The amount of money a certain company makes can be represented by the graph $y=x^2-8x-33$, where $x$ is the quantity produced. What is the smallest quantity at which the company can produce to break even or make a profit?
This means I need to solve the equation $x^2-8x-33=0$. I notice that this is a quadratic equation, so I can try to factor it. I look for two numbers that multiply to -33 and add to -8. I see that -11 and 3 work, so I can write $x^2-8x-33=(x-11)(x+3)$. To make this equal to zero, either $x-11=0$ or $x+3=0$. Solving for $x$, I get $x=11$ or $x=-3$. However, $x=-3$ does not make sense in the context of the problem, because the company cannot produce a negative quantity. Therefore, the smallest quantity at which the company can break even or make a profit is $x=11$. 