1. **State the problem:** We have a parabola $g(x) = ax^{2} + q$ opening downwards with $x$-intercepts $R$ and $S(2,0)$, and $y$-intercept $T(0,8)$. A line $f(x) = mx + c$ passes through $R$ and $T$. We need to find the range of $g$, the $x$-coordinate of $R$, values of $a$ and $q$, equation of $f$, solve $f(x) = g(x)$, find $x$ such that $x imes g(x) \\leq 0$, and find the equation of $h$ which is $g$ reflected about $y=0$. 2. **Find the range of $g$: ** Since $g$ opens downwards and passes through $T(0,8)$ which is the $y$-intercept, $q = g(0) = 8$. The vertex is the maximum point, so the range is all $y$ values less than or equal to 8. **Answer 5.1:** Range of $g$ is $(-\infty, 8]$. 3. **Find the $x$-coordinate of $R$: ** Since $R$ and $S$ are $x$-intercepts, $g(R_x) = 0$ and $g(2) = 0$. Using $g(x) = ax^{2} + q$, and $q=8$: $$ a(2)^2 + 8 = 0 \implies 4a + 8 = 0 \implies 4a = -8 \implies a = -2 $$ Now, $g(x) = -2x^{2} + 8$. Find $R_x$ by solving $g(x) = 0$: $$ -2x^{2} + 8 = 0 \implies -2x^{2} = -8 \implies x^{2} = 4 \implies x = \pm 2 $$ Since $S$ is at $x=2$, $R$ must be at $x = -2$. **Answer 5.2:** $x$-coordinate of $R$ is $-2$. 4. **Calculate $a$ and $q$: ** From above, $a = -2$ and $q = 8$. **Answer 5.3:** $a = -2$, $q = 8$. 5. **Determine equation of $f$: ** $f$ passes through $R(-2,0)$ and $T(0,8)$. Slope $m$: $$ m = \frac{8 - 0}{0 - (-2)} = \frac{8}{2} = 4 $$ Equation using point-slope form: $$ y - 0 = 4(x + 2) \implies y = 4x + 8 $$ **Answer 5.4:** $f(x) = 4x + 8$. 6. **Solve $f(x) = g(x)$: ** Set $4x + 8 = -2x^{2} + 8$: $$ 4x + 8 = -2x^{2} + 8 \implies 4x = -2x^{2} \implies 2x^{2} + 4x = 0 \implies 2x(x + 2) = 0 $$ Solutions: $$ x = 0 \text{ or } x = -2 $$ **Answer 5.5.1:** $x = 0$ or $x = -2$. 7. **Find $x$ such that $x imes g(x) \\leq 0$: ** Recall $g(x) = -2x^{2} + 8$. Find where $g(x) = 0$: $$ -2x^{2} + 8 = 0 \implies x^{2} = 4 \implies x = \pm 2 $$ Analyze sign of $g(x)$: - For $|x| < 2$, $g(x) > 0$ (since parabola opens downwards and vertex at $y=8$). - For $|x| > 2$, $g(x) < 0$. Analyze $x imes g(x) \\leq 0$: - When $x > 0$ and $g(x) \\leq 0$, $x > 0$ and $g(x) \\leq 0$ means $x > 0$ and $x \geq 2$ (since $g(x) \\leq 0$ outside $[-2,2]$), so $x \geq 2$. - When $x < 0$ and $g(x) \\geq 0$, $x < 0$ and $g(x) \\geq 0$ means $x < 0$ and $x \in (-2,2)$, so $x \in [-2,0)$. Combine: $$ x imes g(x) \\leq 0 \implies x \in [-2,0] \cup [2, \infty) $$ **Answer 5.5.2:** $x \in [-2,0] \cup [2, \infty)$. 8. **Equation of $h$ (reflection of $g$ about $y=0$): ** Reflection about $y=0$ changes $y$ to $-y$: $$ h(x) = -g(x) = -(-2x^{2} + 8) = 2x^{2} - 8 $$ **Answer 5.6:** $h(x) = 2x^{2} - 8$.