# Rational Functions

• Oct 8th 2008, 07:09 PM
cheet0face
Rational Functions
Ok, we've been graphing rational functions for about a week now and i've been able to do most of it... until now
Past functions have been factorable or divisble with long division.
But the new problems are different.

ex. y=(x^4-3x^3+x^2-3x+3) / (x^2-3x)

another one is

ex. y=(2x^5-x^3+2) / (x^3-1)

They don't factor and i've tried dividing both with long division and synthetic.
I might just be dumb and dividing wrong but I dont see how it works.
If you could just show me how to do it I would be grateful =)

Thanks
• Oct 8th 2008, 07:20 PM
Prove It
Quote:

Originally Posted by cheet0face
Ok, we've been graphing rational functions for about a week now and i've been able to do most of it... until now
Past functions have been factorable or divisble with long division.
But the new problems are different.

ex. y=(x^4-3x^3+x^2-3x+3) / (x^2-3x)

another one is

ex. y=(2x^5-x^3+2) / (x^3-1)

They don't factor and i've tried dividing both with long division and synthetic.
I might just be dumb and dividing wrong but I dont see how it works.
If you could just show me how to do it I would be grateful =)

Thanks

Long division works.

For the first one you should get

$\displaystyle x^2 + 1 + \frac{3}{x^2 - 3x}$

and for the second you should get

$\displaystyle 2x^2 - 1 + \frac{2x^2 + 1}{x^3 - 1}$.

If what you end up subtracting completely cancels a term, just bring down the second and go from there. If you're trying to add/subtract something two things that are not alike, e.g. $\displaystyle -x^3 + 2x^2$ just leave it as is and go from there.
• Oct 8th 2008, 07:23 PM
icemanfan
You can use long division on both problems. You might have a remainder, but you can still do it.

For the first problem you begin with
$\displaystyle \frac{x^4 - 3x^3 + x^2 - 3x + 3}{x^2 - 3x}$.

$\displaystyle x^2 - 3x$ goes into $\displaystyle x^4$ $\displaystyle x^2$ times. So you subtract $\displaystyle x^2(x^2 - 3x)$ from $\displaystyle x^4 - 3x^3 + x^2 - 3x + 3$, which is:

$\displaystyle x^4 - 3x^3 + x^2 - 3x + 3 - (x^4 - 3x^3)$

$\displaystyle x^4 - 3x^3 + x^2 - 3x + 3 - x^4 + 3x^3$

$\displaystyle x^2 - 3x + 3$.

Save the factor of $\displaystyle x^2$ for the quotient.

Now, $\displaystyle x^2 - 3x$ goes into $\displaystyle x^2$ once, so you subtract $\displaystyle x^2 - 3x$ from $\displaystyle x^2 - 3x + 3$, which gives you 3.

So your quotient is $\displaystyle x^2 + 1$ with a remainder of 3, or exactly $\displaystyle x^2 + 1 + \frac{3}{x^2 - 3x}$.
• Oct 8th 2008, 07:29 PM
cheet0face
Ok, so I divided the first one and got x^2+1 (we're supposed to drop the remainder), but I still don't see how that fits into graphing it?

So would the Root be x=-1?

We have to find the Roots/holes, the Vertical Asymptotes, the Horizontal Asymptotes, and the Y-intercept.

On previous problems they were like so: (x+3)(x-1)(x+2) / (x+2)(x+1)
So I could find the Roots/holes to be x=-3, -2 and 1
VA: x=-1
HA: y=x
y=-3

But I'm not getting anything close to that on these :/
• Oct 8th 2008, 07:33 PM
icemanfan
$\displaystyle x^2 + 1$ has no real roots.

Also, a root and a "hole" are different things.

For the equation $\displaystyle \frac{(x+3)(x-1)(x+2)}{(x+2)(x+1)}$

the roots are -3 and 1, and the "holes" occur at x = -1 and x = -2. -2 is not a root because the function is not defined there.

A vertical asymptote occurs wherever there is a "hole".
• Oct 8th 2008, 07:47 PM
Prove It
Quote:

Originally Posted by icemanfan
$\displaystyle x^2 + 1$ has no real roots.

Also, a root and a "hole" are different things.

For the equation $\displaystyle \frac{(x+3)(x-1)(x+2)}{(x+2)(x+1)}$

the roots are -3 and 1, and the "holes" occur at x = -1 and x = -2. -2 is not a root because the function is not defined there.

A vertical asymptote occurs wherever there is a "hole".

If you're trying to graph the original functions you can't disregard the remainder.

For the first one
$\displaystyle y = x^2 + 1 + \frac{3}{x^2 - 3x}$

To find the x-intercepts, let y equal 0 and solve for x.

To find the y-intercept, let x = 0 and simplify. Is it possible to have a y-intercept in this case?

To find the points of discontinuity, we know that the denominator can't be 0, and so x can't be 0 or 3. Also, there are points of discontinuity where $\displaystyle y=x^2 + 1$. This is because if $\displaystyle x^2 + 1 = x^2 + 1 + \frac{3}{x^2 - 3x}$, how are you supposed to solve for x?
• Oct 9th 2008, 05:22 AM
cheet0face
We'll, she says as x goes farther away, the remainder gets closer and closer to 0 so we're just supposed to drop it.
And what we're supposed to do is draw a graph grid and put in the roots and holes and the x axis and draw dotted lines going up and down for the asymptotes and put in the y-intercept and pretty much just connect the dots without crossing a vertical asymptote.

And there is no y-intercept in this problem so we would have to pick an x value and plug it in to get a y value for an extra point.

What my question is, is I get $\displaystyle x^2+1$ but I don't know what it means. :/ Get what i'm saying?
Like what exactly is that number?
Is it the simplified equation or just the top part?
• Oct 10th 2008, 06:38 PM
Prove It
Quote:

Originally Posted by cheet0face
We'll, she says as x goes farther away, the remainder gets closer and closer to 0 so we're just supposed to drop it.
And what we're supposed to do is draw a graph grid and put in the roots and holes and the x axis and draw dotted lines going up and down for the asymptotes and put in the y-intercept and pretty much just connect the dots without crossing a vertical asymptote.

And there is no y-intercept in this problem so we would have to pick an x value and plug it in to get a y value for an extra point.

What my question is, is I get $\displaystyle x^2+1$ but I don't know what it means. :/ Get what i'm saying?
Like what exactly is that number?
Is it the simplified equation or just the top part?

This means that the entire function $\displaystyle x^2 + 1$ is an asymptote.