Grade 11 highschool Math - Rational exponent laws

[From: ] [author: ] [Date: 12-03-13] [Hit: ]

2^(x)] is I.IThere are no horizontal asymptotes because the limit does not exist.No horizontal asymptote approaching I.A horizontal asymptote can potentially be found by finding the limit of the function as the value approaches -I.L[x:-I,2^(x)]The limit of 2^(x) as x approaches -I is 0L[x:-I,......

All real numbers

A horizontal asymptote can potentially be found by finding the limit of the function as the value approaches I.
L[x:I,2^(x)]

The limit of 2^(x) as x approaches I is I
L[x:I,2^(x)]=I

The value of L[x:I,2^(x)] is I.
I

There are no horizontal asymptotes because the limit does not exist.
No horizontal asymptote approaching I.

A horizontal asymptote can potentially be found by finding the limit of the function as the value approaches -I.
L[x:-I,2^(x)]

The limit of 2^(x) as x approaches -I is 0
L[x:-I,2^(x)]=0

The value of L[x:-I,2^(x)] is 0.
0

The horizontal asymptote is the value of y as x approaches -I.
y=0

Since there is no remainder from the polynomial division, there are no oblique asymptotes.
No Oblique Aymptotes

This is the set of all asymptotes for y=2^(x).
No Vertical Asymptotes_Horizontal Aysmptote:y=0_No Oblique Aysmptotes
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Domain: all real numbers
range: y>0

Here is an on-line graphing calculatorto help you with the rest
http://www.meta-calculator.com/online/

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