Question

can you tell me why i got these wrong?

Answer 1

It's the second option.

This is because of the following reason:\[\frac{ x^2-10x-24 }{ x^2-3x-108 }=\frac{ \cancel{(x-12)}(x+2) }{ \cancel{(x-12)}(x+9) }=\frac{ x+2 }{ x+9 },x \ne 12, -9\] This is because the domain is based on the ORIGINAL FUNCTION, not the simplified one. You get the asymptotes from the simplified version but the domain is still stated from the original. And the choice you selected has x not equal to -2, which is incorrect since 2 is part of the domain. Do you understand?

@AriPotta

Answer 2

no..i don't get it :/

Answer 3

Ok so you know how the function's domain for rational functions is determined by where it does not exist? @AriPotta

Answer 4

And where it does not is the x-value where you get a zero in the denominator correct? @AriPotta

Answer 5

wat

Answer 6

Ok tell me this, how do you find the domain of a rational function and where are its vertical asymptotes.

Answer 7

i have no idea what those are ._. what are asymptotes

Answer 8

Have you ever done rational functions before or know what they are?

Answer 9

i just started learning about them today

Answer 10

K I'll give you a brief introduction to them.

Answer 11

it's ok, i got it now :] thanks for helping me!

this lesson was only on how to simplify rational expressions

Answer 12

Rational functions are functions given as a quotient. For example, y = 1/x is the simplest rational function you can get. These types of functions have different asymptotes. To start you off, these functions have 2 types of asymptotes: Horizontal and Vertical. Horizontal asymptotes are the asymptotes that are given by something like y = 3, which means that the function approaches the y = 3 as it keeps on going. A vertical asymptote is given by an equation like x = 4, which is the x-value that the function doesn't exist at and when it approaches this x-value, it goes to either positive or negative infinity.

The horizontal asymptote is y = 0 when the degree of the denominator is greater than the degree of the numerator. If the degree of the numerator and denominator is the same, then the horizontal asymptote is at y = leading coefficient of numerator/leading coefficient of denominator. If the degree of the numerator is greater than the denominator, then you have to use long division to get it and now you won't have a horizontal asymptote, instead you will have a slant asymptote like y = x or even a parabolic asymptote like y = x^2.

The vertical asymptote is given by wherever the function does not exist. This happens when the denominator of the function is 0 for a certain x value, which would mean that the x-value is not part of the domain. One important thing to remember that the domain is always based on the original function before cancelling. The vertical asymptotes are always based on the simplified version of the function. For example:\[y=\frac{ x^2+4x+4 }{ x^2+5x+6 }=\frac{ \cancel{(x+2)}(x+2) }{ \cancel{(x+2)}(x+3) }=\frac{ x+2 }{ x+3 }\]As you can see, in the above function, the domain will be:\[Domain:x \ne -2, -3\]But the vertical asymptote occurs at:\[V.A:x=-3\]This is because the domain is ALWAYS based on the original function whereas the vertical asymptote comes from the simplified function. When you graph the simplified function, you have to make an open dot at x = -2 to show that the function does not exist at x = -2 because x + 2 got cancelled out. This means that even though there won't be a vertical asymptote at x = -3, we still have to exclude it from the graph because it isn't part of the domain. Do you understand?

And to find the x-intercept, you just set the numerator to 0 and solve for x. This is because x-intercepts occur when y = 0, and that only happens when the numerator is 0, and to get the values at which it is 0, you just take the numerator, set it to 0, and solve for x. If the numerator has no x in it like y = 1/x, then it will have no x-intercepts.

To find the y-intercept, you set x to 0 and whatever you get will be your x-intercept.

There is more that you need to know about these functions and will find out later but those are the basics and most important aspects.

@AriPotta