Question: QUESTION 4 Sketched below are the graphs of $f(x) = -a^x + q$ and $g(x) = -x^2 + 2x + 3$. B and C are the x-intercepts of $g$. D, the turning point of $g$, is also a point on the asymptote of $f$. T is a point on $f$ such that $DT$ is parallel to the y-axis. y x O B C D T $f$ $g$ The graph of $f$ is a decreasing exponential curve with horizontal asymptote through $D$, positioned in the top-right/center of the graph; the graph of $g$ is a downward-opening parabola with turning point at $D$ in the top-center, x-intercepts at $B$ on the bottom-left of the origin and $C$ on the bottom-right; $T$ is directly below $D$ on $f$, so $DT$ is a vertical line in the center. 4.1 Calculate the coordinates of $D$. (3) 4.2 Write down the values of $x$ where $g$ is increasing. (1) 4.3 Calculate the x-intercepts of $g$. (3)

1. **State the problem:** We have two functions: $$f(x) = -a^x + q$$ and $$g(x) = -x^2 + 2x + 3$$ We need to find: - 4.1 The coordinates of the turning point $D$ of $g$. - 4.2 The interval where $g$ is increasing. - 4.3 The x-intercepts of $g$. --- 2. **Find the turning point $D$ of $g(x)$:** The turning point of a quadratic function $g(x) = ax^2 + bx + c$ is at $$x = -\frac{b}{2a}$$ For $g(x) = -x^2 + 2x + 3$, we have $a = -1$, $b = 2$, $c = 3$. Calculate $x$ coordinate of $D$: $$x_D = -\frac{2}{2 \times (-1)} = -\frac{2}{-2} = 1$$ Calculate $y$ coordinate of $D$ by substituting $x_D = 1$ into $g(x)$: $$y_D = -1^2 + 2(1) + 3 = -1 + 2 + 3 = 4$$ So, $$D = (1, 4)$$ --- 3. **Find where $g$ is increasing:** Since $g(x)$ is a downward-opening parabola ($a = -1 < 0$), it increases to the left of the turning point and decreases to the right. Therefore, $g$ is increasing for: $$x < 1$$ --- 4. **Calculate the x-intercepts of $g$:** Set $g(x) = 0$: $$-x^2 + 2x + 3 = 0$$ Multiply both sides by $-1$ to simplify: $$\cancel{-}x^2 + \cancel{2x} + \cancel{3} = \cancel{0} \Rightarrow x^2 - 2x - 3 = 0$$ Use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-2$, $c=-3$. Calculate discriminant: $$\Delta = (-2)^2 - 4(1)(-3) = 4 + 12 = 16$$ Calculate roots: $$x = \frac{-(-2) \pm \sqrt{16}}{2(1)} = \frac{2 \pm 4}{2}$$ Two solutions: $$x_1 = \frac{2 + 4}{2} = \frac{6}{2} = 3$$ $$x_2 = \frac{2 - 4}{2} = \frac{-2}{2} = -1$$ So the x-intercepts are: $$B = (-1, 0), \quad C = (3, 0)$$ --- **Final answers:** - 4.1 $D = (1, 4)$ - 4.2 $g$ is increasing for $x < 1$ - 4.3 x-intercepts are $x = -1$ and $x = 3$