Answer:
The initial (x, y) value of the data is (0, 16)
Step-by-step explanation:
The given data is presented here as follows;
[tex]\left\begin{array}{cc}x&y\\-5&6\\-4&8\\-3&10\\-2&12\\-1&14\end{array}[/tex]
From we data, we have a common first difference in the 'y' values of 2, therefore, we have a straight line graph
The slope of the graph, m = (14 - 6)/(-1 - (-5)) = 2
The equation of the straight line graph in point and slope form is;
y - 14 = 2 × (x - (-1)) = 2 × (x + 1)
Therefore, the equation of the straight line graph in slope and intercept form is given as follows;
y - 14 = 2 × (x + 1)
y = 2·x + 2 + 14 = 2·x + 16
y = 2·x + 16
The initial value of the graph when x = 0 is therefore given as follows;
y = 2·x + 16 = 2 × 0 + 16 = 16
y = 16 when x = 0
The initial value of the graph is (0, 16)
Therefore, the initial value of the given data is x = 0, y = 16.
