# Calculus

posted by Anonymous

If tangents to the curve defined by y=x^2+4 ln x are parallel to the line defined by by
y-6x+3=0, determine the points where the tangents touch the curve.

y’=2x+4/x

2x+4/x=6

What do I do next?

1. Steve

you got the calculus right -- what bothers you about the algebra?
You have the slope of the parallel line: m=6
So, that means you need
2x + 4/x = 6
2x^2-6x+4 = 0
2(x-1)(x-2) = 0
So the lines are tangent at (1,1) and (2,4+4ln2)

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