Asked by Kate
Write an equation that would have a hole at x = 4 and a vertical asympote at x = -2.
My answer was f(x)= (x-4)/(x+2)
If it is not correct can someone please help me solve it and explain it to me? Thanks!
My answer was f(x)= (x-4)/(x+2)
If it is not correct can someone please help me solve it and explain it to me? Thanks!
Answers
Answered by
Scott
a hole is a disruption at a single point
... try [(x - 4) / (x - 4)]
... the division by zero causes the discontinuity, but the equal numerator means no other point is affected
f(x) = (x - 4) / [(x - 4) (x + 2)]
... try [(x - 4) / (x - 4)]
... the division by zero causes the discontinuity, but the equal numerator means no other point is affected
f(x) = (x - 4) / [(x - 4) (x + 2)]
Answered by
Kate
Ohhh I see thanks Scott!
Answered by
Kate
Sorry I actually have one more question. After solving for your equation the vertical asymptote would actually be 2 and not -2?? So can you still help me create the correct equation
Answered by
Kate
Would it be f(x) = (x-4) / (x-4)(x+2)???
Answered by
Reiny
Scott was right, the denominator causing the vertical asymptote at x = -2 is
x+2
final answer:
f(x) = (x-4)/( (x-4)(x+2) )
x+2
final answer:
f(x) = (x-4)/( (x-4)(x+2) )
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.