Question

data (x,y) for the total number (in thousands) of college - bound students who took a test in the year x are approximately (0,1048), (3,1168), and (17,2027), where x = 0 represents 2000. find the values a, b, and c such that the equation y=ax²+bx + c models these data. according to this model, how many students will take the test in 2023?
a = □
b = □
c = □
(simplify your answers. round the final answer to one decimal place as needed. round all intermediate values to three decimal places as needed.)

Explanation:

Step1: Substitute the points into the equation

We have the equation \(y = ax^{2}+bx + c\). Substituting \((x = 0,y = 1048)\) into the equation:
\(1048=a\times0^{2}+b\times0 + c\), so \(c = 1048\).
Substituting \((x = 3,y = 1168)\) into \(y=ax^{2}+bx + 1048\):
\(1168=a\times3^{2}+b\times3+1048\), which simplifies to \(9a + 3b=1168 - 1048=120\), or \(3a + b = 40\) (dividing by 3).
Substituting \((x = 17,y = 2027)\) into \(y=ax^{2}+bx + 1048\):
\(2027=a\times17^{2}+b\times17+1048\), which simplifies to \(289a+17b=2027 - 1048 = 979\).

Step2: Solve the system of equations

From \(3a + b=40\), we have \(b = 40 - 3a\).
Substitute \(b = 40 - 3a\) into \(289a+17b=979\):

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Substitute \(a\approx1.26\) into \(b = 40 - 3a\):
\(b=40-3\times1.26=40 - 3.78 = 36.22\)

Step3: Predict for \(x = 23\) (since \(2023 - 2000=23\))

The equation is \(y = 1.26x^{2}+36.22x + 1048\).
Substitute \(x = 23\) into the equation:

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Answer:

\(a\approx1.3\), \(b\approx36.2\), \(c = 1048\). The number of students in 2023 is approximately \(2547.6\) (in thousands).