Question

A company's profits can be modeled by a quadratic function. In the past, it broke even (made zero profit) when it fulfilled 240 or 880 orders. If the maximum profit earned was $204,800, complete the function that models the profits of the company, where x is the number of orders fulfilled, and f(x) is the profit earned.

Answer 1

Answer:

$f(x)=-2(x-560)^2+204,800$

or

$f(x)=-2x^2+2,240x-422,400$

Step-by-step explanation:

Let

f(x) ----> the profit earned

x ---> is the number of orders fulfilled

we know that

The equation of a quadratic equation in vertex form is equal to

$f(x)=a(x-h)^2+k$

where

a is the leading coefficient of the quadratic equation

(h,k) is the vertex

Remember that

The x-intercepts or roots are the values of x when the value of the function is equal to zero

In this problem

The x-intercepts are

x=240 and x=880

The x-coordinate of the vertex (h) is the midpoint of the roots

so

$h=(240+880)/2=560$

The y-coordinate of the vertex (k) is the maximum profit earned

$k=204,800$ ----> is given

so

The vertex is the point (560,204,800)

substitute in the quadratic equation

$f(x)=a(x-560)^2+204,800$

Find the value of a

we have the ordered pairs (240,0) and (880,0) (the x-intercepts)

take the point (240,0) and substitute the value of x and the value of y in the quadratic equation

$0=a(240-560)^2+204,800$

solve for a

$0=a(102,400)+204,800\\a=-2$

therefore

The function that models the profits of the company is given by

$f(x)=-2(x-560)^2+204,800$ ----> equation in vertex form

Convert to standard form

$f(x)=-2(x^2-1,120x+313,600)+204,800$

$f(x)=-2x^2+2,240x-627,200+204,800$

$f(x)=-2x^2+2,240x-422,400$