Calc
posted by Britt .
Find the equation of a tangent line of a graph pf y=x^2+4lnx, which is parallel to y=6x3?
I know parallel lines have the same slope so the slope is 6 but I do not know how to find the b value.

Well you need to find where on that nasty function the slope is 6. Then at that (x,y) you write the y = 6 x + b that goes through that point.

how do I do this?

y = x^2 + 4 ln x
dy/dx = slope = 2 x +4/x
when is that slope 6?
6 = 2 x + 4/x
I get at x = 1 and at x = 2
well 1 is easy because ln 1 = 0. It just said "a" tangent line, not all tangent lines.
then y = 1
so I am going to use (1,1) 
I understand what you are saying but how do you do the algebra foe 6=2x+4/x I seem to have forgotten my algebra.

actually it's not so bad
the derivative of your function is
2x + 4/x
setting this equal to 6 gave me a quadratic which factored nicely and had roots of
x = 1 or x=2
if x=1 then y = 1+4ln1
y=1 ,
so one case is m=6, point (1,1)
case 2
x=2, then y = 4 + 4ln2
so m=6, point (2,4+4ln2)
etc 
All set now?

No i need help understanding the algebra behind solving 2x+4/x=6

6=2x+4/x
multiply both sides by x
6 x = 2 x^2 + 4
2 x^2  6 x + 4 = 0
1 x^2  3 x + 2 = 0
factor
(x1)(x2) = 0
x = 1
x = 2 are solutions.
I used 1 so the point I could take was (1,1)
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