I need to find the function of this graph in factored form (if possible with solution)

Question

Copy and paste the lik above for pic.

I already know that the function is (?)/(x+2)(x-1) Horizontal Asymptote is y=0 i don't know how to get both intercepts in this since the possible y intercept is the hole (0,2)

Answer 1

There is a vertical asymptote at $x = -2$, which means that there is a factor of $x+2$ in the denominator. In addition, along this asymptote, the graph on the left and right sides go the same direction, so it should have an even power.

There is a vertical asymptote at $x = 1$, which means there is a factor of $x-1$. Here the graph goes in both directions, so the power is odd. It is also narrower than the $-2$ asymptote, so the power is probably lower.

Finally, there is a zero at $x = 2$, which means $x-2$ is a factor of the denominator.

Putting this together, the simplest function we can make is $$ \frac{x-2}{(x+2)^2(x-1)} $$ which does indeed look a lot like your function. However, we need to do something about the undefined point at $x = 0$. That's remedied by multiplying numerator and denominator by $x$, like so: $$ \frac{x(x-2)}{x(x+2)^2(x-1)} $$

---

Edit: The odd and even power.

For a term like $x-1$, if we're just a little to the left of $1$, then it's negative, while if we're a little to the right, it's positive. The same goes for $1/(x-1)$, although instead of having values close to $0$, they're blown up toward their respective infinities.

In other words, a linear factor, either in the numerator or the denominator, changes sign as you pass through where it's $0$, and as such, changes the sign of the entire function.

However, if that factor is squared instead, then it doesn't change sign. For instance, $(x+2)^2$ (or, more relevant to our question, $1/(x+2)^2$) is positive both to the left and to the right of $x = -2$. There are several ways to explain this, the most common is probably to say that squares are never negative. I'm partial to explaining it as the sign changing twice as you pass $x = -2$. What I mean by that is $(x+2)^2 = (x+2)(x+2)$, and as you pass through $x = -2$, both of those factors change sign (as per the paragraph above), so we get a double sign change, which is the same as no sign change.

From this we see that a cubed factor changes sign, while a factor to the fourth power doesn't, and so on.