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Question Find a function with horizontal asymptote at y=5y=5 and vertical asymptote at x=3x=3. Options: a(x)=1x−3+5a(x)=\frac{1}{x-3}+5, b(x)=5x−3b(x)=\frac{5}{x-3}, c(x)=1x+5+3c(x)=\frac{1}{x+5}+3, d(x)=1x+3+5d(x)=\frac{1}{x+3}+5.

Studdy Solution

STEP 1

Assumptions
1. A horizontal asymptote at y=5 y = 5 suggests that as x x approaches infinity, the function approaches the value y=5 y = 5 .
2. A vertical asymptote at x=3 x = 3 suggests that as x x approaches 3, the function's value grows without bound (either positively or negatively).
3. The function must be rational, as rational functions are capable of having both horizontal and vertical asymptotes.

STEP 2

Examine option a, a(x)=1x−3+5 a(x) = \frac{1}{x-3} + 5 .
1. The term 1x−3 \frac{1}{x-3} has a vertical asymptote at x=3 x = 3 because the denominator becomes zero at x=3 x = 3 , which is undefined in real numbers.
2. As x x approaches infinity, the term 1x−3 \frac{1}{x-3} approaches 0, and the function a(x) a(x) approaches 5 5 , which is the horizontal asymptote.

STEP 3

Examine option b, b(x)=5x−3 b(x) = \frac{5}{x-3} .
1. The term 5x−3 \frac{5}{x-3} has a vertical asymptote at x=3 x = 3 .
2. However, as x x approaches infinity, b(x) b(x) approaches 0, not 5, so it does not have a horizontal asymptote at y=5 y = 5 .

STEP 4

Examine option c, c(x)=1x+5+3 c(x) = \frac{1}{x+5} + 3 .
1. The term 1x+5 \frac{1}{x+5} has a vertical asymptote at x=−5 x = -5 , not x=3 x = 3 .
2. As x x approaches infinity, the function c(x) c(x) approaches 3 3 , not 5 5 , so it does not meet the criteria for the horizontal asymptote.

STEP 5

Examine option d, d(x)=1x+3+5 d(x) = \frac{1}{x+3} + 5 .
1. The term 1x+3 \frac{1}{x+3} has a vertical asymptote at x=−3 x = -3 , not x=3 x = 3 .
2. As x x approaches infinity, the function d(x) d(x) approaches 5 5 , but since the vertical asymptote is not at x=3 x = 3 , this function does not meet the criteria.

STEP 6

Determine the correct function.
Based on the analysis in steps 2 through 5, the only function that has both a horizontal asymptote at y=5 y = 5 and a vertical asymptote at x=3 x = 3 is option a, a(x)=1x−3+5 a(x) = \frac{1}{x-3} + 5 .
The function that Sarita examines might be: a(x)=1x−3+5 a(x) = \frac{1}{x-3} + 5

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