1. **State the problem:** Find the equation of the tangent line to the graph of the function $f(x) = 25 - x^2$ at the point $(-5, 0)$. 2. **Recall the formula for the slope of the tangent line:** The slope $m$ at a point $x=a$ is given by the derivative $f'(a)$, which can be found using the limit definition: $$m = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$$ 3. **Calculate the derivative $f'(x)$:** $$f'(x) = \lim_{h \to 0} \frac{(25 - (x+h)^2) - (25 - x^2)}{h} = \lim_{h \to 0} \frac{25 - (x^2 + 2xh + h^2) - 25 + x^2}{h}$$ Simplify the numerator: $$= \lim_{h \to 0} \frac{-2xh - h^2}{h}$$ Cancel $h$ in numerator and denominator: $$= \lim_{h \to 0} \frac{\cancel{h}(-2x - h)}{\cancel{h}} = \lim_{h \to 0} (-2x - h)$$ Taking the limit as $h \to 0$: $$f'(x) = -2x$$ 4. **Evaluate the slope at $x = -5$:** $$m = f'(-5) = -2(-5) = 10$$ 5. **Note the slope given in the problem is $-10$, but our calculation shows $10$.** This suggests a sign error in the problem statement or a misunderstanding. The correct slope from the derivative is $10$. 6. **Find the equation of the tangent line:** The point-slope form is $$y - y_1 = m(x - x_1)$$ Using point $(-5, 0)$ and slope $m=10$: $$y - 0 = 10(x - (-5))$$ $$y = 10(x + 5) = 10x + 50$$ 7. **Final answer:** The equation of the tangent line at $(-5, 0)$ is $$y = 10x + 50$$