Graph the following polynomials functions and label the zeros and y-intercepts: a) f(x)=x^2(x-1)(x-5) b) f(x)=(x-3)(x^2-36)(x+2) Please I need to solve this problem!

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Graph the following polynomials functions and label the zeros and y-intercepts: a) f(x)=x^2(x-1)(x-5) b) f(x)=(x-3)(x^2-36)(x+2) Please I need to solve this problem!      Log On


   

Question 818708: Graph the following polynomials functions and label the zeros and y-intercepts:
a) f(x)=x^2(x-1)(x-5)
b) f(x)=(x-3)(x^2-36)(x+2)
Please I need to solve this problem!
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
I'll do problem "b":
f%28x%29=%28x-3%29%28x%5E2-36%29%28x%2B2%29
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%28x%29=%28x-3%29%28x%2B6%29%28x-6%29%28x%2B2%29

For the x-intercepts we make the y equal to zero and solve for x:
0=%28x-3%29%28x%2B6%29%28x-6%29%28x%2B2%29
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)

For the y-intercept, make the x zero:
f%28x%29=%280-3%29%280%2B6%29%280-6%29%280%2B2%29
which simplifies ...
f%28x%29=%28-3%29%286%29%28-6%29%282%29
f%28x%29=216
So the y-intercept is (0, 216)

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%28600%2C+600%2C+-8%2C+8%2C+-300%2C+300%2C+%28x-3%29%28x%5E2-36%29%28x%2B2%29%29
The problem in part a is already fully factored. So it should be easier than part b.