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In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants a and b, and evaluate the function.
In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an algebraic function N(t) representing the population N of deer over time t.
We let our independent variable t be the number of years after 2006. Thus, the information given in the problem can be written as input-output pairs: (0, 80) and (6, 180). Notice that by choosing our input variable to be measured as years after 2006, we have given ourselves the initial value for the function, a = 80. We can now substitute the second point into the equation [latex]Nleft(tright)=80{b}^{t}[/latex] to find b:
NOTE:Unless otherwise stated, do not round any intermediate calculations. Then round the final answer to four places for the remainder of this section.
The exponential model for the population of deer is [latex]Nleft(tright)=80{left(1.1447right)}^{t}[/latex]. (Note that this exponential function models short-term growth. As the inputs gets large, the output will get increasingly larger, so much so that the model may not be useful in the long term.)
We can graph our model to observe the population growth of deer in the refuge over time. Notice that the graph below passes through the initial points given in the problem, [latex]left(0,text{ 8}0right)[/latex] and [latex]left(text{6},text{ 18}0right)[/latex]. We can also see that the domain for the function is [latex]left[0,infty right)[/latex], and the range for the function is [latex]left[80,infty right)[/latex].
Figure 3. Graph showing the population of deer over time, [latex]Nleft(tright)=80{left(1.1447right)}^{t}[/latex], t years after 2006
A wolf population is growing exponentially. In 2011, 129 wolves were counted. By 2013 the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population N of wolves over time t.
Find an exponential function that passes through the points [latex]left(-2,6right)[/latex] and [latex]left(2,1right)[/latex].
Because we don’t have the initial value, we substitute both points into an equation of the form [latex]fleft(xright)=a{b}^{x}[/latex], and then solve the system for a and b.
Use the first equation to solve for a in terms of b:
Substitute a in the second equation, and solve for b:
Use the value of b in the first equation to solve for the value of a:
Thus, the equation is [latex]fleft(xright)=2.4492{left(0.6389right)}^{x}[/latex].
We can graph our model to check our work. Notice that the graph below passes through the initial points given in the problem, [latex]left(-2,text{ 6}right)[/latex] and [latex]left(2,text{ 1}right)[/latex]. Program for 3d printer. The graph is an example of an exponential decay function.
Figure 4. The graph of [latex]fleft(xright)=2.4492{left(0.6389right)}^{x}[/latex] models exponential decay.
Given the two points [latex]left(1,3right)[/latex] and [latex]left(2,4.5right)[/latex], find the equation of the exponential function that passes through these two points.
Do two points always determine a unique exponential function?
Wondershare mobiletrans 6 6 0 download free. Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in x, which in many real world cases involves time.
Find an equation for the exponential function graphed in Figure 5.
We can choose the y-intercept of the graph, [latex]left(0,3right)[/latex], as our first point. This gives us the initial value, [latex]a=3[/latex]. Next, choose a point on the curve some distance away from [latex]left(0,3right)[/latex] that has integer coordinates. One such point is [latex]left(2,12right)[/latex].
Because we restrict ourselves to positive values of b, we will use b = 2. Substitute a and b into the standard form to yield the equation [latex]fleft(xright)=3{left(2right)}^{x}[/latex].
Find an equation for the exponential function graphed in Figure 6.
Use a graphing calculator to find the exponential equation that includes the points [latex]left(2,24.8right)[/latex] and [latex]left(5,198.4right)[/latex].
Follow the guidelines above. First press [STAT], [EDIT], [1: Edit…], and clear the lists L1 and L2. Next, in the L1 column, enter the x-coordinates, 2 and 5. Do the same in the L2 column for the y-coordinates, 24.8 and 198.4.
Now press [STAT], [CALC], [0: ExpReg] and press [ENTER]. The values a = 6.2 and b = 2 will be displayed. The exponential equation is [latex]y=6.2cdot {2}^{x}[/latex].
Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07).