Question

What is the equation of the line tangent to `f(x)= -3x^5-8x^3+4x^2` at `x=-1`?

Answer 1

`31x+y+38=0`

Explanation:

`f(x)= -3x^5-8x^3+4x^2 implies`

`f^'(x)= -15x^4-24x^2+8x`

We have `f(-1) = -7` and `f^'(-1) = -31`. The tangent that we are seeking is thus a straight line having slope of `-31` going through `(-1,-7)`. So, its equation is

`y-(-7) = -31(x-(-1)) implies`

`31x+y+38=0`