# Find the slope of the line tangent to f(x)= x^2 - 9 at x= 4 and plot the point of tangency.

## Question:

Find the slope of the line tangent to {eq}f(x)= x^2 - 9 {/eq} at x= 4 and plot the point of tangency.

## Tangent Line

We can construct the equation of any line as long as we have both a point and a slope. For a tangent line, we can find both of these pieces of information by examining a function at a single point. This is because the point comes from the function itself and the slope from its derivative.

## Answer and Explanation:

Let's find the tangent line to this function at the given point. To do so, we will need to know two things: the value of this function at this point and the value of the derivative to this function at this point.

{eq}\begin{align*} &f(x) = x^2-9 && f(4) = (4)^2 - 9 = 16-9 = 7\\ &f'(x) = 2x && f'(4) = 2(4) = 8 \end{align*} {/eq}

We can use point slope form to find this tangent line by using the value of the derivative (8) as the slope and (4, 7) as the point.

{eq}y-7 = 8(x-4)\\ y = 8x -25 {/eq}

Both the original function and the tangent line is graphed below. Note that the line touches the function with the same instantaneous slope at that point.