#### Answer

$y=2x+5$, see figure.

#### Work Step by Step

Given the curve as $y=4-x^2$ and a point $(-1,3)$ on the curve, we can find the slope of the tangent line at this point as:
$m=\lim_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}=\lim_{h\to0}\frac{4-(-1+h)^2-(4-(-1)^2)}{h}=\lim_{h\to0}\frac{2h-h^2}{h}=\lim_{h\to0}(2-h)=2$
The equation for the tangent line at $(-1,3)$ can then be written as $y-3=2(x+1)$ or $y=2x+5$, as shown in the figure.