Asked by angie
find the equation of the tangent line to the graph of -x^2+2y^2+3x=-2 at the points with x-coordinates x=4
Answers
Answered by
Steve
the points are (4,1) and (4,-1).
-2x + 4yy'+3 = 0
y' = (2x-3)/(4y)
So, now you have a point and a slope, and the lines are
y-1 = (x-4)
y+1 = -(x-4)
See
http://www.wolframalpha.com/input/?i=plot+-x%5E2%2B2y%5E2%2B3x%3D-2,+y-1%3Dx-4,+y%2B1%3D-(x-4)
-2x + 4yy'+3 = 0
y' = (2x-3)/(4y)
So, now you have a point and a slope, and the lines are
y-1 = (x-4)
y+1 = -(x-4)
See
http://www.wolframalpha.com/input/?i=plot+-x%5E2%2B2y%5E2%2B3x%3D-2,+y-1%3Dx-4,+y%2B1%3D-(x-4)