Posted by **Britt** on Sunday, December 9, 2007 at 6:41pm.

Find the equation of a tangent line of a graph pf y=x^2+4lnx, which is parallel to y=6x-3?

I know parallel lines have the same slope so the slope is 6 but I do not know how to find the b value.

- Calc -
**Damon**, Sunday, December 9, 2007 at 6:47pm
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.

- Calc -
**Britt**, Sunday, December 9, 2007 at 6:50pm
how do I do this?

- Calc -
**Damon**, Sunday, December 9, 2007 at 6:54pm
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)

- Calc -
**Reiny**, Sunday, December 9, 2007 at 6:55pm
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

- Calc -
**Damon**, Sunday, December 9, 2007 at 6:55pm
All set now?

- Calc -
**Damon**, Sunday, December 9, 2007 at 7:02pm
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

(x-1)(x-2) = 0

x = 1

x = 2 are solutions.

I used 1 so the point I could take was (1,1)

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