Questions: Function 1 x y 3 13 4 15 5 20 6 28 7 39 Linear Quadratic Exponential None of the above

Function 1 x y 3 13 4 15 5 20 6 28 7 39 Linear Quadratic Exponential None of the above
Transcript text: Function 1 \begin{tabular}{|c|c|} \hline$x$ & $y$ \\ \hline 3 & 13 \\ \hline 4 & 15 \\ \hline 5 & 20 \\ \hline 6 & 28 \\ \hline 7 & 39 \\ \hline \end{tabular} Linear Quadratic Exponential None of the above
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Solution

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Solution Steps

Step 1: Analyze the Data

We are given a table of values for a function:

\[ \begin{array}{|c|c|} \hline x & y \\ \hline 3 & 13 \\ 4 & 15 \\ 5 & 20 \\ 6 & 28 \\ 7 & 39 \\ \hline \end{array} \]

We need to determine whether the function is linear, quadratic, exponential, or none of the above.

Step 2: Check for Linearity

A function is linear if the difference between consecutive \( y \)-values is constant. Let's calculate the differences:

  • \( y(4) - y(3) = 15 - 13 = 2 \)
  • \( y(5) - y(4) = 20 - 15 = 5 \)
  • \( y(6) - y(5) = 28 - 20 = 8 \)
  • \( y(7) - y(6) = 39 - 28 = 11 \)

The differences are not constant, so the function is not linear.

Step 3: Check for Quadratic Nature

A function is quadratic if the second differences of the \( y \)-values are constant. Let's calculate the second differences:

  • First differences: \( 2, 5, 8, 11 \)
  • Second differences:
    • \( 5 - 2 = 3 \)
    • \( 8 - 5 = 3 \)
    • \( 11 - 8 = 3 \)

The second differences are constant, indicating that the function is quadratic.

Final Answer

The function is quadratic. Therefore, the answer is:

\[ \boxed{\text{Quadratic}} \]

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