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As my question says, how do I find intersection with $x$ or $y$ axis.

For example, if given function is $f(x)=x^3+x^2-x-1$, how do I find the intersection with $x$ and $y$ axis.

Right now, I only know that when we are searching for intersection with $x$, we take $y=0$ and when we search for $y$, we take $x=0$.

However, if the intersection with $y$-axis, $x=0$, then we get, $f(x)=0+0-0-1$, therefore it will be, $(0,-1)$, but when we search for $x$-axis, $y=0$, what do change $0$ for? There is no $y$ in the function.

Kostiantyn Lapchevskyi
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user2041143
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Remember that $y=f(x)$, therefore saying $y=0$ is saying $f(x)=0$. So you just need to find $x$ such that $x^3+x^2-x-1=0$. To do this first inspect for pretty numbers like small (in absolute value) integers.

Below is the graph for your function (though that's probably what you're trying to find, I'm using it just as an illustrative example, I could have chosen a different function).

When does the graph intersect the $y$-axis? Exactly when $x=0$. And when does it intersect the $x$-axis? Exactly when $f(x)=0$.

Also take notice of Brian M. Scott's comment below, which is helpful.

Graph

By looking at the graph you can find that the points in which the graph intersects the $x$-axis are $(-1,0)$ and $(1,0)$ (note that the $y$ coordinate just had to be $0$, otherwise it wouldn't intersect the $x$-axis. Also looking at the graph you see that the point which intersects the $y$-axis is $(0,-1)$ (again, the $x$ component had to be $0$, otherwise it wouldn't intersect the $y$ axis).

Now let's confirm this fact algebraically:

  1. Find the intersection with the $x$-axis. Set $y=0$, that is $f(x)=0$. So you're trying to find $x$ such that $x^3+x^2-x-1=0$, which is equivalent to $(x-1)(x+1)^2=0$. So you get $x=1$ and $x=-1$ as roots of the equation. Since $y=0$, you get the intersecting points $(-1,0)$ and $(1,0)$, as expected.
  2. Find the intersection with the $y$-axis. Set $x=0$. You get $f(0)=y$ which is equivalent to $-1=y$. Therefore the point in which the graph of the function intersects the $y$-axis is $(0, -1)$.
Git Gud
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    Or do the easy factorization: $$x^3+x^2-x-1=(x+1)(x^2-1)=(x-1)(x+1)^2;.$$ Though in this case the rational root test does find them all. – Brian M. Scott Mar 03 '13 at 12:05
  • @BrianM.Scott I think I'm going to delete my answers as I think other answers are more helpful to the OP. Maybe you should post your comment somewhere else so it doesn't get lost. – Git Gud Mar 03 '13 at 12:09
  • Your comment is very useful to be honest. I have one question though, do we have to take our own values for x to find the intersection with x axis? Because y=0, therefore we have to equal the function to 0, and then we add either our own values or follow Brian M.Scott's example for easier way? – user2041143 Mar 03 '13 at 12:10
  • Why not leave the answer and just add my comment to it? I don’t mind. – Brian M. Scott Mar 03 '13 at 12:11
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    @Student: You have to solve the equation $f(x)=0$ in whatever way you can. In this problem factorization works, as does the rational root test, and even trial and error, since the roots are very small integers. In other problems you might only be able to get a numerical approximation, by using some numerical technique. – Brian M. Scott Mar 03 '13 at 12:13
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    @Student There's no recipe in general to find the intersection of the graph of the function (which is what you're doing) with the $x$-axis. It's a case by case thing.When it's a polynomial of "low" degree there are techniques you can apply, like the one Brian mentioned: the rational root test. That's basically what I did. – Git Gud Mar 03 '13 at 12:14
  • Do I have to solve f(x)=0 in both cases when I find intersection for x and y axis? – user2041143 Mar 03 '13 at 12:15
  • @BrianM.Scott I'll just leave it like this, then. Your observation is quite visible. Thanks. – Git Gud Mar 03 '13 at 12:15
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    @Student No! You solve $f(x)=0$ only when you're looking for the intersections with the $x$-axis. If you're looking for the intersection with the $y$-axis, you just let $x=0$, which is what you did. This is easy to understand with a graph, but I'm very bad with such things both here on the computer and in real life. – Git Gud Mar 03 '13 at 12:17
  • That pretty much answered my question. Now I understand! – user2041143 Mar 03 '13 at 12:25
  • @Student Glad I could help. I added some stuff to my answer, though. Hopefully it will make things clearer. – Git Gud Mar 03 '13 at 12:26
  • Ok, from my understanding of the graph, I can only understand that one intersection is when it's (-1,0) for x-axis, where does the y intersect? Edit: The y-intersection is in (1,-1)? – user2041143 Mar 03 '13 at 12:32
  • @user2041143 First off let me say that a graph is just visual aid. Do not use graphs to "prove" things in mathematics. You gotta do it algebraically. Now about your comment, let me answer with a question: at which points does the graph of the function intersect the $x$-axis? – Git Gud Mar 03 '13 at 12:36
  • Yes, my question is at which point does the graph of the function intersects the x-axis and y-axis? – user2041143 Mar 03 '13 at 12:59
  • @user2041143 I added some stuff to the answer. Does that make things clearer? – Git Gud Mar 03 '13 at 13:17
  • Yes, it does. Thank you! – user2041143 Mar 03 '13 at 13:20
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For the Intersection with the $x$-axis you need to find the zeroes of the function, that means finding the $x$ which fulfulls $$0=x^3+x^2-x-1$$ Here x=1 should be one zero, and with polynomial division you should get the rest.

Dominic Michaelis
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You're confused because you were being sloppy (or, because the question was stated sloppily). The question isn't asking about the intersection of a function with the x and y axes; it is asking about the intersection of the graph of that function with the x and y axes.

i.e. the question means to ask "what is the intersection of the curve defined by $y = f(x)$ with the x and y axes".

  • No.. I do understand the question, the problem is that I had to translate it from my native language to English, therefore I might have written it in a wrong way, because I'm not very good in English. – user2041143 Mar 03 '13 at 12:08