1. Asymptote-Mid Terms

Hello,

I have mid terms tomorrow, and i completely blanked on how to do this:

$\displaystyle \frac{x^2-3}{x+2}$

I have to find the domain, range, vertical/horizontal asymptote and discontinuities, if any, what kind; removable to or continuous.

I know the vertical asymptote is -2 but how do i find the horizontal ones. i graphed it and the graph did have multiple horizontal asymptotes.

Thank you.

$\displaystyle \frac{7x}{\sqrt{x^2-10}}$

same things

2. Originally Posted by OnMyWayToBeAMathProffesor
Hello,

I have mid terms tomorrow, and i completely blanked on how to do this:

$\displaystyle \frac{x^2-3}{x+2}$

I have to find the domain, range, vertical/horizontal asymptote and discontinuities, if any, what kind; removable to or continuous.

I know the vertical asymptote is -2 but how do i find the horizontal ones. i graphed it and the graph did have multiple horizontal asymptotes.

Thank you.

$\displaystyle \frac{7x}{\sqrt{x^2-10}}$

same things
Ask yourself this question... is there a horizontal asymptote? Take infinite limits of this function, and you will find that it goes to negative and positive infinity on its ends.

3. We don't even need to take limits, algebra just tells us the answer.

$\displaystyle f(x) = \frac{{x^2 - 3}} {{x + 2}} = \frac{{(x + 2)(x - 2) + 1}} {{x + 2}} = x - 2 + \frac{1} {{x + 2}}.$

We have a slant asymptote, so it doesn't exist horizontal asymptote.

4. Originally Posted by OnMyWayToBeAMathProffesor
Hello,

I have mid terms tomorrow, and i completely blanked on how to do this:

$\displaystyle \frac{x^2-3}{x+2}$

I have to find the domain, range, vertical/horizontal asymptote and discontinuities, if any, what kind; removable to or continuous.

I know the vertical asymptote is -2 but how do i find the horizontal ones. i graphed it and the graph did have multiple horizontal asymptotes.

Thank you.

$\displaystyle \frac{7x}{\sqrt{x^2-10}}$

same things
$\displaystyle y = \frac{7x}{\sqrt{x^2-10}} = \frac{7}{\sqrt{1 - \frac{10}{x^2}}}$ over the domain. What does y do in the limit $\displaystyle x \rightarrow \pm \infty$ ?

as x approaches infinity, y approaches 7. and as x approaches -infinity, y approaches 7. but how? also, how did you get, $\displaystyle y = \frac{7x}{\sqrt{x^2-10}} = \frac{7}{\sqrt{1 - \frac{10}{x^2}}}$

Thank you

6. Originally Posted by Krizalid
We don't even need to take limits, algebra just tells us the answer.

$\displaystyle f(x) = \frac{{x^2 - 3}} {{x + 2}} = \frac{{(x + 2)(x - 2) + 1}} {{x + 2}} = x - 2 + \frac{1} {{x + 2}}.$

We have a slant asymptote [snip]
*Ahem* Krizalid ... how does one know that there's a slant asymptote without considering some sort of limit .....

I know what you mean , but I think it encourages the superficial rote learning of a fact, rather than an understanding of the theory behind that fact .... If a student asked does $\displaystyle y = \frac{1}{x} + 1$, say, have a horizontal asymptote, I would very much hope that the answer they got used the concept of a limit at some stage .....

7. Originally Posted by OnMyWayToBeAMathProffesor
as x approaches infinity, y approaches 7. and as x approaches -infinity, y approaches 7. but how? also, how did you get, $\displaystyle y = \frac{7x}{\sqrt{x^2-10}} = \frac{7}{\sqrt{1 - \frac{10}{x^2}}}$

Thank you
$\displaystyle y = \frac{7x}{\sqrt{x^2-10}} = \frac{7x}{\sqrt{x^2 \left( 1 - \frac{10}{x^2} \right)}} = \frac{7x}{x \sqrt{1 - \frac{10}{x^2}}} = ....$

