1. **Problem statement:** We have two functions: an exponential function $f(x) = a^x + q$ with horizontal asymptote $y=3$, and a linear function $g(x) = mx + c$. They intersect at point $A(2,12)$ and share the same $y$-intercept. 2. **Show that $a=3$ and $q=3$:** - Since $y=3$ is the horizontal asymptote of $f(x)$, as $x \to -\infty$, $f(x) \to q = 3$. - So, $q=3$. - At point $A(2,12)$, $f(2) = a^2 + q = 12$. - Substitute $q=3$: $a^2 + 3 = 12 \implies a^2 = 9 \implies a = 3$ (taking positive root since exponential base is positive). 3. **Determine the equation of $g$:** - $g$ shares the same $y$-intercept as $f$, so $g(0) = f(0)$. - Calculate $f(0) = a^0 + q = 1 + 3 = 4$, so $c = 4$. - $g(2) = m(2) + 4 = 12$ (since $A(2,12)$ lies on $g$). - Solve for $m$: $2m + 4 = 12 \implies 2m = 8 \implies m = 4$. - Therefore, $g(x) = 4x + 4$. 4. **Determine length of $DC$:** - Point $D$ is the $y$-intercept of $g$, so $D = (0,4)$. - Point $C$ is the projection of $A$ onto the $x$-axis, so $C = (2,0)$. - Length $DC = \sqrt{(2-0)^2 + (0-4)^2} = \sqrt{4 + 16} = \sqrt{20} = 2\sqrt{5} \approx 4.47$. 5. **Summary:** - $a=3$, $q=3$. - $g(x) = 4x + 4$. - Length $DC \approx 4.47$. 6. **For 4.2, graphing $p(x) = -x^2 + 4$ and $q(x) = \frac{4}{x} - 2$:** - $p(x)$ is a downward parabola with vertex at $(0,4)$. - $q(x)$ has vertical asymptote at $x=0$ and horizontal asymptote at $y=-2$. - Intercepts and asymptotes should be clearly marked on the graph.