Asked by 645
Which graph represents the function f (x) = StartFraction 3 x minus 2 Over x minus 2 EndFraction?
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 2, and the horizontal asymptote is at y = 0.
On a coordinate plane, a hyperbola is shown. One curve opens up and to the left in quadrant 2, and the other curve opens down and to the left in quadrant 4. A vertical asymptote is at x = 1, and the horizontal asymptote is at y = 1.
On a coordinate plane, a hyperbola is shown. One curve opens up and to the left in quadrant 2, and the other curve opens down and to the left in quadrant 4. A vertical asymptote is at x = negative 2, and the horizontal asymptote is at y = 4.
On a coordinate plane, a hyperbola is shown. One curve opens up and to the right in quadrant 1, and the other curve opens down and to the left in quadrant 3. A vertical asymptote is at x = 2, and the horizontal asymptote is at y = 3.
All Answers 1
Answered by
GPT-5 mini
AI
Pick the graph with vertical asymptote x = 2 and horizontal asymptote y = 3.
Reason: divide to get f(x) = 3 + 4/(x − 2). So the vertical asymptote is x = 2 and the horizontal asymptote is y = 3. For x > 2, 4/(x−2) > 0 so f(x) > 3 (branch in quadrant I); for x < 2 the 4/(x−2) term is negative so f(x) < 3 and the branch goes to −∞ as x → 2−. Thus the correct choice is the graph described with asymptotes x = 2 and y = 3 (the fourth option).
Reason: divide to get f(x) = 3 + 4/(x − 2). So the vertical asymptote is x = 2 and the horizontal asymptote is y = 3. For x > 2, 4/(x−2) > 0 so f(x) > 3 (branch in quadrant I); for x < 2 the 4/(x−2) term is negative so f(x) < 3 and the branch goes to −∞ as x → 2−. Thus the correct choice is the graph described with asymptotes x = 2 and y = 3 (the fourth option).
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.