# Thread: Equation of tangent - Help!

1. ## Equation of tangent - Help!

Hi, I have the problem:

'Find the equation of the tangent to the curve with equation $y=x^3 - x^2 - 6x - 2$ at the point where $x=1$.'

I'm sure you'd have to use synthetic division (polynomials) at some point but I am not sure.

Many thanks.

2. Originally Posted by Stepheennnnn
Hi, I have the problem:

'Find the equation of the tangent to the curve with equation $y=x^3 - x^2 - 6x - 2$ at the point where $x=1$.'

I'm sure you'd have to use synthetic division (polynomials) at some point but I am not sure.

Many thanks.
Nope, you just have to differentiate the function.

$y' = 3x^2 - 2x - 6$

edit: I didn't see $x=1$ as a point.

First get the y co-ordinate at the point where $x=1$

$y = 1^3-1^2-6-2 = -8$ to give the ordered pair (1,-8)

$y'(1) = 3(1^2)-2-6 = -5$

The gradient, m, is therefore -5.

Use the equation of a straight line and the ordered pair above to get the equation.

$y = -(5x+13)$

3. Aah, now I get it.
Thank you!

4. Just noticed there's a second part to the question which has just confused me. It says 'Find the coordinates of the point where this tangent meets the curve again.' Any ideas?

Thanks.