Question

What is the oblique asymptote of the function f(x) = the quantity x squared minus 5x plus 6 over the quantity x minus 4 ?

Answer 1

Kelly tells you that when variables are in the denominator, the equation becomes unsolvable. "There is a value for x that makes the denominator zero, and you can't divide by zero," Kelly explains. Using complete sentences, demonstrate to Kelly how the equation is still solvable. Im confused, can someone help me please? @Jacobbenvenutty, @kewlgeek555 or @phi @KingKhan94

Answer 2

Basically what "Kelly" is saying is that if the variable s in the denominator (1/x) then it is unsolvable. She also says that the value for x makes the denominator 0.
(Which is impossible to solve) Example: 1/0= infinity

What you have to do is prove to Kelly that this statement is infact false meaning it can be solved

Answer 3

sorry it took me so long i was trying to find the divide symbol but couldnt

Answer 4

its fine, lol, but i dont understand hw it can be solvable because I've always been told that it can't due to not being able to divide 0 by a number

Answer 5

True but obviously somewhere her statement is false or they wouldnt give you this question.
Personally i think it's when she says:
"There is a value for x that makes the denominator zero..."
The reason why i say this is there is no value that they have given you that makes any denominator 0?

Answer 6

True i agree that you cannont divide by 0, but nowhere does it state that the denominator if not known (variable) becomes 0

Answer 7

oh thats true, i didnt catch that.

Answer 8

I have to finish my biology test so you finish this problem up... at least i got you started?

Answer 9

so basically she just has to make the denominator, or picka number that doesnt make the denominator zero? and thank you btw.

Answer 10

**
What is the oblique asymptote of the function f(x) = the quantity x squared minus 5x plus 6 over the quantity x minus 4 ?
**
This is difficult to read when typed out in words.
I assume you mean
\[ f(x) = \frac{x^2 -5x+6}{x-4} \]
if you divide x-4 into the the top (Do you know how to do this ??) you get
x-1 with a remainder of 2. We put the remainder over the divisor x-4, and write f(x) as
\[ f(x) = x -1 + \frac{2}{x-4} \]

notice that when x becomes "large" the fraction 2/(x-4) becomes 2/(big number) example: 2/2000 = 0.001
the fraction approaches zero, and the equation approaches
\[ f(x) = x-1\]
the equation of a line. It is the asymptote
See attached.

Answer 11

If you "solve for f(x)" which usually means find where
f(x)= 0
you have
\[ \frac{x^2 -5x+6}{x-4} = 0 \]
if x is not 4, we can multiply both sides by (x-4) and get
\[ x^2 -5x+6 = 0 \\
(x-2)(x-3) = 0 \\
x= 2 \text{ or } x= 3 \]
the solution does not have x=4 (good!), so they are valid.
It is always good to check with the original equation.

Let's check x=2:
\[ \frac{x^2 -5x+6}{x-4} = 0 \\\text{ with x=2 we get}\\
= \frac{2^2 -5(2)+6}{2-4} \\
= \frac{4 -10+6}{-2} = \frac{0}{-2}= 0
\]
It works. And so will x= 3