Question

Find the equation of the tangent line to the curve at the given point. \\ y=1+2x-x^3, (1,2)
An explanation would be appreciated.

Answer 1

If the gradient and a point is given, we can find the equation of the line. A point on the tangent line is already given. All you have to find now is the gradient. Apply differentiation to the given curve for that.

Answer 2

I always puzzle over this kind of presentation. Were you given this problem with no explanation whatsoever? Something is seriously wrong with that! I blame your teacher or instructor if you have one - whoever wrote the curriculum if you don't.

Anyway...

y=1+2x-x^3, (1,2)

First, make sure the point is ON the curve.

2 = 1+2(1)-(1)^3 = 1 + 2 - 1 = 2 -- Okay, we're good.

We need the equation of a line passing through (1,2)

y - k = m(x-h) -- In Point-Slope Form

We know the point.

y - 2 = m(x-1)

Where shall we find the slope at x = 1? The answer for now is the 1st Derivative.

y' = 2 -3x^2

Evaluating at x = 1, y'(1) = 2 - 3 = -1 <== This is the required slope.

y - 2 = -(x-1) -- Done!