# Tangents

• Feb 5th 2009, 02:42 PM
Cursed
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? (Thinking)
• Feb 5th 2009, 03:32 PM
Cursed
Sorry for the bump, but I really need help.
• Feb 5th 2009, 04:11 PM
Jen
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.

Attachment 9995

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.