Below is an input-output table for the function $f(x)$:




$x$
$f ( x )$

0
-3

1
6

2
-9






Complete the table for $g(x)$:



\begin{table}

\captionsetup{labelformat=empty}

\caption{Table 2}


$x$
$g (x)$

0

\begin{tabular}[t]{l}
□

-3
\\

\hline 1 & □ \\

\hline 2 & □ \\

\hline

\end{tabular}

\end{table}

Verified Answer :

  • The points \((0, -3), (1, 6), (2, -9)\) define a unique quadratic function \(f(x) = ax^2 + bx + c\).
  • Solving the system of equations derived from these points, we find \(c = -3\), \(a = -12\), and \(b = 21\).
  • The resulting polynomial function is \(f(x) = -12x^2 + 21x - 3\).
  • The final expression is \(\boxed{f(x) = -12x^2 + 21x - 3}\).

Explanation

  1. Analyzing the Problem Data

The problem provides a table of values for a function \(f(x)\) at three points: \((0, -3)\), \((1, 6)\), and \((2, -9)\). Our objective is to determine a polynomial expression that represents this function. Given three distinct points, we can determine a unique quadratic polynomial of the form \(f(x) = ax^2 + bx + c\) that passes through all three points.

  1. Determining the Function f(x)

To find the constants \(a\), \(b\), and \(c\), we substitute the given points into the general form \(f(x) = ax^2 + bx + c\):

  1. For \(x = 0\), \(f(0) = a(0)^2 + b(0) + c = -3\), which implies \(c = -3\).
  2. For \(x = 1\), \(f(1) = a(1)^2 + b(1) + c = a + b - 3 = 6\), which simplifies to \(a + b = 9\).
  3. For \(x = 2\), \(f(2) = a(2)^2 + b(2) + c = 4a + 2b - 3 = -9\), which simplifies to \(4a + 2b = -6\), or \(2a + b = -3\).

By solving the system of linear equations \(a + b = 9\) and \(2a + b = -3\), we subtract the first equation from the second: \((2a + b) - (a + b) = -3 - 9\), yielding \(a = -12\). Substituting \(a = -12\) back into \(a + b = 9\), we get \(-12 + b = 9\), which gives \(b = 21\). Therefore, the quadratic polynomial is \(f(x) = -12x^2 + 21x - 3\).

  1. Verifying the Result

We now verify the function \(f(x) = -12x^2 + 21x - 3\) for each given value of \(x\):

  • For \(x = 0\), \(f(0) = -12(0)^2 + 21(0) - 3 = -3\). (Matches)
  • For \(x = 1\), \(f(1) = -12(1)^2 + 21(1) - 3 = -12 + 21 - 3 = 6\). (Matches)
  • For \(x = 2\), \(f(2) = -12(2)^2 + 21(2) - 3 = -12(4) + 42 - 3 = -48 + 42 - 3 = -9\). (Matches)

The polynomial accurately models the given table of values.

Examples

Quadratic functions are fundamental in modeling real-world phenomena where rates of change are not constant, such as the path of a projectile. If a ball is thrown into the air, its height at various times follows a quadratic model similar to the one we just derived. By identifying points through observation, engineers and scientists can reconstruct the governing formula to predict future behavior, such as where the object will land or its maximum height, which is essential in physics and sports analytics.