Find the equation of the tangent to the curve...

Find the equation of the tangent to the curve y= (2+x)(4x-3) which is parallel to the line y=9x+2

I've only just worked out the gradient as m=1/2

Help?

Hello

How can the gradient be = m = 1/2, if the tangent is parallel to the y = 9x+2??
The gradient is m = 9.

y = (2+x)(4x-3) = 4x^2+5x-6
y' = 8x + 5
and this is the gradient of the curve, which is supposed to be 9, because the gradient of the tangent is 9.
So set y' = 9, and find the point (x, y), where the tangent touches the curve:
9 = 8x + 5
--> x = 1/2
put x = 1/2 into equation of the curve for y:
y = 4*(1/2)^2 + 5*1/2 - 6
y = - 2.5
The point where the tangent touches the curve is (1/2, - 2.5)

Now put this point into the equation of the tangent:
y = 9x + b
- 2.5 = 9*1/2 + b
b = - 7
And the equation of the tangent is
y = 9x - 7 <--- ans

Regards