Question

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

Answer 1

`y=-31x+24`

Explanation:

To find the equation of the line tangent to the graph of `f(x) =-3x^5-8x^3+4x^2`, we need a point on the line and the slope of the line.

If the line is to be tangent to the curve at the point where `x=1`, then the tangent line includes `(1,f(1))`.

`f(1) = -3-8+4 = -7`, so the point `(1,-7)` is on the line.

We find the slope of the line using the derivative:

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

At the point where `x=1`, the slope of the tangent line is `f'(1) = -15-24+8--31`

The slope intercept equation of the line through `(1,-7)` with slope `m=-31` is

`y=-31x+24`