QuestionDetermine if the function represented by the points (1, 5), (2, 9), (3, 13), (4, 17), (5, 21) is linear, quadratic, cubic, or exponential.
Studdy Solution
STEP 1
Assumptions1. The given table represents a function with as the input and as the output. . We are to determine whether this function is linear, quadratic, cubic, or exponential.
STEP 2
To determine the type of function, we need to look at the differences between consecutive -values and -values. For a linear function, the differences between -values should be constant when the differences between -values are constant. For a quadratic function, the differences between the differences of -values (second differences) should be constant. For a cubic function, the third differences should be constant. An exponential function would show a constant ratio between -values.
STEP 3
First, let's calculate the differences between consecutive -values.
\begin{align*}
\Delta y1 &= y2 - y1 =9 -5 = \\
\Delta y2 &= y3 - y2 =13 -9 = \\
\Delta y3 &= y - y3 =17 -13 = \\
\Delta y &= y5 - y =21 -17 = \\
\end{align*}
STEP 4
The differences between consecutive -values are all equal to4, which is constant. Therefore, the function could be linear. However, we need to check the other possibilities as well.
STEP 5
Next, let's calculate the second differences to check if the function could be quadratic. The second differences are the differences between consecutive first differences.
\begin{align*}
\Delta^2 y1 &= \Delta y2 - \Delta y1 =4 -4 =0 \\
\Delta^2 y2 &= \Delta y3 - \Delta y2 =4 -4 =0 \\
\Delta^2 y3 &= \Delta y4 - \Delta y3 =4 -4 =0 \\
\end{align*}
STEP 6
The second differences are all equal to0, which is constant. However, a quadratic function would have a non-zero constant second difference. Therefore, the function is not quadratic.
STEP 7
We can skip checking for a cubic function, as a cubic function would have a linear function as a subset, and we already determined that the function could be linear.
STEP 8
Finally, let's check if the function could be exponential. For an exponential function, the ratio between consecutive -values should be constant.
\begin{align*}
\frac{y2}{y1} &= \frac{}{5} =1.8 \\
\frac{y3}{y2} &= \frac{13}{} \approx1.44 \\
\frac{y4}{y3} &= \frac{17}{13} \approx1.31 \\
\frac{y5}{y4} &= \frac{21}{17} \approx1.24 \\
\end{align*}
STEP 9
The ratios between consecutive -values are not constant. Therefore, the function is not exponential.
STEP 10
Based on our analysis, the function is linear because the differences between consecutive -values are constant and the function does not meet the criteria for being quadratic, cubic, or exponential.
The function is linear.
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