what is the rate of change of the function described in the table?\n-1 $\frac{1}{10}$\n0 $\frac{1}{2}$\n1…

Explanation:

Step1: Recall rate - of - change formula

The rate of change of a function between two points $(x_1,y_1)$ and $(x_2,y_2)$ is $\frac{y_2 - y_1}{x_2 - x_1}$. Let's take two consecutive points, say $(x_1 = 0,y_1=\frac{1}{2})$ and $(x_2 = 1,y_2=\frac{5}{2})$.

Step2: Calculate the rate of change

Substitute the values into the formula: $\frac{y_2 - y_1}{x_2 - x_1}=\frac{\frac{5}{2}-\frac{1}{2}}{1 - 0}$. First, simplify the numerator: $\frac{5}{2}-\frac{1}{2}=\frac{5 - 1}{2}=\frac{4}{2}=2$. Then, $\frac{\frac{5}{2}-\frac{1}{2}}{1 - 0}=\frac{2}{1}=2$. Let's check with another pair, say $(x_1 = 1,y_1=\frac{5}{2})$ and $(x_2 = 2,y_2=\frac{25}{2})$. Then $\frac{y_2 - y_1}{x_2 - x_1}=\frac{\frac{25}{2}-\frac{5}{2}}{2 - 1}=\frac{\frac{25 - 5}{2}}{1}=\frac{\frac{20}{2}}{1}=10$. Let's use the general formula for an exponential - like function $y = ab^x$. Substitute $(x = 0,y=\frac{1}{2})$ into $y = ab^x$, we get $y(0)=ab^0=a$, so $a=\frac{1}{2}$. Then substitute $(x = 1,y=\frac{5}{2})$ into $y=\frac{1}{2}b^x$, we have $\frac{5}{2}=\frac{1}{2}b^1$, so $b = 5$. The function is $y=\frac{1}{2}\times5^x$. The rate of change between two points $(x_1,y_1)$ and $(x_2,y_2)$ for an exponential function $y = ab^x$ is $\frac{ab^{x_2}-ab^{x_1}}{x_2 - x_1}=a\frac{b^{x_2}-b^{x_1}}{x_2 - x_1}$. In general, for an exponential function $y = ab^x$, the ratio of consecutive $y$ - values gives the common ratio. $\frac{y(1)}{y(0)}=\frac{\frac{5}{2}}{\frac{1}{2}} = 5$, $\frac{y(2)}{y(1)}=\frac{\frac{25}{2}}{\frac{5}{2}} = 5$, $\frac{y(3)}{y(2)}=\frac{\frac{125}{2}}{\frac{25}{2}} = 5$. The rate of change of an exponential function $y = ab^x$ is given by the common ratio $b$ when considering the ratio of consecutive $y$ - values for unit - increment in $x$.

Answer:

5