Questions: For each function, state whether it is linear, quadratic, or exponential. Function 1: - x: 1, y: -56 - x: 2, y: -39 - x: 3, y: -25 - x: 4, y: -14 - x: 5, y: -6 Function 2: - x: 2, y: -9 - x: 3, y: -12 - x: 4, y: -9 - x: 5, y: 0 - x: 6, y: 15 Function 3: - x: 5, y: 5 - x: 6, y: 15 - x: 7, y: 45 - x: 8, y: 135 - x: 9, y: 405 - Linear - Quadratic - Exponential - None of the above

For each function, state whether it is linear, quadratic, or exponential.

Function 1:
- x: 1, y: -56
- x: 2, y: -39
- x: 3, y: -25
- x: 4, y: -14
- x: 5, y: -6

Function 2:
- x: 2, y: -9
- x: 3, y: -12
- x: 4, y: -9
- x: 5, y: 0
- x: 6, y: 15

Function 3:
- x: 5, y: 5
- x: 6, y: 15
- x: 7, y: 45
- x: 8, y: 135
- x: 9, y: 405

- Linear
- Quadratic
- Exponential
- None of the above
Transcript text: For each function, state whether it is linear, quadratic, or exponential. \begin{tabular}{|c|c|c|c|c|c|} \hline \multicolumn{2}{|r|}{Function 1} & \multicolumn{2}{|r|}{Function 2} & \multicolumn{2}{|r|}{Function 3} \\ \hline $x$ & $y$ & $x$ & $y$ & $x$ & $y$ \\ \hline 1 & -56 & 2 & -9 & 5 & 5 \\ \hline 2 & -39 & 3 & -12 & 6 & 15 \\ \hline 3 & -25 & 4 & -9 & 7 & 45 \\ \hline 4 & -14 & 5 & 0 & 8 & 135 \\ \hline 5 & -6 & 6 & 15 & 9 & 405 \\ \hline \multicolumn{2}{|l|}{OLinear} & \multicolumn{2}{|l|}{OLinear} & \multicolumn{2}{|l|}{OLinear} \\ \hline \multicolumn{2}{|l|}{Oquadratic} & \multicolumn{2}{|l|}{Quadratic} & \multicolumn{2}{|l|}{Quadratic} \\ \hline \multicolumn{2}{|l|}{Oexponential} & \multicolumn{2}{|l|}{Exponential} & \multicolumn{2}{|l|}{Exponential} \\ \hline \multicolumn{2}{|l|}{None of the above} & \multicolumn{2}{|l|}{None of the above} & \multicolumn{2}{|l|}{Onone of the above} \\ \hline \end{tabular}
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Solution

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Solution Steps

Step 1: Analyze Function 1

To determine the type of function, we need to analyze the pattern of changes in \(y\) as \(x\) increases.

For Function 1:

  • \(x = 1\), \(y = -56\)
  • \(x = 2\), \(y = -39\)
  • \(x = 3\), \(y = -25\)
  • \(x = 4\), \(y = -14\)
  • \(x = 5\), \(y = -6\)

Calculate the first differences:

  • \(-39 - (-56) = 17\)
  • \(-25 - (-39) = 14\)
  • \(-14 - (-25) = 11\)
  • \(-6 - (-14) = 8\)

Calculate the second differences:

  • \(14 - 17 = -3\)
  • \(11 - 14 = -3\)
  • \(8 - 11 = -3\)

Since the second differences are constant, Function 1 is quadratic.

Step 2: Analyze Function 2

For Function 2:

  • \(x = 2\), \(y = -9\)
  • \(x = 3\), \(y = -12\)
  • \(x = 4\), \(y = -9\)
  • \(x = 5\), \(y = 0\)
  • \(x = 6\), \(y = 15\)

Calculate the first differences:

  • \(-12 - (-9) = -3\)
  • \(-9 - (-12) = 3\)
  • \(0 - (-9) = 9\)
  • \(15 - 0 = 15\)

Calculate the second differences:

  • \(3 - (-3) = 6\)
  • \(9 - 3 = 6\)
  • \(15 - 9 = 6\)

Since the second differences are constant, Function 2 is quadratic.

Step 3: Analyze Function 3

For Function 3:

  • \(x = 5\), \(y = 5\)
  • \(x = 6\), \(y = 15\)
  • \(x = 7\), \(y = 45\)
  • \(x = 8\), \(y = 135\)
  • \(x = 9\), \(y = 405\)

Calculate the ratios of consecutive terms:

  • \(\frac{15}{5} = 3\)
  • \(\frac{45}{15} = 3\)
  • \(\frac{135}{45} = 3\)
  • \(\frac{405}{135} = 3\)

Since the ratios are constant, Function 3 is exponential.

Final Answer

  • Function 1: \(\boxed{\text{Quadratic}}\)
  • Function 2: \(\boxed{\text{Quadratic}}\)
  • Function 3: \(\boxed{\text{Exponential}}\)
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