Solution — 2023 Question 3 Graphs| Quadratic Equations & Inequalities| Integration & Applications

(a)

The coordinates of the points of intersection are $(2,0) ∎$ and $(6,2) ∎ .$

(b)

\begin{matrix} Area of region \\ = Area under curve - Area of triangle \\ = \int_{2}^{6} \sqrt{x - 2} d x - \frac{1}{2} (6 - 2) (2) \\ = \int_{2}^{6} \sqrt{x - 2} d x - 4 \\ = \int_{2}^{6} {(x - 2)}^{\frac{1}{2}} d x - 4 \\ = {[\frac{2 {(x - 2)}^{\frac{3}{2}}}{3}]}_{2}^{6} - 4 \\ = \frac{2 {(6 - 2)}^{\frac{3}{2}}}{3} - \frac{2 {(2 - 2)}^{\frac{3}{2}}}{3} - 4 \\ = \frac{2 {(4)}^{\frac{3}{2}}}{3} - \frac{2 {(0)}^{\frac{3}{2}}}{3} - 4 \\ = \frac{4}{3} ∎ {units}^{2} \end{matrix}

(c)

Equating $y$ for the curve and the line,

\begin{matrix} \sqrt{x - 2} & = \frac{1}{2} x + k \\ 2 \sqrt{x - 2} & = x + 2 k \\ 4 (x - 2) & = {(x + 2 k)}^{2} \\ 4 x - 8 & = x^{2} + 4 k x + 4 k^{2} \\ x^{2} + (4 k - 4) x + 8 + 4 k^{2} & = 0 \end{matrix}

Given that they intersect, the discriminant must be non-negative

\begin{matrix} b^{2} - 4 a c & \geq 0 \\ {(4 k - 4)}^{2} - 4 (4 k^{2} + 8) (1) & \geq 0 \\ 16 k^{2} - 32 k + 16 - 16 k^{2} - 32 & \geq 0 \\ - 32 k - 16 & \geq 0 \\ 32 k & \leq - 16 \\ k & \leq - \frac{1}{2} ∎ \end{matrix}