according to one companys profit model, the company has a profit of 0 when 10 units are sold and a maximum…
according to one companys profit model, the company has a profit of 0 when 10 units are sold and a maximum profit of $18,050 when 105 units are sold. what is the function that represents this companys profit $f(x)$ depending on the number of items sold, $x$?\n$f(x)=-2(x + 105)^2+18,050$\n$f(x)=-2(x - 105)^2+18,050$\n$f(x)=-10(x + 105)^2+18,050$\n$f(x)=-10(x - 105)^2+18,050$
Answer
Explanation:
Step1: Recall vertex - form of quadratic function
The vertex - form of a quadratic function is $f(x)=a(x - h)^2+k$, where $(h,k)$ is the vertex of the parabola. Since the profit has a maximum, $a<0$. The maximum profit occurs at the vertex of the parabola. We know that the maximum profit of $18050$ occurs when $x = 105$, so the vertex of the parabola is $(h,k)=(105,18050)$. So the function has the form $f(x)=a(x - 105)^2+18050$.
Step2: Use the given point to find $a$
We know that when $x = 10$, $f(x)=0$. Substitute $x = 10$ and $f(x)=0$ into $f(x)=a(x - 105)^2+18050$: [ \begin{align*} 0&=a(10 - 105)^2+18050\ 0&=a(-95)^2+18050\ 0&=9025a+18050\
- 18050&=9025a\ a&=- 2 \end{align*} ]
Step3: Write the profit function
Substitute $a=-2$ into $f(x)=a(x - 105)^2+18050$, we get $f(x)=-2(x - 105)^2+18050$.
Answer:
$f(x)=-2(x - 105)^2+18050$