1. **Problem 1: Find the slant asymptote of** $k(x) = \frac{3x^2 + 5x - 26}{x + 4}$. 2. To find the slant asymptote of a rational function where the degree of the numerator is exactly one more than the degree of the denominator, perform polynomial long division. 3. Divide $3x^2 + 5x - 26$ by $x + 4$: $$\begin{aligned} &\quad 3x - 7 \\ x + 4 &\overline{) 3x^2 + 5x - 26} \\ &- (3x^2 + 12x) \\ \hline &\quad\quad -7x - 26 \\ &- (-7x - 28) \\ \hline &\quad\quad\quad 2 \end{aligned}$$ 4. The quotient is $3x - 7$ and the remainder is 2, so the slant asymptote is: $$y = 3x - 7$$ 5. **Answer for Problem 1:** Option B: $y = 3x - 7$. 6. **Problem 2: Identify the transformations from** $h(x)$ **to** $k(x) = -h(x + 4) + 3$. 7. The function $k(x) = -h(x + 4) + 3$ means: - Inside the function argument, $x + 4$ means a horizontal translation left by 4 units. - The negative sign outside, $-h(\cdot)$, means a reflection over the x-axis. - The $+3$ outside means a vertical translation up by 3 units. 8. So the transformations are: - Reflection over the x-axis - Horizontal translation by -4 (left 4 units) - Vertical translation by 3 (up 3 units) 9. **Answer for Problem 2:** Option C. **Final answers:** - Slant asymptote: $y = 3x - 7$ - Transformations: Reflection over x-axis, horizontal translation by -4, vertical translation by 3