# Homework Help: Need to determine function based on graph

1. Oct 18, 2014

### TheExibo

There is this graph: http://i.imgur.com/BdwhN3b.jpg where the equation of the function must be determined. Down below is some thinking I've done, where I've found that the oblique asymptote is -0.5x+2, and the v-asymptote=6. I know how to solve something similar with 2 x-int's, but I'm having difficulty understanding what to do with 3 x-int's. Can anyone help me please? Thanks!

2. Oct 19, 2014

### zoki85

plugin 3 values of (x,f(x)) from the graph of the eq. in order to get system of equations and solve it to find a,b,c.

Last edited: Oct 19, 2014
3. Oct 19, 2014

### NTW

I believe I'm saying the same as zoki85, but I'll try to elaborate a bit more...

It seems that you have come to the conclusion that your function has the form

y = 0,2x + 50 + (ax2 + bx + c)/-(x+b2)

In my opinion, and to find a, b, c you should take three values of x, y from the drawing, reasonably chosen among the left and right branches. For the left branch, one point. For the right branch, and always IMHO, one for the maximum, and another to the right, away from the maximum.

Thus, you'll have a 3 x 3 system and solve it for the coefficients a, b, c...

4. Oct 19, 2014

### TheExibo

I got it using only two of the intercepts: (-40,0) and (0.5,0) and putting them into -0.2x+50+(ax+b)/-(x+6)2

Graphing it in desmos.com, it shows the exact same graph. I must explain how I got this function, so how can I explain the third x-intercept? Is it true that I only need two points, one on each side of the asymptote, to find the exact function of the graph that is shown?

5. Oct 19, 2014

### NTW

I now see that you needed only to find two coefficients, a and b. I saw the numerator in x2, but it's true that I misquoted the denominator in your notes, writing (y = 0,2x + 50 + (ax2 + bx + c)/-(x+b2)

6. Oct 19, 2014

### TheExibo

But by doing that, I've come across the problem where I can't explain how the graph ended up with all 3 correct x-intercepts, when I only used the two. Anyone have any ideas?

7. Oct 20, 2014

### ehild

You also used the asymptote at infinity where the function behaves as 50-0.2x.

ehild

8. Oct 20, 2014

### tycoon515

Being that your graph has an oblique asymptote, an improper rational function should be the kind of function that comes to mind. By the looks of the graph, as we let $x$ tend toward either $\infty$ or $-\infty$, the function gets closer and closer in value to a linear function, so we should consider an improper rational function whose numerator is $1$ greater in degree than its denominator. Also, a rational function is equal to $0$ wherever its numerator, but not its denominator, is equal to $0$. Since you have three $x$-intercepts, the numerator should be a polynomial that is the product of three factors, each one corresponding to a different $x$-intercept. A preliminary construction might look like the following:

$f(x)=k\frac{(x+40)(x-.5)(x-277.4)}{?}$

The numerator, when expanded, appears to be a polynomial whose dominant term is $x^3$. Therefore, we should consider a denominator that is a polynomial of degree $2$. Also, the graph is undefined and has a vertical asymptote at only one value of $x$: $-6$. Notice also that the graph of $f(x)$ tends toward $-\infty$ as $x$ approaches $-6$ from either side. Therefore, the denominator should be a quadratic with at least one $(x+6)$ factor. To find the other factor, observe either that there is no sign change of the function across the vertical asymptote at $x=-6$, or that for any factor other than $(x+6)$ to be present in the denominator, the graph would have to have another vertical asymptote that isn't at $x=-6$, so the denominator clearly has repeated $(x+6)$ factors. Our construction now takes the following form:

$f(x)=k\frac{(x+40)(x-.5)(x-277.4)}{(x+6)^2}=k\frac{(x+40)(\frac{2x-1}{2})(\frac{5x-1387}{5})}{(x+6)^2}=k\frac{(x+40)(2x-1)(5x-1387)}{10(x+6)^2}$

$=k\frac{(2x^2+79x-40)(5x-1387)}{10(x^2+12x+36)}=k\frac{10x^3-2379x^2-109773x+55480}{10x^2+600x+1800}$

A function of this form is EXACTLY equal to $0$ at all three of the intercepts shown on the graph provided. And, if we let $k=-\frac{1}{5}$, it has an oblique asymptote of $-\frac{1}{5}x+\frac{2379}{50}$, which has the same slope as the asymptote on the graph but NOT the same $y$-intercept. Also, the equation you came up with has the correct slope and $y$-intercept of the oblique asymptote, but it is NOT EXACTLY equal to $0$ when $x=277.4$, so one of two things must be going on here: either the $x$-intercepts displayed on the graph are slightly off (rounded), or the actual equation of the oblique asymptote is slightly different than $-\frac{1}{5}x+50$.

9. Oct 20, 2014

### TheExibo

Awesome! Thank you to everyone for the help!