Question

How do you find the equation of a line tangent to the function `y=x^3-5x` at x=1?

Answer 1

`y=-2x-2`

Explanation:

Given -

`y=x^3-5x`

Slope of the curve at any point on the curve is given by its
first derivative

`dy/dx=3x^2-5`

Slope of the curve exactly at `x=1`

Substitute `x=1` in the first derivative.

`dy/dx=3(1^2)-5=3-5 = -2`

The slope of the tangent is `m=-2`

The tangent is passing through the point `x=1`

To find the equation of the tangent , we must know the
y-coordinate at point `x=1`

For this substitute `x=1` in the given function

`y=1^3-5(1)=1-5=-4`

Point `(1, -4)` is on the line as well as on the tangent.

We know the slope of the tangent `(m=-2)` and the point
through which it passes `(1, -4)`

`mx+c=y`

`(-2)(1)+c=-4`
`c=-4+2=-2`

Now we have Y- intercept `c=-2` and slope `m=-2`

The equation of the tangent is -

`y=mx+c`

`y=-2x-2`