Question

which of the following equations most closely models her data?
a ( y = 3x + 7 )
b ( y = 3 + 7 ^ { x } )
c ( y = 3 ( 7 ^ { x } ) )

Explanation:

Step1: Analyze the general trend of the data points

The data points show an exponential - like behavior. A linear function \(y = mx + b\) (where \(m\) is the slope and \(b\) is the y - intercept) has a constant rate of change. The function \(y=3x + 7\) is a linear function. For a linear function, if \(x=- 5\), \(y=3\times(-5)+7=-15 + 7=-8\); if \(x = 0\), \(y=3\times0+7 = 7\); if \(x=5\), \(y=3\times5+7=15 + 7 = 22\). The values from the linear function do not match the trend of the data points which grow (or decay) at a non - constant rate.

Step2: Analyze the function \(y = 3+7^{x}\)

When \(x = 0\), \(y=3 + 7^{0}=3+1=4\). When \(x=-5\), \(y=3+7^{-5}=3+\frac{1}{7^{5}}\approx3\). When \(x = 5\), \(y=3+7^{5}=3 + 16807=16810\). The growth of \(y = 3+7^{x}\) is not in line with the general trend of the data points.

Step3: Analyze the function \(y = 3(7^{x})\)

For an exponential function of the form \(y = a\cdot b^{x}\) (\(a\) is the initial value and \(b>1\) is the base for growth). When \(x=-5\), \(y = 3\times7^{-5}=\frac{3}{7^{5}}\approx\frac{3}{16807}\approx0.00018\). When \(x = 0\), \(y=3\times7^{0}=3\times1 = 3\). When \(x=5\), \(y=3\times7^{5}=3\times16807 = 50421\). The general shape of the exponential function \(y = a\cdot b^{x}\) (in this case \(a = 3\) and \(b = 7\)) is more consistent with the non - linear, exponential - like trend of the data points compared to the linear function \(y=3x + 7\) and the function \(y = 3+7^{x}\).

Answer:

C. \(y = 3(7^{x})\)