A line with equation intersects the axes at the points P and Q. A parabola of the form , where …

The line is $y = -3x + 9$. Setting $x = 0 \Rightarrow y = 9$, so $P(0, 9)$. Setting $y = 0 \Rightarrow -3x + 9 = 0 \Rightarrow x = 3$, so $Q(3, 0)$. The parabola is $y = ax^2 + c$. Since it passes through $P(0, 9)$: $9 = a(0)^2 + c \Rightarrow c = 9$. Since it passes through $Q(3, 0)$: $0 = a(3)^2 + 9 \Rightarrow 9a + 9 = 0 \Rightarrow a = -1$. Therefore, the equation of the parabola is $y = -x^2 + 9$.

The shaded region is bounded between the parabola $y = 9 - x^2$ and the line $y = 9 - 3x$ from $x = 0$ to $x = 3$. Notice that for $x \in [0, 3]$, $9 - x^2 \ge 9 - 3x$. $\text{Area} = \int_0^3 [(9 - x^2) - (9 - 3x)] dx = \int_0^3 (3x - x^2) dx$ $= \left[\frac{3x^2}{2} - \frac{x^3}{3}\right]_0^3 = \frac{3(9)}{2} - \frac{27}{3} = \frac{27}{2} - 9 = \frac{9}{2} \text{ sq. units}$.