Tangents

**Cursed** · February 5th 2009, 02:42 PM

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

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My attempt:

$f'(x) = -2x$

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

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

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

How would one go about doing it correctly?

**Cursed** · February 5th 2009, 03:32 PM

Sorry for the bump, but I really need help.

**Jen** · February 5th 2009, 04:11 PM

Consider what you know about the derivative.

You know that you are looking for a tangent line at the point $(k, f(k))$ . 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 $(k,f(k))$ and $(4,0)$ where it intersects the x-axis.

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

Hope these hints help.

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