Question

Algebraically find the intercepts and the equations of the asymptotes of the rational function.
f(x)= (x^2+2x-3)/(x^2+x-6)

Answer 1

Can someone please help me? :)

Answer 2

x-intercept implies that y =0 , y =0 iff numerator =0, solve for x
y-intercept implies that x =0 , just replace x =0 into the function to find y
asympotes of the function happen when denominator =0, let it =0 and solve for x.
that's it

Answer 3

I believe the y-intercept = .5
I am still stuck on how to make the numerator (x^2+2x-3)=0 to find the x-intercept and then solve for the asymptotes.
Somebody please help. :)

Answer 4

the y -intercept is x^2 +2x-3 =0 which give you 2 roots, not as you said. Redo, please

Answer 5

think about factoring

Answer 6

Is the factored form (x-2)(x+3)
So the y-intercept would be 2 and -3?

Answer 7

that is for the denominator

Answer 8

So I need to factor (x^2+x-6)? That is that is also (x-2)(x+3) = 2 and -3 right?

Answer 9

remember factoring is your friend, you can always check it with FOIL

Answer 10

When I Foil out (x^2+x-6) my answer is (x-2)(x+3) so is 2 and -3 my y-intercept?

Answer 11

those are in the denominator...are you allowed to divide by 0?

Answer 12

No you can't divide by zero. So you need to solve for the numerator which is (x^2+2x-3). The numerator can be factored to (x-1)(x+3). So the answers for the y-intercepts are 1 and -3.
Does that look right?

Answer 13

but look -3 is common to both

Answer 14

This function factors as follows:

(x+3)(x-1)
f(x) = ----------
(x+2)(x-3)

To find the horizontal intercepts, let f(x) = y = 0. Write the horizontal intercepts as points (x, y).

Answer 15

To find the vertical intercept (there's only one), set x = 0 and calculate the y value. Write the vertical intercept as a point (x,y).

Answer 16

Doesn't the denominator factor to this instead?
(x+3)(x-1)
f(x) = ----------
(x-2)(x+3)

Answer 17

Let's review how to determine vertical asymptotes. We first factor the denominator of this rational function (already done, above) and set each factor = to 0 separately. We must write equations in the form x = a to represent these vertical asymptotes. Try it.

Answer 18

Apologies. You are correct in asking whether the denominator doesn't factor to (x-2)(x+3). My mistake. So we'll have to redo the work of finding the vertical asymptotes.

Answer 19

Finding horiz. asympt. is a bit more involved, in that we have to determine what happens to y as x grows larger and larger. In other words, we take the limit of f(x) as x goes to infinity. Are you familiar with this terminology?

Answer 20

yes

Answer 21

However you find it, the limit of f(x) as x approaches infinity (a very large quantity) is
x^2
y = ---- , or 1. Your HA is the horizontal line y = 1.
x^2

Answer 22

Please take a moment to review what we've discussed. Anything still unclear for you?

Answer 23

I am confused why you took x^2/x^2 to get the horizontal asymptote.

Answer 24

Since you're apparently familiar with limits, consider what happens to f(x) as x grows larger and larger. The x^2 terms dominate in both numerator and denominator (they grow so large that we can essentially ignore all the terms in x or constants).

A more sophisticated way of evaluating the limit involves dividing every term in your rational function by the highest power of x (which is x^2).
1 + 2/x - 3/x^2
You'll end up with f(x) = ---------------.
1 + 1/x - 6/x^2

What happens to the variable terms of this expression as x approaches infinity?

Answer 25

They get larger. Thank you so much. I think I understand the problem well enough now to figure the rest out on my own. Thanks again!

Answer 26

My great pleasure. I do need to point out that your "They get larger" response is correct if you're discussing the denominators of the fractions, but my point is that each of these fractions goes to zero as x grows without limit. We're then left with y = 1/1, or 1.
All the best to you.