Question

Can someone please help me graph this rational function:
x^2+x-2
---------
x^2-3x-4

Also need the
Domain:
x and y intercepts:
horizontal asymptotes:
vertical asymptotes:

*This is a practice problem, if someone can help me find the answers to this problem i can help myself do the actual problem that i need.
Thanks!

Answer 1

Start by factoring both the num'r and den'r. You need to see all the factors of each, both to make sure that the expression can't be reduced,a nd also to do the other parts of the problem.

Answer 2

\[\Large \dfrac {x^2+x-2}{x^2-3x-4}=\dfrac{(~~?~~)(~~?~~)}{(~~?~~)(~~?~~)}\]

Answer 3

(x-2)?
(x-3)(x-4)?

Answer 4

You need 2 factors in each, so I'm not sure what you mean by the (x-2)....? If that's your factoring of the num'r, you need another factor.

As for (x-3)(x-4), if that's your factoring of the den'r, you should check it by multiplication.

Answer 5

Always, always, always, check factoring with multiplication. :) So when you factor a trinomial, FOIL your result and make sure you get back to what you started with.

Does \(\Large (x-3)(x+4)=x^2-3x-4\) ??

Answer 6

Can you please factor them for me

Answer 7

I could, but if you are expected to do this problem, then surely you are expected to know how to factor a trinomial. Do you remember learning to factor?

Do you know how to multiply with FOIL?

Factoring is just "undoing" multiplication. Use some trial and error if necessary.

You need: \(\Large (x\pm ?)(x\pm?)=x^2-3x-4\)
So figure out the ?'s and the signs in the middle that make the FOIL product = the expression on the right

Answer 8

And the same for the numerator:
\(\Large (x\pm ?)(x\pm?)=x^2+x-2\)

Answer 9

I understand, but shouldnt you have 4 terms that come out after you foil?

Answer 10

When you foil 2 linear binomials - meaning that each (-----) has an x term and a constant - you will get

F: first = an x^2 term
O: outside = an x term
I: inside = an x term
L: last = a constant term

The 2 x terms from the inside and outside products combine. So you will get a 2nd degree trinomial.

Answer 11

E.g.

\(\Large (x-3)(x+4)=x^2-3x+4x-12=x^2+x-12\)

Answer 12

When the trinomial you are trying to factor is of the form of the ones you have here - namely that the \(x^2\) term has a coefficient of 1 - you just need to think about PRODUCTS that give you the last term, where the pair of factors in the product also ADD UP to the coefficient on the middle term.

So for \(\Large x^2-3x-4\) you are looking for 2 numbers where the product is -4, and the sum is -3.

\(\Large (??)(??)=-4 ~~\text {and} ~~(??)+(??)=-3\)

Answer 13

(x+1)(x-4)!?

Answer 14

and

Answer 15

(x-1)(x+2)

Answer 16

Good... so now you have:

\[\Large \dfrac {x^2+x-2}{x^2-3x-4}=\dfrac{(x+2)(x-1)}{(x+1)(x-4)}\]

Nice job. Now start thinking about the properties of the function, and how they relate to the graph. It IS in lowest form (important to know that). So in terms of domain, with a rational function you really just need to make sure den'r is not =0. So you need to solve
\[\Large (x+1)(x-4)=0\]
Those solutions tell you what to leave OUT of the domain.

Answer 17

And the x values that are not in the domain, are also where you will see a certain kind of asymptote - do you know which kind?

Answer 18

So they domain is all real numbers besides -1 and 4?

Answer 19

Domain: everything except where den'r=0

x intercepts: x-values where y=0... for a rational function, the only way y=0 is if the num'r=0, so set num'r=0 and solve. This will give you 2 points on the graph

y intercepts: y-value when x=0, so plug in x=0 and solve. This will give you one point on the graph

horizontal asymptotes: There is a "rule" for the horizontal asymptote, when the degree of the num'r = degree of the den'r, as here. Do you know what that rule is?

vertical asymptotes: Related to the domain questions above.

Answer 20

Yes - to your domain! Exactly! good job.

Answer 21

is the intercept 1 and -2?

Answer 22

x intercept

Answer 23

Right! And those are both in the domain (sometimes not - important to check) - so those are the two x-intercepts.

Answer 24

is the y intercept -4?

Answer 25

And how do i find the horizontal asymptote and the range?

Answer 26

Anyone????

Answer 27

y-intercept is when x=0

\[\Large \dfrac{(0+2)(0-1)}{(0+1)(0-4)}=\dfrac{(2)(-1)}{(1)(-4)}=??\]

Answer 28

For horizontal asymptotes, it depends first on the degree of the num'r vs. the degree of the den'r. Here they are equal (both num'r and den'r have degree=2). In that case, the horizontal asymptote is the line

\[y=\frac{ N }{ D }\]where M is the leading coefficient of the num'r and D is the leading coefficient of the den'r.

Answer 29

Can someone please help me find the range and horizontal asymptote?