Find the domain, $x$ intercepts, $y$ intercepts Let $f(x)=\dfrac{x^2+1}{x^2-9}$
So to find the $y$ intercept I take $f(0)$ correct? So when I substitute $0$ for $x$ I got $-\frac{1}{9}$ so is the $y$ intercept $(0,-\frac{1}{9})$
Also to find the $x$ intercept I set the numerator equal to $0$. So then I got $x^2+1=0$ but wouldn't that make $x^2=-1$ which is imaginary? I'm a little confused what I am doing wrong..
 A: Yes your $y$-intercept is correct, and you are correct that the only solution is where $x^2=-1$ which would mean there are no real solutions, only imaginary solutions. So you would simply say that you're function has no $x$-intercept! 
A: Yes. This is correct. To find the $y$ intercept of a function, $f(x)$, find $f(0)$. 
Your approach for the $x$ intercept is also correct. Assuming that you are only trying to find intercepts in the real numbers, taking $x^2+1=0$ means that there are no $x$ intercepts. 
You calculated your derivative incorrectly. The derivative of $f(x)$ is given by: $$\frac{df}{dx}=\frac{(x^2-9)(2x)-(x^2+1)(2x)}{(x^2-9)^2}=\frac{-20x}{(x^2-9)^2}$$
Setting $\frac {df}{dx}=0$ yields $0=-20x$. However, you must also consider the points at which the derivative is undefined. To do this, solve $0=x^2-9$. Finally, consider where $\frac {df}{dx}$ is positive and negative using various values.
A: What you have done so far is exactly right. In some sense you could say it is a trick question.
If you plot the function you'll find there is no $x$-intercept. The fact that $x^2+1$ has no roots in $\mathbb{R}$ tells you exactly that.
A: You are correct about the $y$-intercept, though the terminology is a little ambiguous. It can mean the intersection point on the $y$-axis with the graph of the function, in which case it is $(0,-1/9)$ as you said. It can also mean just the $y$-coordinate of that point, which is $-1/9$.
Your reasoning about the $x$-intercepts is also basically correct. You have shown that there is no $x$-intercept. That happens for many functions, so don't worry about this function being an exception. A function may have no $x$-intercept, one, two, many, countable infinitely many, uncountable infinitely many.
A function can have only one or no $y$-intercepts, however, due to the definition of a function.
