Re: Tangent Line in Point..

Rewrite the tangent line $\displaystyle \displaystyle \begin{align*} 2y + 4x - 6 = 0 \end{align*}$ as $\displaystyle \displaystyle \begin{align*} y = -2x + 3 \end{align*}$, 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.

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 $\displaystyle \displaystyle \begin{align*} 2y + 4x - 6 = 0 \end{align*}$ as $\displaystyle \displaystyle \begin{align*} y = -2x + 3 \end{align*}$, 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.

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.

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??

$\displaystyle y'= 6x^2-x+c$

$\displaystyle 6x^2-2x+c+2$

$\displaystyle x'= (2+ \sqrt{-44-24c})/12$ and $\displaystyle x''= (2- \sqrt{-44-24c})/12$