a small company shows the profits from their business with the function p(x)= - 0.01x² + 60x - 500, where x…

Explanation:

Step1: Identify the function type

The profit function $P(x)= - 0.01x^{2}+60x - 500$ is a quadratic function in the form $y = ax^{2}+bx + c$, where $a=-0.01$, $b = 60$ and $c=-500$.

Step2: Find the x - value of the vertex

For a quadratic function $y = ax^{2}+bx + c$, the x - value of the vertex (which gives the number of units for maximum profit) is $x=-\frac{b}{2a}$. Substitute $a=-0.01$ and $b = 60$ into the formula: $x=-\frac{60}{2\times(-0.01)}=\frac{-60}{-0.02}=3000$.

Step3: Find the break - even points

To find the break - even points, set $P(x)=0$, so we have the quadratic equation $-0.01x^{2}+60x - 500 = 0$. Multiply through by - 100 to get $x^{2}-6000x + 50000=0$. Using the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a = 1$, $b=-6000$ and $c = 50000$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-6000)^{2}-4\times1\times50000=36000000 - 200000=35800000$. Then $x=\frac{6000\pm\sqrt{35800000}}{2}=\frac{6000\pm100\sqrt{3580}}{2}=3000\pm50\sqrt{3580}\approx3000\pm50\times59.83$. $x_1\approx3000 - 2991.5=8.5$ and $x_2\approx3000 + 2991.5=5991.5$.

Answer:

a. 3000 b. The company shows a profit when the number of units sold $x$ is between approximately 9 and 5991 (rounding the break - even points to the nearest whole numbers).