Question

How do you find the slope of a tangent line to the graph of the function `f(x)=(x^3 + 1)(x^2 -2)` at (2, 18)?

Answer 1

60x-y-102=0. See the tangent-inclusive graph.

Explanation:

The slope of the tangent at P(2, 18) is

`f'=5x^4-6x^2+2x=60`

So, the equation of the tangent at P is

#y-18=60(x-2) giving

`60x-y-102=0`

graph{(x^5-2x^3+x^2-2-y)(60x-y-102)=0 [-3, 3, -20, 20]}