find the equation of the tangent to the curve y=2x^2+3x that is parallel to the line y=14x-6?

Question

its kind of hard, since i have only learned how to work with a point, havent learn the prallel or perpendicular part.
Find the equation of the tangent to the curve y=2x^2+3x that is parallel to the line y=14x-6? this question really beats me. SOME1 Help!!!

Answer 1

ok, first get the derivitive of the curve...4x+3..in order for it to be parallel to y = 14x - 6, set the derivitive equal to the slope of the line, 14. and solve for x... 4x+3 = 14, 4x = 11, x = 11/4. plug this solution into the origional equation to obtain y = 2(11/4)^2 +3(11/3) to obtain y, which is 187/8. so now we know that the line passes through the point (11/4, 187/8) with a slope of 14, now lets solve this linear equation...
y = mx +b
187/8 = 14(11/4) + b
implies b = -121/8
therefor, y = 14x - (121/8)

Answer 2

You need to find the place where the curve has the same slope as the line. Take the derivative of the function.

y = 2x² + 3x
dy/dx = 4x + 3 = 14
4x = 11
x = 11/4

y = 2x² + 3x = 2(11/4)² + 3(11/4) = 187/8

So the point of tangency to the curve is (x,y) = (11/4,187/8).

The equation of the tangent line is:

y - 187/8 = 14(x - 11/4)
y = 14x - 77/2 + 187/8
y = 14x - 121/8