1. **State the problem:** We have a quadratic function $g(x) = ax^2 + q$ with x-intercepts $R$ and $S(2,0)$, and y-intercept $T(0,8)$. A linear function $f(x) = mx + c$ passes through points $R$ and $T$. We need to find the equations of $g$ and $f$. 2. **Find $q$ using the y-intercept of $g$:** Since $T(0,8)$ lies on $g$, substitute $x=0$: $$g(0) = a(0)^2 + q = q = 8$$ So, $q = 8$. 3. **Use the x-intercept $S(2,0)$ on $g$:** Substitute $x=2$, $g(2) = 0$: $$0 = a(2)^2 + 8 = 4a + 8$$ Solve for $a$: $$4a = -8 \Rightarrow a = -2$$ 4. **Find the other x-intercept $R$ of $g$:** The quadratic is $$g(x) = -2x^2 + 8$$ Set $g(x) = 0$ to find x-intercepts: $$-2x^2 + 8 = 0 \Rightarrow -2x^2 = -8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$$ We already have $S(2,0)$, so $R$ is at $(-2,0)$. 5. **Find the equation of $f(x) = mx + c$ passing through $R(-2,0)$ and $T(0,8)$:** Calculate slope $m$: $$m = \frac{8 - 0}{0 - (-2)} = \frac{8}{2} = 4$$ Use point-slope form with point $T(0,8)$: $$y - 8 = 4(x - 0) \Rightarrow y = 4x + 8$$ **Final answers:** $$g(x) = -2x^2 + 8$$ $$f(x) = 4x + 8$$