# Tangent Line in Point..

• November 6th 2012, 02:37 PM
Chipset3600
Tangent Line in Point..
Hello MHF, i hv a new problemn, the question is:

knowing that 2y+4x-6=0 is the equation which is one of the lines tangent to the curve y= 2x^3-x^2+cx+d.
Find the derivative of this function in one of the points of the curve.

How can i find the common point?
• November 6th 2012, 03:08 PM
Prove It
Re: Tangent Line in Point..
Rewrite the tangent line \displaystyle \begin{align*} 2y + 4x - 6 = 0 \end{align*} as \displaystyle \begin{align*} y = -2x + 3 \end{align*}, so you can see that the tangent line has gradient = -2.

This means the derivative of your curve must be -2. So take the derivative, set it equal to -2, solve for x, then substitute this value into the equation for the curve to find y.
• November 6th 2012, 03:13 PM
Chipset3600
Re: Tangent Line in Point..
But now how can i find the value of "C"?
Quote:

Originally Posted by Prove It
Rewrite the tangent line \displaystyle \begin{align*} 2y + 4x - 6 = 0 \end{align*} as \displaystyle \begin{align*} y = -2x + 3 \end{align*}, so you can see that the tangent line has gradient = -2.

This means the derivative of your curve must be -2. So take the derivative, set it equal to -2, solve for x, then substitute this value into the equation for the curve to find y.

• November 6th 2012, 03:14 PM
Prove It
Re: Tangent Line in Point..
You don't need to find the value of c. It is a parameter. Your point of tangency will be in terms of c and d.
• November 6th 2012, 03:31 PM
Chipset3600
Re: Tangent Line in Point..
Quote:

Originally Posted by Prove It
You don't need to find the value of c. It is a parameter. Your point of tangency will be in terms of c and d.

This would be my X??
$y'= 6x^2-x+c$
$6x^2-2x+c+2$
$x'= (2+ \sqrt{-44-24c})/12$ and $x''= (2- \sqrt{-44-24c})/12$