Question

in 1995, the sports league introduced a salary cap that limits the amount of money spent on players salaries. the quadratic model ( y = 0.231x^{2}+2.600x + 35.17 ) approximates this cap in millions of dollars for the years 1995 - 2006, where ( x = 0 ) represents 1995, ( x = 1 ) represents 1996, and so on. complete parts a and b.

a. approximate the sports league salary cap in 2004.

the approximate sports league salary cap in 2004 is ( $77.3 ) million.

(round to the nearest tenth as needed.)

b. according to the model, in what year did the salary cap reach 60 million dollars?

according to the model, in ( square ) the salary cap reached 60 million dollars.

(round down to the nearest year.)

Explanation:

Step1: Substitute \(y = 60\) into the quadratic equation

We have the equation \(60=0.2313x^{2}+2.600x + 35.17\).
Rearrange it to the standard quadratic form \(ax^{2}+bx + c=0\):
\(0.2313x^{2}+2.600x+35.17 - 60=0\)
\(0.2313x^{2}+2.600x - 24.83 = 0\)
Here \(a = 0.2313\), \(b = 2.600\), \(c=- 24.83\)

Step2: Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\)

First, calculate the discriminant \(\Delta=b^{2}-4ac=(2.600)^{2}-4\times0.2313\times(-24.83)\)
\(\Delta = 6.76+22.937\)
\(\Delta=29.697\)

Then \(x=\frac{-2.600\pm\sqrt{29.697}}{2\times0.2313}\)
\(x=\frac{-2.600\pm5.45}{0.4626}\)

We have two solutions for \(x\):
\(x_{1}=\frac{-2.600 + 5.45}{0.4626}=\frac{2.85}{0.4626}\approx6.16\)
\(x_{2}=\frac{-2.600 - 5.45}{0.4626}=\frac{-8.05}{0.4626}\approx - 17.4\)

Since \(x\) represents the number of years after 1995 and \(x\geq0\), we take \(x\approx6.16\)

Step3: Find the year

Since \(x = 0\) represents 1995, and \(x\approx6.16\), rounding down \(x = 6\)
The year is \(1995+6=2001\)

Answer:

2001