1. ## Tangents

Let $\displaystyle f(x) = 12 - x^2$ for $\displaystyle x > 0$ and $\displaystyle f(x)$ is greater than or equal to 0. The line tangent to the graph of $\displaystyle f$ at the point $\displaystyle (k,f(k))$ intercepts the x-axis at $\displaystyle x = 4$. What is the value of k?

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My attempt:

$\displaystyle f'(x) = -2x$

$\displaystyle y = -2(4) + b = 0, b= 8$

$\displaystyle y = -2x +8$ --> tangent line

I know it's wrong because the tangent line is $\displaystyle y = -4x + 16$ and is tangent at $\displaystyle x = 2$.

How would one go about doing it correctly?

2. Sorry for the bump, but I really need help.

3. Consider what you know about the derivative.

You know that you are looking for a tangent line at the point $\displaystyle (k, f(k))$. You also know that the derivative of your function is linear so it will have a constant slope. Consider what is happening between the points $\displaystyle (k,f(k))$ and $\displaystyle (4,0)$ where it intersects the x-axis.

Use what you know about the slope between $\displaystyle (k,f(k))$ and $\displaystyle (4,0)$ and the fact that the derivative represents a slope to help you solve for k.

Hope these hints help.