# Graphs of Fractional Functions

1. Aug 10, 2013

### xpeteyzx

Hello,
Question:
Find the equations of the asymptotes of the following fractional functions., and then, draw the graph and label x- and y-axis points of intersection.

The equation:
y= x/(x + 1)

My try at it:
I first divided x by (x + 1) to get the equations of the asymptotes:
x=-1, y=1 from y= 1 + 1/(x+1).

Then I drew the asymptotes equations on the graph:
https://docs.google.com/drawings/d/1p2Lfkqab9qAVxzx9z0LFyWAm8-WeoP30yBSn9bHsDKM/edit (red is the asmptotes equations, black is the graph).

Then I got stumped at the points of intersection with the x-axis:
x-axis: (?, 0)
y-axis: (0, 2)

I got the y-axis by substituting 0 in for x in the previous equation:
y = 1 + 1/(x + 1)
y = 1 + 1/(0 + 1)
y = 1 + 1/1
y = 1 + 1
y = 2

But for x:
y = 1 + 1/(x+1)
because if I do negative one:
0 = 1 + 1/(-1+1)
0 = 1 + 1/0
0 = 1 is incorrect.

2. Aug 10, 2013

### SteamKing

Staff Emeritus
These are usually called 'rational functions' instead of 'fractional functions'.

In your attempt at finding the x-intercept, you have used y = 1 + 1/(x+1) instead of the original function.

y = 1 + 1/(x+1) is not equal to y = x / (x + 1)

3. Aug 10, 2013

### xpeteyzx

but if I use the original equation, wouldn't it be a no-solution?

4. Aug 10, 2013

### Mentallic

It wouldn't be a "no solution", but it would be difficult to find the solution. Your technique is correct, but your answer is wrong. If you want to find the x-intercepts of the equation $y=x^2-1$, you wouldn't want to be solving the incorrectly factored form $y=(x-2)(x-3)$ would you? You'd want to factor it correctly, and then go from there.

Similarly in this case, you want to find the correct values of A and B such that

$$\frac{x}{x+1}=A+\frac{B}{x+1}$$

And once you do, check your answer! For example, using the values A=1 and B=1 that you gave,

$$1+\frac{1}{x+1}=\frac{1(x+1)+1}{x+1}=\frac{x+2}{x+1}\neq \frac{x}{x+1}$$

Hence those values of A and B were not correct.

5. Aug 10, 2013

### xpeteyzx

I understand your explanation, but then how would I get it into the y= A + B/(x+1) format?

6. Aug 10, 2013

### vanhees71

Just write
$$\frac{x}{x+1}=\frac{(x+1)-1}{x+1}=\ldots$$

7. Aug 10, 2013

### Mentallic

vanhees71's method is the easiest and quickest, and with a little practice, you'll quickly learn to master it.

The alternative method of finding A and B is called partial fraction decomposition and it may arise in your later math studies, so understanding its technique could be useful to you.

$$\frac{x}{x+1}=A+\frac{B}{x+1}$$

the right hand side needs to be all on one fraction, so that we can compare it to the left,

$$=\frac{A(x+1)+B}{x+1}$$

Now, since both sides have the same denominator, we just need to equate the numerators,

$$x=A(x+1)+B$$

$$x=Ax+A+B$$

for these equations to be equal, the coefficient (number in front of) the x terms must be equal and the constants must be equal. It might help to look at the equality like this

$$1x+0=Ax+(A+B)$$

So we have that A=1 and A+B=0, hence B=-1.

If there is anything you don't understand, just ask.