Thread: 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?

Rewrite the tangent line as , 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.

But now how can i find the value of "C"?

Originally Posted by Prove It

Rewrite the tangent line as , 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.

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.

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??


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