In terms of k, what is the y-intercept of the line tangent to the curve f(x)=4x−x 2 at the point on the curve where x=k?
+33666 See the following reasoning:
$$The curve: $f(x)=4x-x^2$ has gradient $4-2x$\\
At the point $x=k$ the gradient is $4-2k$\\
The straight line that is tangent to the curve at $x=k$ therefore has the form $y=(4-2k)x+c$ where $c$ is the y-intercept.\\
At the tangent point the values of $y$ and $f(x=k)$ must be the same so $$(4-2k)k+c=4k-k^2$$
Rearranging this we find that, $c$, the y-intercept of the straight line is $$c=k^2$$$$
+33666 See the following reasoning:
$$The curve: $f(x)=4x-x^2$ has gradient $4-2x$\\
At the point $x=k$ the gradient is $4-2k$\\
The straight line that is tangent to the curve at $x=k$ therefore has the form $y=(4-2k)x+c$ where $c$ is the y-intercept.\\
At the tangent point the values of $y$ and $f(x=k)$ must be the same so $$(4-2k)k+c=4k-k^2$$
Rearranging this we find that, $c$, the y-intercept of the straight line is $$c=k^2$$$$
