# Curve Sketching: Exponential Functions

• Jan 5th 2008, 02:27 PM
pbfan12
Curve Sketching: Exponential Functions
I'm really having trouble sketching the following curve:

y = e^x^-2

I need help
- finding the x and y intercepts
- how to find if there are any asymptotes
- how to find the critical values for the intervals of inc/dec and concavity

If you could tell me anything, or just start me in the right direction, that would be very helpful. I don't understand the exponential/logarithmic functions very well but my assignment is due in two days and my teacher never gave me an example on how to do it.

• Jan 5th 2008, 02:49 PM
mr fantastic
Quote:

Originally Posted by pbfan12
I'm really having trouble sketching the following curve:

y = e^x^-2

I need help
- finding the x and y intercepts
- how to find if there are any asymptotes
- how to find the critical values for the intervals of inc/dec and concavity

If you could tell me anything, or just start me in the right direction, that would be very helpful. I don't understand the exponential/logarithmic functions very well but my assignment is due in two days and my teacher never gave me an example on how to do it.

Well alright. But there's something I want you to do first. Tell me:

How do you get a y-intercept? Starter: what's the value of x at any y-intercept.
How do you get an x-intercept? Starter: What's the value of y at any x-intercept
How do you find critical points? Starter: It's got something to do with dy/dx .......
What is the definition of concavity? Starter: It's got something to do with $\frac{d^2y}{dx^2}$ .......

As for asymptotes:

Vertical asymptotes are vertical lines passing through the value of x for which your function is not defined. y = e^(-x^2) is defined for all values of x and therefore does not have any vertical asymptotes.

Horizontal asymptotes are the horizontal lines that your curve approaches as x gets really really big either postively or negatively. That is, you look at the value that y approaches as x --> +oo or as x --> -oo.

As x --> +oo, y = e^(-x^2) --> 0. Ditto for x --> -oo. Therefore y = e^(-x^2) has the horizontal asymptote y = 0, that is, the x-axis.

Did you mean $y = e^{-x^2} = \frac{1}{e^{x^2}}$ or $y = e^{1/x^2}$??

If the last, then $y = e^{1/x^2} \rightarrow e^0 = 1$ as $x \rightarrow \pm \infty$ and so the horizontal asymptote is y = 1 ...... And it's not defined for x = 0 .......
• Jan 5th 2008, 02:52 PM
galactus
Here's the sketch. Can you find the asymptotes and domain?.

Where's the horizontal asymptote?. $\lim_{x\rightarrow{\infty}}\frac{1}{e^{x^{2}}}=0$
• Jan 5th 2008, 06:31 PM
mr fantastic
Quote:

Originally Posted by galactus
Here's the sketch. Can you find the asymptotes and domain?.

Where's the horizontal asymptote?. $\lim_{x\rightarrow{\infty}}\frac{1}{e^{x^{2}}}=0$

Ahem ...... galactus, you've interpretted the same rule, $y = \frac{1}{e^{x^{2}}}$, for the function as I did but you've plotted $y = e^{1/x^{2}}$ (and for what it's worth, both the y = 1 horizontal asymptote and the x = 0 (y-axis) vertical asymptote are evident).

In fact, your reply is what prompted my edit :) (which is why I thanked it as a useful post)
• Jan 5th 2008, 08:09 PM
pbfan12
Oh, yes, I did mean, the last one.

Thank you so much mr fantastic and galactus for taking the time to reply to my post, you've help me a lot! If I have any more questions, I know who to ask.
:)