Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have a graph image that the line touches on the x-axis the values $-4, -2, 0$, and $1$.

I believe that part of the answer is $f(x) = (x+4)(x+2)(x-1)$, but what is the forth zero?

Edit: On the graph image, the line doesn't actually go through the point -4, but hovers directly above it

share|cite|improve this question
up vote 0 down vote accepted

"Touching the $x$-axis" occurs when $f(x) = 0$.

How can you modify your function so that $f(0) = 0$?

share|cite|improve this answer
@B I tried adding a (x+0), am I getting closer? – Tyler Zika Oct 30 '12 at 6:01
What do you mean by "adding"? What's your new $f(x)$? And what happens when you try $f(x)$ for $x = -4, -2, 0$ and $1$? Do you get $0$? – Benjamin Dickman Oct 30 '12 at 6:02
@B The new function is f(x)=(x+4)(x+2)(x-1)(x-0). That doesn't work though, but I am hoping I'm going in the right direction.. – Tyler Zika Oct 30 '12 at 6:04
That new function will certainly give you all the $0$'s you described. Though more often one would use the notation $x$ instead of $(x-0)$, so that it would be: $f(x) = (x+4)(x+2)(x-1)x$. Can you check the value of your graph at any point besides $x = -4, -2, 0$ or $1$? If so, what is the point, what should the output be, and what output are you getting with your current $f(x)$? – Benjamin Dickman Oct 30 '12 at 6:09
@B Just realized something, on the graph, the line doesn't actually go through the point -4 on the x-axis, it almost hovers right over it. The line does go through 1, 0, and 2. Would that effect my answer? – Tyler Zika Oct 30 '12 at 6:15

Answer would be $f(x) = x(x-1)(x+2)(x+4) $. This will give you all the zeros you are looking for. Trick is to make sure your function is zero at each of the values 0,1,-2, and -4.

share|cite|improve this answer
this probably would be the right answer if that was the question that my homework was telling me :) – Tyler Zika Oct 30 '12 at 6:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.