Find the equation of a higher-degree polynomial from graph

**boyle96** · July 21st 2012, 01:43 PM

hey everybody, this is my first post on the forum. I'm on my trig+ summer assignment and I only need help on one problem. The gist of it is that there's a graph, and you need to find the equation for the polynomial on the graph (I've attached a picture of the graph). Here's what I've done so far:

I put all the x intercepts in the numerator, vertical asymptotes in the denominator, and (x-5) in both because it's a hole. Then, to get it to equal the point (5, -756/25), I put in "a" and set the equation equal to -756/25. Then I got "a", factored, factored, factored, and got an answer (numerator over denominator), but it didn't work. It seems like the polynomial is fifth degree because of the turning points and end behavior, but I can't find enough factors to "fill up" the numerator and/or denominator to said fifth degree. SO, I'm positive that I've done everything that I know how to do...anybody have any thoughts? Thanks again.

Oh, and PS: My best answer was (-72x^3+936x^2+576x-17280)/(x^3-10x^2-125x+750)

**HallsofIvy** · July 21st 2012, 02:02 PM

That looks good to me.

You understand, don't you, that this is NOT a polynomial? It is a rational function having polynomials as numerator and denominator.

**boyle96** · July 21st 2012, 02:07 PM

sorry man, been on summer break too long, would have said polynomial if this was may. But when I plug my final answer into the calculator, it doesn't match up with the graph in the attachment...

**Soroban** · July 21st 2012, 05:32 PM

Hello, boyle96!

Your function seems to be correct, but . . .

$\text{What is that diagonal dotted line?}$

$\text{If that is an oblique asymptote (I suspect that it is }y = -x),$
. . $\text{we have much more work to do.}$

**boyle96** · July 22nd 2012, 03:28 AM

Yeah, it's a slant asymptote. I'm not sure if it's y=-x because it doesn't pass through the origin.

**Soroban** · July 22nd 2012, 06:25 AM

Hello again, boyle96!

I thought I had it solved . . . but I've hit a wall.

$\text{Yeah, it's a slant asymptote. \;I'm not sure if it's }y=-x$
$\text{because it doesn't pass through the origin.}$

Well, the diagram was made by someone (not from a textbook).
. . Note that the parentheses are not closed.
So I will assume that that the slant asymptote was intended to include the origin.

I thought that inserting an additional factor of $x$ would suffice.

. . that is: . $f(x) \;=\;a\,\frac{{\color{red}x}(x+4)(x-12)(x-5)}{(x+10)(x-15)(x-5)}$

We want: . $\lim_{x\to5}f(x) = \text{-}\tfrac{756}{25}$ , so we have:

. . $a\cdot\frac{(5)(9)(\text{-}7)}{(15)(\text{-}10)} \:=\:\text{-}\frac{756}{25} \quad\Rightarrow\quad a \:=\:\text{-}\frac{72}{5}$

Hence: . $f(x) \;=\;\text{-}\frac{72x(x+4)(x-12)(x-5)}{5(x+10)(x-15)(x-5)}$

But this has a slant asymptote of: . $y \:=\:\text{-}\tfrac{72}{5}x$

Now what?

**skeeter** · July 22nd 2012, 08:26 AM

the factor (x+4) in the numerator should be squared since the curve "touches" the zero at x = -4

$f(x) = \frac{k(x+4)^2(x-5)(x-12)}{(x-5)(x+10)(x-15)}$

removing the factors for the "hole" location ...

$\lim_{x \to 5} \frac{k(x+4)^2(x-12)}{(x+10)(x-15)} = -\frac{756}{25} \implies k = -8$

so ...

$f(x) = -\frac{8(x+4)^2(x-5)(x-12)}{(x-5)(x+10)(x-15)}$

Attachment 24329

**boyle96** · July 22nd 2012, 08:40 AM

This looks great skeeter, thanks so much dude!

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