Question

How do you find the equation of the tangent line to the curve `f(x)=(x-2)^8` at the point where x=3?

Answer 1

`y=8x-23`

Explanation:

The first step in finding tangent lines is to find the slope, which can be done by taking the derivative and evaluating it at some `x`-value (in this case, we will evaluate `f'(x)` at `x=3`):
`f(x)=(x-2)^8`
`f'(x)=8(x-2)^7->` using power rule
`f'(1)=8((3)-2)^7=8(1)^7=8`

This is the slope of the tangent line.

Now, we can find the actual equation. Tangent line equations are of the form `y=mx+b`, where `x` and `y` are points on the line, `m` is the slope, and `b` is the `y`-intercept. We have the slope, `8`, and we can find `x` and `y` by evaluating `f(3)`:
`f(3)=((3)-2)^8=(1)^8=1`

We have just identified the point `(3,1)`. We will now use `m`, `x`, and `y` to find `b`:
`1=(8)(3)+b`
`1=24+b`
`-23=b`

Since this is the `y`-intercept, we can say our equation is `y=8x-23`.