
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 xaxis at $\displaystyle x = 4$. What is the value of k?

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)

Sorry for the bump, but I really need help.

1 Attachment(s)
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 xaxis.
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.