Maryann is tracking the change in her vertical jump over 6 months. Use the table to write a linear function that models her jump distance.

\begin{tabular}{|l|l|}
\hline
Month & Vertical Jump (inches) \\
\hline
[tex]$0$[/tex] & [tex]$16$[/tex] \\
\hline
[tex]$2$[/tex] & [tex]$17$[/tex] \\
\hline
[tex]$4$[/tex] & [tex]$18$[/tex] \\
\hline
[tex]$6$[/tex] & [tex]$19$[/tex] \\
\hline
\end{tabular}

A. \( f(x) = \frac{1}{2} x + 16 \)

B. \( f(x) = \frac{1}{2} x + 19 \)

C. \( f(x) = 2x + 16 \)

D. [tex]\( f(x) = 2x + 19 \)[/tex]



Answer :

To determine the linear function that models Maryann's vertical jump over 6 months, we need to find the slope (m) and the y-intercept (b) of the line that fits the provided data points: (0, 16), (2, 17), (4, 18), and (6, 19).

1. Calculate the slope (m):
The slope is calculated using the formula for the slope between two points, \((x_1, y_1)\) and \((x_2, y_2)\):
[tex]\[ m = \frac{y_2 - y_1}{x_2 - x_1} \][/tex]
Let's use the first two data points (0, 16) and (2, 17):
[tex]\[ m = \frac{17 - 16}{2 - 0} = \frac{1}{2} \][/tex]

2. Find the y-intercept (b):
The y-intercept is found by using one of the points and the slope in the linear equation \(y = mx + b\). Using the point (0, 16):
[tex]\[ 16 = \frac{1}{2} \cdot 0 + b \Rightarrow b = 16 \][/tex]

3. Write the linear function:
With the slope and y-intercept determined, the linear function is:
[tex]\[ f(x) = \frac{1}{2}x + 16 \][/tex]

Therefore, the correct linear function that models Maryann's vertical jump over the 6 months is:
[tex]\[ f(x) = \frac{1}{2} x + 16 \][/tex]

Thus, the correct answer is:
[tex]\( f(x)=\frac{1}{2} x+16 \)[/tex]