8. Originally Posted by OnMyWayToBeAMathProffesor
as x approaches infinity, y approaches 7. and as x approaches -infinity, y approaches 7. but how? also, how did you get, $\displaystyle y = \frac{7x}{\sqrt{x^2-10}} = \frac{7}{\sqrt{1 - \frac{10}{x^2}}}$

Thank you
$\displaystyle \frac{10}{x^2} \rightarrow 0$ as $\displaystyle x \rightarrow \pm \infty$. So $\displaystyle y \rightarrow \frac{7}{\sqrt{1 - 0}} = ....$

9. so the numerator basically defines the end behavior? so:

$\displaystyle y \rightarrow \frac{7}{\sqrt{1 - 0}} = ....$ as $\displaystyle x \rightarrow \pm \infty$

10. Originally Posted by OnMyWayToBeAMathProffesor
so the numerator basically defines the end behavior? so:

$\displaystyle y \rightarrow \frac{7}{\sqrt{1 - 0}} =$ 7 as $\displaystyle x \rightarrow \pm \infty$ Mr F edit: added the 7.
Not quite. It only seems that way since the denominator --> 1 .....

To understand what I'm saying, try finding the horizontal asymptote of $\displaystyle y = \frac{7x}{\sqrt{3x^2-10}}$, say ......

11. Originally Posted by OnMyWayToBeAMathProffesor

$\displaystyle \frac{7x}{\sqrt{x^2-10}}$

same things
this question is not as simple as you are all making it out to be

first note that the 10 in the denominator is immaterial as x goes to infinity or negative infinity.

also note, that an alternative definition for |x| is $\displaystyle \sqrt{x^2}$

thus we have: $\displaystyle \lim_{x \to \pm \infty} \frac {7x}{\sqrt{ x^2 - 10}} = \lim_{x \to \pm \infty} \frac {7x}{\sqrt{x^2}} = \lim_{x \to \pm \infty} \frac {7x}{|x|} =$ $\displaystyle \left \{ \begin{array}{lr} 7 & \mbox {as } x \to \infty \\ & \\ -7 & \mbox{ as } x \to - \infty \end{array} \right.$

so there are two horizontal asymptotes here

for the vertical asymptotes, find where the function is undefined, that is, where the denominator is zero. such x's are vertical asymptotoes. these are also the x's that are NOT in the domain (among other x's perhaps, here for instance, we have to find out what x's causes the square root to mess up)

are you good on everything else, MathProfessor?

12. Good get.

Originally Posted by Jhevon
this question is not as simple as you are all making it out to be
[snip]
also note, that an alternative definition for |x| is $\displaystyle \sqrt{x^2}$

thus we have: $\displaystyle \lim_{x \to \pm \infty} \frac {7x}{\sqrt{ x^2 - 10}} = \lim_{x \to \pm \infty} \frac {7x}{\sqrt{x^2}} = \lim_{x \to \pm \infty} \frac {7x}{|x|} =$ $\displaystyle \left \{ \begin{array}{lr} 7 & \mbox {as } x \to \infty \\ & \\ -7 & \mbox{ as } x \to - \infty \end{array} \right.$

so there are two horizontal asymptotes here
Quite right - good pick up.

13. Originally Posted by mr fantastic
$\displaystyle y = \frac{7x}{\sqrt{x^2-10}} = \frac{7x}{\sqrt{x^2 \left( 1 - \frac{10}{x^2} \right)}} = \frac{7x}{x \sqrt{1 - \frac{10}{x^2}}} = ....$
Correction:

$\displaystyle y = \frac{7x}{\sqrt{x^2-10}} = \frac{7x}{\sqrt{x^2 \left( 1 - \frac{10}{x^2} \right)}} = \frac{7x}{|x| \sqrt{1 - \frac{10}{x^2}}} = \frac{7 \, \text{sgn}(x)}{\sqrt{1 - \frac{10}{x^2}}}$

The appropriate corrections in my subsequent posts are easy to see and simple to make.