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When graphing exponential functions like \(y=250(1.1)^x\) and \(y=200(1.2)^x\), the goal is to visually represent the way these functions grow. Exponential functions have the general form \(f(x) = a\cdot b^x\), where \(a\) is a constant multiplier and \(b\) determines the growth rate. In our exercise, you are given specific points to plot: \(x=1, 2, 3, 4, 5\).
For each of these \(x\)-values, calculate the respective \(y\)-values using the formula. For example, when \(x=1\), the computations are:\

• For \(y=250(1.1)^x\), calculate \(250 \cdot (1.1)^1 = 275\).
• For \(y=200(1.2)^x\), calculate \(200 \cdot (1.2)^1 = 240\).

Once you have all the points, plot them on a graph. Make sure to label each curve clearly so you can distinguish between them. You'll notice that the function with a higher base (in this case, \(1.2\)) rises more steeply compared to the one with a lower base. Graphs are valuable tools in seeing how different functions behave and comparing their growth patterns visually.

Inequalities in mathematics tell us about the relative sizes of two expressions. In this exercise, you need to find the \(x\)-values for which \(250(1.1)^x > 200(1.2)^x\) and when \(250(1.1)^x < 200(1.2)^x\). These inequalities can be visualized by comparing the heights of the two curves on your graph.
As you observe the plotted graph, identify where one curve is higher than the other. The \(x\) values for these segments tell you where one function's value surpasses or falls below the other. For instance, one section of the graph might show where the function \(y=250(1.1)^x\) is greater, meaning for those \(x\)-values, \(y=250(1.1)^x > y=200(1.2)^x\).
To solve these inequalities, you can either use algebra or precise graphing tools. Algebraically, solving them might involve rearranging the terms to form a comparison like solving \((1.1/1.2)^x > (200/250)\). However, when the growth paths do not cross or aren't easily visible, numerical methods and graphing tools can help achieve more accurate results.

Finding the intersection points of two functions involves identifying the \(x\)-values at which they yield the same \(y\)-value. Graphically, these are the points where the two curves meet. Initially, you might estimate these intersection points by looking at where the plotted lines or curves cross each other.
To find precise intersection points, you can set the two functions equal to each other: \(250(1.1)^x = 200(1.2)^x\). By solving this equation, you would typically apply logarithms to isolate \(x\):

• Take natural logarithms of both sides: \(\ln(250) + x\ln(1.1) = \ln(200) + x\ln(1.2)\).
• Rearrange to solve for \(x\): \(x = (\ln(200) - \ln(250)) / (\ln(1.1) - \ln(1.2))\).

This method gives you an exact solution for \(x\). Alternatively, using graphing calculators, many of which have features for computing intersection points, simplifies the task. These tools refine approximations and confirm your algebraic solutions effectively.