Asked by Greg
find an equation of the line tangent to y= sqrt(25-x^2) at the point (3,4)
i got y= x/8 + 29/8
this is wrong. but im not sure why. can you explain thank you
i got y= x/8 + 29/8
this is wrong. but im not sure why. can you explain thank you
Answers
Answered by
Steve
y=√(25-x^2)
y' = -x/√(25-x^2)
y'(3) = -3/4
so, now you have a point and a slope. The line is thus
y-4 = -3/4 (x-3)
y' = -x/√(25-x^2)
y'(3) = -3/4
so, now you have a point and a slope. The line is thus
y-4 = -3/4 (x-3)
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.