1. **State the problem:** The light intensity $I$ varies inversely with the square of the distance $d$ from the projector, so $I = \frac{k}{d^2}$ where $k$ is a constant. 2. **Given:** $k = 16$, when $d = 3$, $I = 24$ units. 3. **Find $k$ using the given values:** $$I = \frac{k}{d^2} \implies 24 = \frac{k}{3^2} = \frac{k}{9}$$ Multiply both sides by 9: $$k = 24 \times 9 = 216$$ 4. **a) Find intensity when $d = 6$:** $$I = \frac{216}{6^2} = \frac{216}{36} = 6$$ 5. **b) Effect when distance is halved:** If $d$ becomes $\frac{d}{2}$, then $$I_{new} = \frac{216}{(\frac{d}{2})^2} = \frac{216}{\frac{d^2}{4}} = 4 \times \frac{216}{d^2} = 4I$$ So, the intensity becomes 4 times greater when distance is halved. 6. **c) Effect when distance increases by 25%:** New distance: $$d_{new} = 1.25d$$ New intensity: $$I_{new} = \frac{216}{(1.25d)^2} = \frac{216}{1.5625 d^2} = \frac{1}{1.5625} I = 0.64 I$$ Intensity decreases to 64% of original. 7. **d) Graph description:** The function is $$I = \frac{216}{d^2}$$ which is a rectangular hyperbola in the first quadrant, steeply decreasing as $d$ increases. **Final answers:** - a) $I = 6$ units at $d=6$ - b) Intensity quadruples when distance is halved - c) Intensity decreases to 64% when distance increases by 25%