Question

For `f(x)=-x/3` what is the equation of the tangent line at `x=-3`?

Answer 1

`y-1 = -1/3(x+3)` or `y=-x/3`

Explanation:

First we need to find the slope at the given point so we take the derivative of `f(x)`: `d/dx[f(x) = -x/3] -> f'(x) = -1/3`

This tells us the slope is `-1/3` at any given point, now that we have a slope we need to find a point. The question tells us our `x` value is `-3` so plugging that value into `f(x)` gives us `1`.

So our point is `(-3,1)` and our slope is `-1/3` now we can put this into slope point from which is:

`y-y_1 = m(x-x_1)` where `m` is slope and `(x_1, y_1)`

Plugging all values gives us:

`y-1 = -1/3(x- (-3))` which is also `y=-x/3`, we get the original equation because the graph is linear and so the function can represent the value at any given point since the slope does not change