1. **State the problem:** We need to find the break-even function $t(p)$ relating ticket sales $t$ to price $p$, then solve for the ticket price $p$ that meets the student attendance goal given $n(p) = 600 - 20p$. 2. **Write the break-even function rule:** The break-even point means revenue equals cost. If $t$ is the number of tickets sold and $p$ is the price per ticket, then revenue $R = p \times t$. Assuming the break-even revenue is a fixed amount $R_0$, then: $$R_0 = p \times t(p) \implies t(p) = \frac{R_0}{p}$$ Since the problem does not specify $R_0$, we assume the break-even revenue is the total number of tickets sold times price, so the inverse function is: $$t(p) = \frac{C}{p}$$ where $C$ is a constant representing total revenue needed to break even. 3. **Given $n(p) = 600 - 20p$, solve for $p$ such that $t(p) = n(p)$:** Set: $$t(p) = n(p)$$ $$\frac{C}{p} = 600 - 20p$$ Multiply both sides by $p$: $$C = p(600 - 20p) = 600p - 20p^2$$ Rearranged: $$20p^2 - 600p + C = 0$$ 4. **Solve quadratic for $p$:** $$20p^2 - 600p + C = 0$$ Divide entire equation by 20: $$p^2 - 30p + \frac{C}{20} = 0$$ Use quadratic formula: $$p = \frac{30 \pm \sqrt{(-30)^2 - 4 \times 1 \times \frac{C}{20}}}{2} = \frac{30 \pm \sqrt{900 - \frac{4C}{20}}}{2} = \frac{30 \pm \sqrt{900 - \frac{C}{5}}}{2}$$ 5. **Interpretation:** To find exact $p$, we need $C$, the break-even revenue. Since $C$ is not given, we cannot compute numeric values. 6. **Graph description:** Graph $t(p) = \frac{C}{p}$ and $n(p) = 600 - 20p$ on $p \in [0,50]$, $t,n \in [0,600]$. Intersections are solutions. **Final answers:** - Break-even function: $$t(p) = \frac{C}{p}$$ - Ticket price(s) solving $$\frac{C}{p} = 600 - 20p$$ satisfy quadratic: $$20p^2 - 600p + C = 0$$ - Exact numeric solutions require $C$ value. **Note:** If you provide $C$, I can compute exact $p$ values.