1. **Problem:** Find the limit $\lim_{x \to -3} h(x)$ using the graph. From the graph, at $x = -3$, $h(x) = 1$. 2. **Problem:** Find the right-hand limit $\lim_{x \to 2^+} h(x)$. From the graph, as $x$ approaches 2 from the right, $h(x)$ approaches 2. 3. **Problem:** Find points of discontinuity of $f(x)$ from the given options. Discontinuities occur where the function is not continuous. Based on the graph and options, the correct points are $-3,1,2$. 4. **Problem:** Calculate $\lim_{x \to 2} \sqrt[3]{\frac{5x + 6}{x}}$. Substitute $x=2$: $$\sqrt[3]{\frac{5(2) + 6}{2}} = \sqrt[3]{\frac{10 + 6}{2}} = \sqrt[3]{8} = 2$$ 5. **Problem:** Calculate $\lim_{x \to -3} \frac{x^2 + 5x + 6}{2x + 6}$. Factor numerator: $$x^2 + 5x + 6 = (x+2)(x+3)$$ Denominator: $$2x + 6 = 2(x+3)$$ Simplify: $$\frac{(x+2)(x+3)}{2(x+3)} = \frac{x+2}{2}, x \neq -3$$ Substitute $x = -3$: $$\frac{-3 + 2}{2} = \frac{-1}{2} = -\frac{1}{2}$$ 6. **Problem:** Given $$f(x) = \begin{cases} x^2 - 2x, & x \leq 3 \\ 2x - 1, & x > 3 \end{cases}$$ Find $\lim_{x \to 3^+} f(x)$. For $x > 3$, use $2x - 1$: $$\lim_{x \to 3^+} f(x) = 2(3) - 1 = 6 - 1 = 5$$ 7. **Problem:** Given $f(x) = x^3 - 3x^2$, find $f'(2)$. Derivative: $$f'(x) = 3x^2 - 6x$$ Evaluate at $x=2$: $$f'(2) = 3(2)^2 - 6(2) = 12 - 12 = 0$$ (Note: None of the options match 0, so likely a typo; closest is 2.) 8. **Problem:** Given $y = \frac{x^5 - 8x^3}{4x}$, find $\frac{dy}{dx}$. Rewrite: $$y = \frac{x^5}{4x} - \frac{8x^3}{4x} = \frac{x^4}{4} - 2x^2$$ Differentiate: $$\frac{dy}{dx} = \frac{4x^3}{4} - 4x = x^3 - 4x$$ 9. **Problem:** Given $f(x) = x^2 - 3x + 6$, find the tangent line at $(4,2)$. Find $f'(x)$: $$f'(x) = 2x - 3$$ Slope at $x=4$: $$m = 2(4) - 3 = 8 - 3 = 5$$ Equation of tangent line: $$y - 2 = 5(x - 4) \Rightarrow y = 5x - 20 + 2 = 5x - 18$$ None of the options match exactly; closest is $y = 7x + 3$ (likely a typo). 10. **Problem:** Given $f(x) = 3x^2 + 3x$, find points where the tangent slope is zero. Derivative: $$f'(x) = 6x + 3$$ Set slope zero: $$6x + 3 = 0 \Rightarrow x = -\frac{1}{2}$$ Find $f(-\frac{1}{2})$: $$3(-\frac{1}{2})^2 + 3(-\frac{1}{2}) = 3(\frac{1}{4}) - \frac{3}{2} = \frac{3}{4} - \frac{3}{2} = -\frac{3}{4}$$ Point is $(-\frac{1}{2}, -\frac{3}{4})$, not listed in options. **Final answers:** 1) a) 1 2) c) 2 3) b) -3,1,2 4) c) 2 5) a) -1/2 6) b) 5 7) (closest) b) 2 8) d) $x^3 - 4x$ 9) (none exact) 10) (none exact)