How do I graph this? f(x) = (x-a)(x+b)^2 for 0<a<b<c

[WTF?]

I'm given the equation \(\displaystyle f(x)=(x-a)(x+b)^2\)

Assuming \(\displaystyle 0<a<b<c\)

Since the 0s in the equation are \(\displaystyle (a,0)\) and \(\displaystyle (-b,0)\) (with multiplicity of 2, so the line touches the point of \(\displaystyle (-b,0)\). Anyway, how can I assume the vertical intercept? At what value of y would there be an intercept in this confusing plane?!

I'm assuming that it would be \(\displaystyle -a*(b^2)\) but that makes no sense...

Other equations making my life miserable include:

\(\displaystyle f(x)=(x^2-a^2)(x^2-b^2)\)
\(\displaystyle f(x)=(x-a)(x+b)^2(x-c)^3\)

Any direction would be appreciated.

 

[WTF?]

Re: How do I graph this?!

Ohhhh neverind. It doesn't really matter where they intersect the y-axis, as long as the graph shows that its even, or odd, or negative, etc....
Hmmm...

 

[stapel]

I'm sorry, but I don't understand what you're supposed to be doing with the listed function...? What were the instructions for this exercise?

 

[WTF?]

Re:

I'm sorry, but I don't understand what you're supposed to be doing with the listed function...? What were the instructions for this exercise?

I'm suppose to graph it on a coordinate plane where 0<a<b<c

Anyway, I think i figured this sucker out.

HOWEVER, i have a new problem that seems even more confusing...

\(\displaystyle f(x)=(x^2-a^2)/(x+b)^2(x-c)\)

So So I factor out the top and bottom so that i can figure out the x-intercepts and vertical asymptotes, \(\displaystyle (x-a)(x+a)/(x+b)(x+b)(x-c)\)


So the x-intercepts are (a,0) and (-a, 0)

and the vertical asymptotes are x=-b and x=c

But that's as far as I get, do you suppose I could substitute real numbers for a, b, and c? say 1, 2, and 3 respectively? Otherwise it just seems kinda weird..