rational functions

[relapse]

Hello, I have these 3 questions for a homework assignment but we've never discussed in class how to determine the answer yet. I'm sure someone could probably just glance at them and get the correct answer. I wanted to know if someone could tell me the answer and explain why briefly.

I uploaded the questions to here: http://img125.imageshack.us/img125/2599 ... ons3ts.jpg

 

[stapel]

For the benefit of other users, the questions are as follows:

1) Determine which of the following rational functions has an x-intercept at 5.

. . . . .\(\displaystyle \large{\mbox{A) }\,f(x)\,=\,\frac{x\,+\,5}{x\,+\,9}}\)

. . . . .\(\displaystyle \large{\mbox{B) }\,f(x)\,=\,\frac{x\,+\,9}{x\,-\,5}}\)

. . . . .\(\displaystyle \large{\mbox{C) }\,f(x)\,=\,\frac{x\,-\,5}{x\,+\,9}}\)

. . . . .\(\displaystyle \large{\mbox{D) }\,f(x)\,=\,\frac{5x\,-\,1}{x\,+\,9}}\)


2) Determine which of the following rational functions has a vertical asymptote at x = 4.

. . . . .\(\displaystyle \large{\mbox{A) }\,f(x)\,=\,\frac{x\,+\,9}{x\,+\,4}}\)

. . . . .\(\displaystyle \large{\mbox{B) }\,f(x)\,=\,\frac{4x\,-\,1}{x\,+\,9}}\)

. . . . .\(\displaystyle \large{\mbox{C) }\,f(x)\,=\,\frac{x\,-\,4}{x\,+\,9}}\)

. . . . .\(\displaystyle \large{\mbox{D) }\,f(x)\,=\,\frac{x\,+\,9}{x\,-\,4}}\)


3) Determine which of the following rational functions has the x-axis as its horizontal asymptote.

. . . . .\(\displaystyle \large{\mbox{A) }\,f(x)\,=\,\frac{4x\,-\,1}{x\,+\,9}}\)

. . . . .\(\displaystyle \large{\mbox{B) }\,f(x)\,=\,\frac{x\,-\,4}{x}}\)

. . . . .\(\displaystyle \large{\mbox{C) }\,f(x)\,=\,\frac{9}{x\,-\,4}}\)

. . . . .\(\displaystyle \large{\mbox{D) }\,f(x)\,=\,\frac{x\,+\,4}{x\,+\,9}}\)

To the original poster:

1) To find an x-intercept, set y equal to zero, and solve.

2) To find a vertical asymptote, find where the rational function is not defined; that is, find the zeroes of the denominator.

3) To find the horizontal asymptote, compare the leading terms of the numerator and denominator, according to the listing of rules they gave you in the text.

Eliz.

 

[relapse]

I'm terrible at algebra, but I got:

1). A

2). D

3). D

when you say set y to zero, do you mean set f(x) to f(0) ?

 

[daon]

1. No
2. Yes
3. No

...

To set "y" to zero (that is, find the x value s.t. y is zero), substitute 0 for f(x), do not evaluate f(0).

For 1, plug in 5 to each of the choices. Whichever ends up with a value of zero is the correct answer. Remember zero over anything except for zero is zero.

For 3- for a horizontal asymptote, the coefficients of the two leading terms must be zero. Which ratio of coefficients is zero over something? If this is not obvious, as staple said, follow the rules in your text. One of them says something about "if the power in the numerator is less than the power in the denominator then..."

 

[stapel]

when you say set y to zero, do you mean set f(x) to f(0) ?

"Evaluating y at x = 0" and "setting y equal to zero" are very different things. Evaluating at x = 0 is how one finds the y-intercept; you're needing to find the x-intercept.

I apologize for the confusion.

Eliz.

 

[relapse]

1). C

2). D

3). A ?

 

[stapel]

1). C

2). D

3). A ?

Please don't play guessing games. If you're not sure of your answers, kindly reply showing your work; we'll be glad to check it.

Thank you.

Eliz.

 

[relapse]

1). C

2). D

3). A ?

Please don't play guessing games. If you're not sure of your answers, kindly reply showing your work; we'll be glad to check it.

Thank you.

Eliz.

for 1). I did 5 - 5 over r + 9 = zero, so thats how i got C.

3). 4 - 4 over 4 which makes a zero on top so the answer is B.

 

[Gene]

That's the way!
-------------
Gene

 

[daon]

relapse, your justification for #3 is wrong. I won't tell you if it is right or wrong until you give a reason for your decision.

Read your rules! Then state your justification!

 

[Gene]

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G