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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

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2 Answers 2

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

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

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@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.

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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

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