1. The problem gives two height functions of an archer’s arrow, in terms of horizontal distance $x$, and asks to understand the translations modeling the arrow’s paths. 2. For part (a), the function is $f(x) = 1 + \sqrt{3}x + 2$. 3. We simplify $f(x)$ by combining constant terms: $$f(x) = \sqrt{3}x + (1 + 2) = \sqrt{3}x + 3$$ 4. This is a linear function representing a straight, slanting path with slope $\sqrt{3}$ and $y$-intercept $3$. 5. For part (b), the function is written as $f(x) = \left(\frac{1}{2}x - 1\right)^2 + 3$. 6. This is a quadratic function representing a parabola. 7. The base function is $x^2$ and it has undergone transformations: - Horizontal stretch by a factor of 2 (due to $\frac{1}{2}x$ inside the square) - Horizontal shift right by 2 units (because $\frac{1}{2}x - 1 = 0$ when $x = 2$) - Vertical shift upward by 3 units (added $+3$ outside the square) 8. To visualize the translation of the parabola: The vertex is at $x=2$, $y=3$. The parabola opens upward, similar in shape to $x^2$, but wider due to the stretch. Final answers: a) $f(x) = \sqrt{3}x + 3$, a line with slope $\sqrt{3}$ and $y$-intercept 3. b) $f(x) = \left(\frac{1}{2}x - 1\right)^2 + 3$, a parabola with vertex at $(2,3)$ opened upward with horizontal stretch by 2.