1. **Problem 15:** Given $f(x - 1) = (x + 2)^2 - 3$, find the vertex coordinates of $y = f(x - 2) + 1$. 2. First, rewrite $f(x)$ from $f(x - 1)$: $$f(x - 1) = (x + 2)^2 - 3$$ Let $u = x - 1$, then $f(u) = (u + 3)^2 - 3$. So, $$f(x) = (x + 3)^2 - 3$$ 3. Now find $y = f(x - 2) + 1$: $$y = f(x - 2) + 1 = ((x - 2) + 3)^2 - 3 + 1 = (x + 1)^2 - 2$$ 4. The vertex form of a parabola $y = (x - h)^2 + k$ has vertex at $(h, k)$. Here, $y = (x + 1)^2 - 2 = (x - (-1))^2 - 2$, so vertex is at $(-1, -2)$. 5. **Answer for 15:** A. $(-1, -2)$ --- 6. **Problem 16:** Given $f(x) = 3x^2 + kx + 2$, find $k$ if the vertex's $y$-coordinate is $-10$. 7. The vertex $x$-coordinate for $f(x) = ax^2 + bx + c$ is: $$x_v = -\frac{b}{2a} = -\frac{k}{2 \times 3} = -\frac{k}{6}$$ 8. The $y$-coordinate of the vertex is: $$y_v = f(x_v) = 3\left(-\frac{k}{6}\right)^2 + k\left(-\frac{k}{6}\right) + 2 = 3\frac{k^2}{36} - \frac{k^2}{6} + 2 = \frac{k^2}{12} - \frac{k^2}{6} + 2$$ 9. Simplify: $$\frac{k^2}{12} - \frac{2k^2}{12} + 2 = -\frac{k^2}{12} + 2$$ 10. Set $y_v = -10$: $$-\frac{k^2}{12} + 2 = -10$$ $$-\frac{k^2}{12} = -12$$ $$k^2 = 144$$ 11. So, $$k = \pm 12$$ 12. **Answer for 16:** C. $12$ or $-12$