Before solving the inequality, find the values of $n$ where the profit is exactly $0$ by solving $-3n^{2} + 42n - 120 = 0$. Factoring may help.

Once you know the two $n$-values where the profit is $0$, think about where the quadratic is positive or negative: on the intervals to the left, between, and to the right of those two values. Remember the parabola opens downward because the coefficient of $n^{2}$ is negative.

In Desmos, enter the function y = -3x^2 + 42x - 120. This graph represents profit (in thousands of dollars) as a function of $x$ (which stands in for $n$).

Click on the points where the graph crosses the $x$-axis; Desmos will show their $x$-coordinates. These are the values of $n$ where $P(n) = 0$.

Look at the part of the graph that lies on or above the $x$-axis (where $y \ge 0$). Note the range of $x$-values between the two intercepts where this happens; that interval of $x$-values corresponds to all $n$ that give at least $0$ profit.

"At least $0$ profit" means the profit is $0$ or greater. So we want all values of $n$ such that

This is a quadratic inequality to solve for $n$.

First, find the values of $n$ where the profit is exactly $0$ by solving

Factor out $-3$:

Now factor the quadratic inside the parentheses:

So the equation becomes

which gives the two solutions $n = 4$ and $n = 10$. These are the $n$-values where profit is exactly $0$.

Now go back to the inequality

Divide both sides by $-3$ (a negative number), which reverses the inequality sign:

The product $(n - 4)(n - 10)$ is:

• Positive when $n$ is less than $4$ (both factors negative) and when $n$ is greater than $10$ (both factors positive).
• Negative when $4 < n < 10$ (one factor positive, the other negative).
• Zero at $n = 4$ and $n = 10$.

So $(n - 4)(n - 10) \le 0$ holds for all $n$ between $4$ and $10$, including $4$ and $10$.

From the sign analysis, $P(n) \ge 0$ for all $n$ with $4 \le n \le 10$.

This matches answer choice A) For all $n$ such that $4 \le n \le 10$.