1. **State the problem:** We want to find the Instantaneous Rate of Change (IROC) of the function $y = \frac{t^2}{4} - 2$ at $t = 15$ without using derivatives. 2. **Recall the formula for IROC:** $$\text{IROC} = \lim_{h \to 0} \frac{y(t+h) - y(t)}{h}$$ This represents the slope of the tangent line at $t=15$. 3. **Calculate $y(15)$:** $$y(15) = \frac{15^2}{4} - 2 = \frac{225}{4} - 2 = 56.25 - 2 = 54.25$$ 4. **Calculate $y(15+h)$:** $$y(15+h) = \frac{(15+h)^2}{4} - 2 = \frac{225 + 30h + h^2}{4} - 2 = \frac{225}{4} + \frac{30h}{4} + \frac{h^2}{4} - 2$$ 5. **Form the difference quotient:** $$\frac{y(15+h) - y(15)}{h} = \frac{\left(\frac{225}{4} + \frac{30h}{4} + \frac{h^2}{4} - 2\right) - 54.25}{h}$$ Since $54.25 = \frac{225}{4} - 2$, substitute: $$= \frac{\cancel{\frac{225}{4}} + \frac{30h}{4} + \frac{h^2}{4} - 2 - \cancel{\left(\frac{225}{4} - 2\right)}}{h} = \frac{\frac{30h}{4} + \frac{h^2}{4}}{h}$$ 6. **Simplify the fraction:** $$= \frac{\cancel{h}\left(\frac{30}{4} + \frac{h}{4}\right)}{\cancel{h}} = \frac{30}{4} + \frac{h}{4}$$ 7. **Take the limit as $h \to 0$:** $$\lim_{h \to 0} \left(\frac{30}{4} + \frac{h}{4}\right) = \frac{30}{4} + 0 = 7.5$$ **Final answer:** The instantaneous rate of change of $y = \frac{t^2}{4} - 2$ at $t=15$ is $7.5$.