Question

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

Answer 1

`y = -16x + 25`

Explanation:

The tangent is the line that just includes one point of the curve. That line is the first derivative of the curve at that point.
`f(x)=(x−1)^2−3x^2 = x^2 - 2x +1 - 3x^2`
`f(x) = -2x^2 - 2x +1`
`f'(x) = -4x - 2`

at x = 4, `f'(x) = -16` This is the slope of the desired line.

The equation of a line can be written in the form
`(y−y_0)=m(x−x_0)`, where m represents the slope of the line, and `(x_0,y_0)` represents a point on the line.

Our original curve yields the point (4, -39)
`f(4) = -2x^2 - 2x +1 = -39`

The line equation is thus:
`(y− (-39)) = -16(x−4)` ; `y + 39 = -16x + 64`
`y = -16x + 25`