Question

How do you find the equation of a line tangent to the function `y=6-x^2` at (2,2)?

Answer 1

The equation of the line is `y = -4x + 10`

Explanation:

The function `6-x^2` is a polynomial, which is definitely differentiable, with `y' = -2x`. That `-2x` is also the slope of the line tangent to the graph of `y` at any given point `x`.

Since we are trying to find the tangent at `(2,2)`, we know that the slope will be `-2 * 2 = -4`. So, the equation of the line is:

`y = -4x + a`, where `a` is the point at which the line intersects the `y` axis. To find `a`, simply plug in the `x` and `y` coordinates of the point:

`2 = -8 + a => a = 10`. The final equation is:

`y = -4x + 10`.