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Graph the function \( f(x)=\left(\frac{1}{5}\right)^{x} \) by moving the key points. You can only graph integer points, for example you cannot plot the point \( \left(-1, \frac{1}{4}\right) \)

Ask by Ruiz Gibbs. in the United States
Feb 11,2025

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Answer

To graph \( f(x) = \left(\frac{1}{5}\right)^x \) using only integer points: 1. **Plot Key Points:** - \( x = 0 \): \( f(0) = 1 \) → (0, 1) - \( x = -1 \): \( f(-1) = 5 \) → (-1, 5) - \( x = -2 \): \( f(-2) = 25 \) → (-2, 25) 2. **Draw the Graph:** - Plot the points (-2, 25), (-1, 5), and (0, 1). - For \( x > 0 \), the function approaches 0 as x increases. - For \( x < 0 \), the function increases rapidly. 3. **Connect the Points:** - Draw a smooth curve through the plotted points. - As x increases beyond 0, the graph approaches the horizontal asymptote y = 0. Summary: - The graph starts at (0, 1) and decreases to the right, approaching y = 0. - To the left of 0, the graph increases rapidly, passing through (-1, 5) and (-2, 25).

Solution

We wish to sketch the graph of   f(x) = (1/5)^x    or equivalently  f(x) = 5^(–x) using a “key‐point” method based on integer coordinates only. (Recall that for an exponential function the rule “multiply by the same fixed number when you move 1 unit horizontally” applies.) Notice that when moving to the right the y‐values are divided by 5, and when moving to the left they are multiplied by 5. Step 1. Choose a starting point. A natural point is x = 0 because   f(0) = (1/5)^0 = 1. So mark the point (0, 1). Step 2. Find points by “shifting” by integer amounts. • Move one unit to the left (x from 0 to –1). In an exponential graph the change in the y‐value is multiplicative. Since x = 0 gave 1, moving left multiplies 1 by 5:   f(–1) = (1/5)^(–1) = 5.   Plot the point (–1, 5). (Both coordinates are integers.) • Move another unit to the left (x from –1 to –2). Now multiply the y‐value 5 by 5:   f(–2) = (1/5)^(–2) = 25.   Plot the point (–2, 25). • Now move one unit to the right from the starting point (x from 0 to 1). Going in the positive direction divides the previous y‐value by 5:   f(1) = (1/5)^1 = 1/5.   While 1/5 is not an integer, you can still mark an approximate location along the grid (if you so desire). (Remember: the problem only allows you to “plot” points with both coordinates integers. In our case, the key “movable” points that land exactly on grid intersections are at x ≤ 0.) • Similarly, at x = 2 we have f(2) = (1/5)^2 = 1/25, which is very close to 0. Step 3. Draw the graph. • Plot the key points with integer coordinates: (–2, 25), (–1, 5), and (0, 1). • Even though f(1) and f(2) are not integer–valued, note that as x increases the function values rapidly drop toward 0. Thus, for x > 0 the graph is very flat, approaching the horizontal asymptote y = 0 (draw a dashed horizontal line along the x–axis to indicate the asymptote). Step 4. Connect the points with a smooth curve. • Starting from the left at (–2, 25), the graph comes steeply downward through (–1, 5) to (0, 1). • For x > 0, the curve continues to decrease smoothly, approaching y = 0 without ever touching it. Summary of Key Points (with available integer coordinates):   (–2, 25), (–1, 5), (0, 1)   and the behavior:   • For x increasing beyond 0, f(x) shrinks toward 0.   • For x decreasing (becoming more negative), f(x) grows very large. Graphing Tip: Even though f(1) and f(2) (and other points for positive x) are not conveniently located on the integer grid, the points for x ≤ 0 give you strong “anchor points” for the graph. Use the rule:   “Moving one unit to the left multiplies f(x) by 5,    and one unit to the right divides f(x) by 5.” This will help you “move” your key points correctly. By following these steps, you have sketched the graph of f(x) = (1/5)^x using only integer key points where possible.

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To graph the function \( f(x)=\left(\frac{1}{5}\right)^{x} \) using key integer points, let's start with a few calculations for integer values of \( x \): - For \( x = -2 \): \( f(-2) = \left(\frac{1}{5}\right)^{-2} = 25 \) (point: \( (-2, 25) \)) - For \( x = -1 \): \( f(-1) = \left(\frac{1}{5}\right)^{-1} = 5 \) (point: \( (-1, 5) \)) - For \( x = 0 \): \( f(0) = \left(\frac{1}{5}\right)^{0} = 1 \) (point: \( (0, 1) \)) - For \( x = 1 \): \( f(1) = \left(\frac{1}{5}\right)^{1} = \frac{1}{5} \approx 0.2 \) (point: \( (1, 0) \)) (not an integer point, so we can skip) - For \( x = 2 \): \( f(2) = \left(\frac{1}{5}\right)^{2} = \frac{1}{25} \approx 0.04 \) (also not an integer point) Thus, the key integer points to graph will be \( (-2, 25) \), \( (-1, 5) \), and \( (0, 1) \). Plot these on a coordinate plane, noting that the function decreases sharply as \( x \) increases, approaching the x-axis but never touching it, creating an exponential decay curve. To better visualize, connect the points smoothly, making sure to reflect the rapid decrease toward the right of the graph while keeping it positive. The y-values become smaller as the x-values increase.

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