• Oct 14th 2007, 03:22 PM
ffezz
Hello again people. I need a tad of help with my homework. Here are the questions I am having trouble with.

$f(x) = \frac {7x - m}{2 - nx}$

So, I need to find the values of m and n such that the vertical asymptote is x = 6 and the x intercept is 5. Im guessing for the value to be an assymptote, we have to look at just the bottom and that has to = to 0. So I get
2 - n(6) = 0
-2 = -n(6)
-2/6 = -n = -1/3
or
n = 1/3
but im not sure if thats right.

Ok now the other question.
Find the value if p for which f(x) would have a removable discontinuity and then define f(p) so that $f(x) = \frac {x^2 - 6x +9}{x-p}$

I have no clue where I am even supposed to start with this question. Someone explain this to me because wouldn't there always be a value that would give you 0 on the bottom?

Thanks
Chris
• Oct 14th 2007, 04:29 PM
topsquark
Quote:

Originally Posted by ffezz
Hello again people. I need a tad of help with my homework. Here are the questions I am having trouble with.

$f(x) = \frac {7x - m}{2 - nx}$

So, I need to find the values of m and n such that the vertical asymptote is x = 6 and the x intercept is 5. Im guessing for the value to be an assymptote, we have to look at just the bottom and that has to = to 0. So I get
2 - n(6) = 0
-2 = -n(6)
-2/6 = -n = -1/3
or
n = 1/3
but im not sure if thats right.

That is correct. Now, what condition do you have to make happen to make the x-intercept be 5?

Quote:

Originally Posted by ffezz
Ok now the other question.
Find the value if p for which f(x) would have a removable discontinuity and then define f(p) so that $f(x) = \frac {x^2 - 6x +9}{x-p}$

To have a "removable" discontinuity means that we would have to cancel the factor in the denominator by a factor from the numerator. So what are the factors of the numerator?

-Dan
• Oct 14th 2007, 05:58 PM
ffezz
ok so umm for the first question because its an x intercept, the y value has to = 0 so i get 7(5) - n which gives me n as -35. So the value of n is -35 and m is 1/3?

and for the 2nd question

$(x-3)(x-3)$ so we can factor out and x-3 which means the value of p is 3

how would i find a value which would make the functions continuous for all values?
• Oct 14th 2007, 06:42 PM
topsquark
Quote:

Originally Posted by ffezz
ok so umm for the first question because its an x intercept, the y value has to = 0 so i get 7(5) - n which gives me n as -35. So the value of n is -35 and m is 1/3?

Yes. :)

Quote:

Originally Posted by ffezz
and for the 2nd question

$(x-3)(x-3)$ so we can factor out and x-3 which means the value of p is 3

how would i find a value which would make the functions continuous for all values?

The given function cannot be made continuous for all values of x. The function $\frac{x^2 - 6x + 9}{x - 3}$ is not defined at x = 3, and thus the function is not continuous there. The best you can do is the function with p = 3.

-Dan
• Oct 14th 2007, 07:03 PM
ffezz
hehe alright =) i missed a questions though somehow =\

Find constants a and b such that the function is continuous for all of x
$f(x) = ax+3,$ if $x>5$
$8$ if $x=5$
$x^2 + bx + a$ if $x<5$
• Oct 14th 2007, 08:49 PM
ffezz
• Oct 15th 2007, 05:19 AM
topsquark
Quote:

Originally Posted by ffezz
hehe alright =) i missed a questions though somehow =\

Find constants a and b such that the function is continuous for all of x
$f(x) = ax+3,$ if $x>5$
$8$ if $x=5$
$x^2 + bx + a$ if $x<5$

New questions should go in new threads.

It looks to me like you simply need to find an a and b such that
$ax + 3 = x^2 + bx + a = 8$ when x = 5.

-Dan
• Oct 15th 2007, 05:21 AM
topsquark
Quote:

Originally Posted by ffezz