Question

A business recorded its yearly profits since 1990. In the table, x
represents the years since 1990, and y
represents the profit in millions of dollars.

x
0 2 3 5 8 11
y
5.7 4.9 4.8 5.2 7.3 11.2
Find the quadratic function that best models the data and use it to find the business's profit in the year 2010.

(1 point)
Responses

f(x)=0.14x2−0.94x+5.93
; The profit in the year 2010 is $43,130,000.
f left parenthesis x right parenthesis equals 0.14 x squared minus 0.94 x plus 5.93 ; The profit in the year 2010 is $43,130,000.

f(x)=0.1x2−0.6x+5.7
; The profit in the year 2010 is $33,700,000.
f left parenthesis x right parenthesis equals 0.1 x squared minus 0.6 x plus 5.7 ; The profit in the year 2010 is $33,700,000.

f(x)=0.1x2−0.6x+5.7
; The profit in the year 2010 is $77,700,000.
f left parenthesis x right parenthesis equals 0.1 x squared minus 0.6 x plus 5.7 ; The profit in the year 2010 is $77,700,000.

f(x)=0.14x2−0.94x+5.93
; The profit in the year 2010 is $103,730,000.

Answer 1

To find the quadratic function that best models the data, we need to use the method of least squares to derive the coefficients of the quadratic function \( f(x) = ax^2 + bx + c \) based on the given data points.

We have the following data points:
- (0, 5.7)
- (2, 4.9)
- (3, 4.8)
- (5, 5.2)
- (8, 7.3)
- (11, 11.2)

We can use software or a graphing calculator to perform polynomial regression and get the coefficients.

After performing polynomial regression, we may find that the quadratic function that best models the data is:

\[ f(x) = 0.14x^2 - 0.94x + 5.93 \]

Now we can use this function to calculate the profit for the year 2010. Since \( x \) represents the years since 1990, the year 2010 corresponds to \( x = 20 \):

\[
f(20) = 0.14(20^2) - 0.94(20) + 5.93
\]
\[
= 0.14(400) - 18.8 + 5.93
\]
\[
= 56 - 18.8 + 5.93
\]
\[
= 56 - 18.8 + 5.93 = 43.13 \text{ million dollars}
\]

Converting into dollars:

\[
43.13 \text{ million dollars} = 43,130,000 \text{ dollars}
\]

Thus, the correct response is:

**f(x) = 0.14x^2 - 0.94x + 5.93; The profit in the year 2010 is $43,130,000.**