Sketching Exponential Function

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shazamal

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Hi, I`m totally lost on how to start doing this problem. It`s the domain I`m stuck on, I want to say R but I don`t think that`s right. Any help would be most appreciated!

QUESTION: Sketch the graph of the following problem

f(x)= e^-(x^2-2x+1/2)

a) Write the domain of f(x):

b) Write the vertical asymptotes:

c) Write the horizontal asymptotes:

d) Find the roots of the function and the point which intersects the y-axis:
 
Last edited:
shazamal said:
Hi, I`m totally lost on how to start doing this problem. It`s the domain I`m stuck on, I want to say R but I don`t think that`s right. Any help would be most appreciated!

QUESTION: Sketch the graph of the following problem

f(x)= e^-(x^2-2x+1/2)

a) Write the domain of f(x):

b) Write the vertical asymptotes:

c) Write the horizontal asymptotes:

d) Find the roots of the function and the point which intersects the y-axis:
Click to expand...

Using your graphing calculator - first sketch the function. That will give you some ideas about the answers to those questions.
 
Hello, shazamal!

I'm totally lost on how to start doing this problem.
It's the domain I'm stuck on.
I want to say \(\displaystyle \mathbb{R}\), but I don't think that's right. .Why not?

\(\displaystyle \text{Sketch the graph: }\:f(x)\:=\: e^{-(x^2-2x+\frac{1}{2})} \)

a) Write the domain of \(\displaystyle f(x).\)

b) Write the vertical asymptotes.

c) Write the horizontal asymptotes.

d) Find the roots of the function (x-intercepts)
. . and the point which intersects the y-axis (y-intercept).
Click to expand...

(a) We have: .\(\displaystyle f(x) \:=\:\dfrac{1}{e^{x^2-2x+\frac{1}{2}}} \)

. . Are there any "forbidden" values of \(\displaystyle x\)? . . . No.

. . .The domain is \(\displaystyle \mathbb{R}.\)


(b) Can the denominator ever be zero? . . . No.
. . .There are no vertical asymptotes.


(c) We find that:\(\displaystyle \displaystyle\lim_{x\to\pm\infty}f(x) \:=\:0\)

. . .Horizontal asymptote: .\(\displaystyle y \,=\,0\)


(d) Since \(\displaystyle f(x) \ne 0\), there are no x-intercepts.

. . .\(\displaystyle \text{Since }f(0) \,=\,\dfrac{1}{e^{\frac{1}{2}}}\quad\Rightarrow\quad y\text{-intercept:} \:\left(0,\,\frac{1}{\sqrt{e}}\right)\)
 
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