Asked by Calc
Determine equations for 2 lines that pass through (-1,-3) and are tangent to the graph of y=x^2+1
Answers
Answered by
oobleck
For x>0, we have a line though (-1,-3) (h,h^2+1) with slope 2h
That means that
(h^2+1+3)/(h+1) = 2h
h = √5 - 1
So the line is tangent to the curve at (√5-1, 7-2√5)
See the graphs at
https://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2%2B1%2C+y%3D%282%E2%88%9A5-2%29%28x%2B1%29-3+for+x%3D0..2
now do the same for x < 0
That means that
(h^2+1+3)/(h+1) = 2h
h = √5 - 1
So the line is tangent to the curve at (√5-1, 7-2√5)
See the graphs at
https://www.wolframalpha.com/input/?i=plot+y%3Dx%5E2%2B1%2C+y%3D%282%E2%88%9A5-2%29%28x%2B1%29-3+for+x%3D0..2
now do the same for x < 0
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.