# 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..

• Why do you think a function has to have an x-intercept? Consider the graph of $f(x)=x^2+1$ Dec 14, 2015 at 1:19

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!

• Ok so now I am going about trying to find the open intervals on which f is increasing or decreasing. I went about this by finding the derivative which I got to be -16x/(x^2-9)^2. I then set the derivative equal to 0 and got x^4-18x^2+20x+81=0. What should I do after this and am I on the right track?
– Lil
Dec 14, 2015 at 1:30
• Yes you're on the right track, however the derivative should be $f'(x)= \frac {-20x}{(x^2-9)^2}$. Then use the fact that $f$ is increasing when the derivative is positive and decreasing when the derivative is negative. Keep in mind also that $f$ is not defined at $x=3$ and $x=-3$. The denominator of $f'$ is always positive and the numerator will be positive for $x<0$ and negative for $x>0$ so those will be your intervals of increase/decrease! More specifically you can say $f$ is increasing on $(- \infty,-3) \cup (-3,0)$ and decreasing on $(0,3) \cup (3, \infty)$
– user275377
Dec 14, 2015 at 1:44
• since f is undefined at x=3 and x=-3 are those the vertical asymptotes?
– Lil
Dec 14, 2015 at 2:12
• yes that's correct, so those two points cannot be in the domain of $f$
– user275377
Dec 14, 2015 at 3:26
• alright. Now how do i find out where f is concave up or concave down? I take the second derivative then what?
– Lil
Dec 14, 2015 at 3:28

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.

• Ok so now I am going about trying to find the open intervals on which f is increasing or decreasing. I went about this by finding the derivative which I got to be -16x/(x^2-9)^2. I then set the derivative equal to 0 and got x^4-18x^2+20x+81=0. What should I do after this and am I on the right track?
– Lil
Dec 14, 2015 at 1:37

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.

• Ok so now I am going about trying to find the open intervals on which f is increasing or decreasing. I went about this by finding the derivative which I got to be -16x/(x^2-9)^2. I then set the derivative equal to 0 and got x^4-18x^2+20x+81=0. What should I do after this and am I on the right track?
– Lil
Dec 14, 2015 at 1:30

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.

• Ok so now I am going about trying to find the open intervals on which f is increasing or decreasing. I went about this by finding the derivative which I got to be -16x/(x^2-9)^2. I then set the derivative equal to 0 and got x^4-18x^2+20x+81=0. What should I do after this and am I on the right track?
– Lil
Dec 14, 2015 at 1:37
• ok. how does solving x^2-9=0 and getting x=+/-3 tell me the intervals where the graph is increasing or decreasing or relative extrema?
– Lil
Dec 14, 2015 at 1:48
• Now, you have shown that the critical values of your function are at $x=-3,0, 3$. The critical values are where the function can change from increasing to decreasing or vice versa. To show that a function is increasing, we must show that $\frac {df}{dx}>0$. Likewise, a function is decreasing over an interval when $\frac {df}{dx}<0$. For example, $f'(-4)=\frac {17}{7}$. Because this value is greater than zero, you can conclude that the interval $(-\infty,-3)$ is increasing. Dec 14, 2015 at 1:50
• Lil, you asked about the concavity of your function, and since I don't have enough rep. to comment, I'll just put it here. You are correct in determining the concavity via the second derivative, which should be $$\frac {d^2f}{dx^2}= \frac {60(x^2+3)}{(x^2-9)^3}$$. Now you must determine where this is zero or undefined. Using more values will show where the function is concave up or down. Dec 14, 2015 at 13:38