In 2009, 80 frogs were introduced into a marsh. By 2015, the population had grown to 180 frogs. Write an algebraic function N(t) representing the population of frogs N over time t.

1 answer

To create an algebraic function \( N(t) \) representing the population of frogs over time, we can assume a linear growth model since we have two specific data points:

  1. In 2009 (which we can take as \( t = 0 \)), the population \( N(0) = 80 \).
  2. In 2015 (which is \( t = 6 \)), the population \( N(6) = 180 \).

Step 1: Find the slope \( m \)

The slope \( m \) of a line tells us how much the population changes per unit time (in this case, per year). The slope can be calculated using the formula:

\[ m = \frac{N(t_2) - N(t_1)}{t_2 - t_1} \]

Using our data points:

\[ m = \frac{180 - 80}{6 - 0} = \frac{100}{6} \approx 16.67 \]

So, the population increases by approximately \( 16.67 \) frogs per year.

Step 2: Write the equation of the line

Using the point-slope form of the equation for a line:

\[ N(t) = mt + b \]

where \( b \) is the y-intercept. We can substitute \( m \) and the value of \( N(0) \):

\[ N(0) = 80 \implies b = 80 \]

Step 3: Write the final function

Putting this all together, we get:

\[ N(t) = \left(\frac{100}{6}\right)t + 80 \]

For simplicity, converting \( \frac{100}{6} \) to a decimal gives approximately \( 16.67 \). Therefore, the function can also be written as:

\[ N(t) \approx 16.67t + 80 \]

Step 4: Proper mathematical representation

However, to maintain precision, you should keep the fraction:

\[ N(t) = \frac{50}{3}t + 80 \]

Final answer

Thus, the algebraic function representing the population of frogs over time \( t \) (where \( t \) is the number of years since 2009) is:

\[ N(t) = \frac{50}{3}t + 80 \]