Suppose that $f(x)$ has $(x-2)^2$ and $(x+1)$ as its only factors. How do I sketch the graph of $f$?
So far what I've done is determine (hopefully correctly) that the x-intercepts will be at $-1$ and $+2$. But what about the $y-$intercept?? If y-intercept is 4, is my sketch below correct (excuse the poor drawing)?
EDIT: Made a potentially incorrect assumption that $f(x) = (x-2)^2 (x+1)$, ignoring the possibility that $f(x) = c (x-2)^2 (x+1)$ for some $c \neq 0$. Hopefully you can see how a different value of $c$ would change things.
The y-intercept can be a found in a few ways.
First way: We know that the y-intercept is when the value of $x=0$, so just set $x = 0$ and compute! $f(0) = (-2)^2 \times 1 = 4$, so the y-intercept is $(0, 4)$.
Second way: We can multiply out $f(x)$, to get that $f(x) = (x^2 - 4x + 4)(x+1)$, or that $f(x) = x^3 - 3x^2 + 4$. In general, the constant term is going to be the y-intercept (this makes sense, given that the constant term is all that's left when you plug in $x=0$). Hence, the y-intercept is $(0, 4)$.
Note that these two ways are actually the same thing - just a question of how you want to think about it.
As a side note, when you're sketching this, note that there is a double root at $x = 2$. This means that the graph will "bounce" off of the x-axis at this point. Also, basic differentiation will help give you a better idea of the curvature of the graph at given points. This is, of course, just a question of how detailed your sketch needs to be. I can elaborate on this if you would like.
Hope this helps. Cheers!