Find the equation of the tangent to the curve x^3-2y-8x-3=0 at the point (1,2)

Answer 1

We will need to grab a few things from our toolbox here. 1 is implicit differentiation and the other is point slope form of a line.

Recall that the derivative can be thought of as the slope of the tangent line.

I will first rearrange tge equation for convenience, so we have

2y = x^3-8x-3.

Now taking the derivative with respect to x and using implicit differentiation (assuming y is a function of x)

2dy/dx = 3x^2 -8

Which means that

dy/dx = (3/2)x^2 -4

This is the slope of the tangent line at every x, however we want to know the slope at the point (1,2), so substitute 1 in for x and we have

dy/dx = (3/2)-4 = -(5/2).

So -(5/2) is the slope.

Now point slope form of a line states that

(y-y1) = m(x-x1)

So substituting 1 in for x and 2 in for y we have

(y-2) = -(5/2)(x-1)

Which means that

(y-2) = -(5/2)x -(5/2)

AND finally

y = -(5/2)x -(1/2)

Which is the equation of the tangent line at the point (1,2)