Question

the data in the table represents a companys profit based on the number of items produced.

profit based on number of items produced
\

$$\begin{array}{|c|c|} \\hline \\text{items produced } (x) & \\text{dollars of profit } (y) \\\\ \\hline 100 & -70,500 \\\\ \\hline 200 & 50 \\\\ \\hline 300 & 50,100 \\\\ \\hline 400 & 80,300 \\\\ \\hline 500 & 90,400 \\\\ \\hline 600 & 78,000 \\\\ \\hline \\end{array}$$

which equation best represents the data?

\\(y = -1.026x^2 + 1016.402x - 162075\\)
\\(y = -1.036x^2 + 1024.771x - 163710\\)
\\(y = 298.214x - 66317.667\\)
\\(y = 196.2x - 18710\\)

Explanation:

Analyze the trend of the data

The data points \((x, y)\) are:

• \((100, -70500)\)
• \((200, 50)\)
• \((300, 50100)\)
• \((400, 80300)\)
• \((500, 90400)\)
• \((600, 78000)\)

The profit \(y\) increases from \(x = 100\) to \(x = 500\) and then decreases at \(x = 600\). This indicates a downward-opening quadratic relationship, so the equation must be of the form \(y = ax^2 + bx + c\) with \(a < 0\).

Test the first quadratic equation

Substitute \(x = 300\) into \(y = -1.026x^2 + 1016.402x - 162075\):

$$
LATEXBLOCK0
$$

This is extremely close to the table value of \(50100\).

Test the second quadratic equation

Substitute \(x = 300\) into \(y = -1.036x^2 + 1024.771x - 163710\):

$$
LATEXBLOCK1
$$

Now substitute \(x = 100\) into both equations to compare:
For the first equation:

$$
LATEXBLOCK2
$$

For the second equation:

$$
LATEXBLOCK3
$$

The first equation provides a significantly closer fit to the data.

Answer:

• (A) \(y = -1.026x^2 + 1016.402x - 162075\) (Correct answer)
• (B) \(y = -1.036x^2 + 1024.771x - 163710\)
• (C) \(y = 298.214x - 66317.667\)
• (D) \(y = 196.2x - 18710\)