Slopes and tangent lines

Question

Find an equation for the tangent to the curve at the given point.

$y=4-x^2, (2,5)$

I believe this involves limit functions and derivatives, but I have no idea how to work it out with them. PLEASE HELP.

Answer 1

Assume $y_1=4-x^2$ and $y_2$ is the line(s) we are looking for.

The general form of a line is:

$y-y_0=m(x-x_0)$, where $m$ is the slope and $(x_0,y_0)$ is a given point.

$y_2$ originates from $(2,5)$ so the line's equation is like: $y_2-5=m_2(x-2)$

$y_2$ is tangent to $y_1$ so their slopes should be equal (i.e. be parallel) and coincide (i.e. have same $x$ and $y$), as a result:

If $m_1=m_2$ and $y_1=y_2$ the $x$ founded is where $y_2$ is tangent to $y_1$.

On the other hand, $m_1=\frac{dy_1}{dx}=-2x$

Finally:

$4-x^2=m_2(x-2)+5\to 4-x^2=-2x(x-2)+5 \to x^2-4x-1=0\to x=2\pm\sqrt{5} $

So, we have 2 lines to have the properties we are looking for.

$m_2=-2x|_{x=2+\sqrt{5}}=-4-2\sqrt5 \to y_2=-(4+2\sqrt5)(x-2)+5$

AND

$m_2=-2x|_{x=2-\sqrt{5}}=-4+2\sqrt5 \to y_2=(-4+2\sqrt5)(x-2)+5$

These lines pass $(2,5)$ point, and are tangent to $y=4-x^2$