1. The problem is to analyze the function $x(t) = (t - 6)^2 - 9$ which represents profit over time. 2. This is a quadratic function in vertex form, where the vertex is at $(6, -9)$. 3. The parabola opens upwards because the coefficient of the squared term is positive. 4. To find the x-intercepts (times when profit is zero), set $x(t) = 0$: $$ (t - 6)^2 - 9 = 0 $$ $$ (t - 6)^2 = 9 $$ $$ t - 6 = \pm 3 $$ $$ t = 6 \pm 3 $$ So, the roots are $t = 3$ and $t = 9$. 5. The y-intercept (profit at time $t=0$) is: $$ x(0) = (0 - 6)^2 - 9 = 36 - 9 = 27 $$ 6. Summary: - Vertex at $(6, -9)$ is the minimum profit point. - Profit is zero at $t=3$ and $t=9$. - Initial profit at $t=0$ is 27 (hundreds of dollars). Final answer: The parabola has vertex $(6, -9)$, crosses the time axis at $t=3$ and $t=9$, and starts with profit 27 at $t=0$.