Logarithmic Vertical Asymptote (which has asymptote x = -b ?)

[njbeep]

Hi there,

I am hoping for an explanation as to why the answer to the following question is (A).

Question: Which one of the following functions has a graph with a vertical asymptote with
equation x = –b?

(A) y = log₂(x+b) - According to my teacher and the back of the book this is the correct answer.
(B) y = ((1)/(x-b))
(C) y = ((1)/(x-b))-b
(D) y = 2^x-b
(E) y = 2^(x-b)

Through a process of elimination i disregarded C and D because they appear linear. Whe ni plot D and E on my calculator i get vertical asymptotes. When i plot A i get an horizonstal asymptote and my teacher (who is not very good admittedly) has only told me it "probably has to do with x = -b" but she's not entrely sure.

Any insight or explanation would be greatly appreciated. I have attached the original multiple choice question.

Many thanks,
njbeep

 

[tkhunny]

A: x + b = 0 ==> x = -b
B: x - b = 0 ==> x = b
C: x - b = 0 ==> x = b. It's not linear.
D: Horizontal. It's not linear.
E: Horizontal

I think I might be suggesting learning how to use the view window on your calculator.
I have no diagnosis for the erroneous vertical asymptotes for D and E. There are none.

Is this your neighbor trying to home-school you in mathematics? Better get someone competent or go back to the public schools.

 

[mmm4444bot]

… When i plot D and E on my calculator i get vertical asymptotes. When i plot A i get an horizonstal asymptote …

When you graphed those functions, what number(s) did you substitute for symbol b?

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You ought to graph each of the following and compare the results.

The graph of y=log(x) has vertical asymptote x=0.

The graph of y=log(x-3) has vertical asymptote x=3.

The graph of y=log(x+5) has vertical asymptote x=-5.

Have you studied graph shifts?

Subtracting a (positive) constant from x causes the graph to shift to the right, by the same amount. When the graph shifts, so does its vertical asymptote.

Adding a (positive) constant to x causes the graph to shift to the left, by the same amount.

Therefore, the graph of y=log(x+b) will have the same shape as the graph of y=log(x), but it will be shifted b units. This is why the asymptote changes from x=0 to x=-b. (Think of b as a positive constant, to start. Once you understand it, then consider that b is negative and convince yourself that the asymptote is x=-b either way.)