Question

The line tangent to the graph of function f at the point (8,1) intersects the y-axis at y=3. How do you find f'(8)?

Answer 1

Write the equation of the line tangent to the graph at `(8,1)`. Substitute for `x` and `y`, then solve for `f'(8)`

Explanation:

You have choices for how to write the equation of the tangent line.

Point Slope Form solution

The tangent line contains point `(8,1)` and has slope `f'(8)`, so its equation is

`y-1=f'(8)(x-8)`

The line contains `(0,3)`, so we get

`3-1=f'(8)(0-8)` which leads to

`f'(8) = -1/4`

Slope-Intercept Form solution

The tangent line has slope `f'(8)` and `y`-intercept `3`, so the equation of the tangent line is

`y=f'(8)x+3`

We know that the point `(8,1)` is on the line, so we get

`1=f'(8)*(8)+3`.

This leads to `f'(8) = -1/4`

Answer 2

`f'(8)` is the slope of the tangent line at `x=8`.

Explanation:

We know that the tangent line at `x=8` passes through the points `(8,1)` and `(0,3)`.

The slope of the line that passes through these points is `f'(8)=(3-1)/(0-8)=2/(-8)=-1/4`.