Question
Determine equations for 2 lines that pass through (-1,-3) and are tangent to the graph of y=x^2+1
Answers
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
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