1. **State the problem:** We have a parabola $g(x) = ax^2 + q$ with x-intercepts $R(x_R,0)$ and $S(2,0)$, and y-intercept $T(0,8)$. A line $f(x) = mx + c$ passes through points $R$ and $T$. We need to find the range of $g$, coordinates of $R$, values of $a$ and $q$, equation of $f$, solve $f(x) = g(x)$, find $x$ such that $x \cdot g(x) \leq 0$, and find the equation of $h$ which is $g$ reflected about $y=0$. 2. **Find the range of $g$:** Since $g$ is a downward-opening parabola (because it passes through two x-intercepts and has a positive y-intercept), its maximum value is at the vertex. The y-intercept is $g(0) = q = 8$, which is the maximum value. The parabola opens downward, so the range is all $y$ such that $$y \leq 8.$$ 3. **Find the x-coordinate of $R$:** Since $R$ and $S$ are the roots of $g(x) = 0$, and $S$ is at $x=2$, let $R = r$. The quadratic can be factored as $$g(x) = a(x - r)(x - 2).$$ 4. **Calculate $a$ and $q$:** Using the y-intercept $T(0,8)$: $$g(0) = a(0 - r)(0 - 2) = a(-r)(-2) = 2ar = 8 \implies 2ar = 8 \implies ar = 4.$$ Also, the quadratic in standard form is $$g(x) = ax^2 - a(r + 2)x + 2ar.$$ Comparing to $ax^2 + q$, the constant term $q = 2ar = 8$ (already known). 5. **Find $r$ and $a$ using vertex form:** The vertex $x$-coordinate is midpoint of roots: $$x_v = \frac{r + 2}{2}.$$ Since the vertex is at $x=0$ (because maximum at $x=0$), $$0 = \frac{r + 2}{2} \implies r = -2.$$ Using $ar = 4$ and $r = -2$: $$a(-2) = 4 \implies a = -2.$$ So, $$a = -2, \quad q = 8.$$ 6. **Equation of $g$:** $$g(x) = -2x^2 + 8.$$ 7. **Equation of $f$:** Line $f$ passes through $R(-2,0)$ and $T(0,8)$. Slope: $$m = \frac{8 - 0}{0 - (-2)} = \frac{8}{2} = 4.$$ Equation: $$y - 0 = 4(x + 2) \implies y = 4x + 8.$$ So, $$f(x) = 4x + 8.$$ 8. **Solve $f(x) = g(x)$:** Set $$4x + 8 = -2x^2 + 8.$$ Simplify: $$4x + 8 - 8 = -2x^2 \implies 4x = -2x^2 \implies 2x^2 + 4x = 0 \implies 2x(x + 2) = 0.$$ Solutions: $$x = 0 \text{ or } x = -2.$$ 9. **Find $x$ such that $x \cdot g(x) \leq 0$:** Recall: $$g(x) = -2x^2 + 8 = -2(x^2 - 4) = -2(x - 2)(x + 2).$$ The roots are at $x = -2$ and $x = 2$. The parabola opens downward, so $g(x) \geq 0$ between $-2$ and $2$, and $g(x) < 0$ outside this interval. We want: $$x \cdot g(x) \leq 0.$$ Analyze intervals: - For $x < -2$: $x$ is negative, $g(x)$ is negative (since outside roots), product positive $>0$ (not satisfy). - For $-2 \leq x \leq 0$: $x$ negative or zero, $g(x) \geq 0$, product $\leq 0$ (satisfy). - For $0 \leq x \leq 2$: $x$ positive, $g(x) \geq 0$, product $\geq 0$ (not satisfy). - For $x > 2$: $x$ positive, $g(x)$ negative, product negative $<0$ (satisfy). So, $$x \cdot g(x) \leq 0 \text{ for } x \in [-2,0] \cup (2, \infty).$$ 10. **Equation of $h$ (reflection of $g$ about $y=0$):** Reflection about $y=0$ changes $y$ to $-y$: $$h(x) = -g(x) = -(-2x^2 + 8) = 2x^2 - 8.$$ So, $$h(x) = 2x^2 - 8,$$ where $p=2$ and $k=-8$. **Final answers:** - Range of $g$: $$y \leq 8.$$ - $x$-coordinate of $R$: $$-2.$$ - $a = -2$, $q = 8.$ - Equation of $f$: $$f(x) = 4x + 8.$$ - Solutions to $f(x) = g(x)$: $$x = 0, -2.$$ - Values of $x$ for $x \cdot g(x) \leq 0$: $$x \in [-2,0] \cup (2, \infty).$$ - Equation of $h$: $$h(x) = 2x^2 - 8.$$