Answer :
To draw the graph of the function [tex]\( f(x) = -\left(\frac{1}{5}\right)^{x-5} \)[/tex], let's follow these steps:
1. Understand the function form:
- Notice that [tex]\(\left(\frac{1}{5}\right)^{x-5}\)[/tex] is an exponential function with base [tex]\(\frac{1}{5}\)[/tex] and it is shifted 5 units to the right.
- The negative sign in front of the exponential changes the function from positive to negative.
2. Identify key points:
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = -\left(\frac{1}{5}\right)^{5-5} = -\left(\frac{1}{5}\right)^0 = -1 \][/tex]
- When [tex]\( x > 5 \)[/tex], since the base [tex]\( \frac{1}{5} \)[/tex] is less than 1, [tex]\(\left(\frac{1}{5}\right)^{x-5}\)[/tex] will get smaller as [tex]\( x \)[/tex] increases. This means [tex]\( f(x) \)[/tex] will approach 0 from the negative side.
- When [tex]\( x < 5 \)[/tex], [tex]\(\left(\frac{1}{5}\right)^{x-5}\)[/tex] becomes larger (because raising a fraction to a negative power yields a value greater than 1, for example [tex]\( \left(\frac{1}{5}\right)^{-1} = 5 \)[/tex]). Thus, [tex]\( f(x) \)[/tex] will become more negative as [tex]\( x \)[/tex] decreases further from 5.
3. Asymptotic behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( \left(\frac{1}{5}\right)^{x-5} \to 0 \)[/tex], therefore [tex]\( f(x) \to 0 \)[/tex]. This indicates a horizontal asymptote at [tex]\( y = 0 \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( \left(\frac{1}{5}\right)^{x-5} \to \infty \)[/tex], so [tex]\( f(x) \to -\infty \)[/tex].
4. Plotting points and drawing the graph:
- Plot the point [tex]\( (5, -1) \)[/tex].
- Choose a few more points around [tex]\( x = 5 \)[/tex] to see how the function behaves. For instance:
[tex]\[ \text{For } x = 4, \quad f(4) = -\left(\frac{1}{5}\right)^{4-5} = -\left(\frac{1}{5}\right)^{-1} = -5 \][/tex]
[tex]\[ \text{For } x = 6, \quad f(6) = -\left(\frac{1}{5}\right)^{6-5} = -\left(\frac{1}{5}\right)^1 = -\frac{1}{5} \][/tex]
[tex]\[ \text{For } x = 3, \quad f(3) = -\left(\frac{1}{5}\right)^{3-5} = -\left(\frac{1}{5}\right)^{-2} = -25 \][/tex]
[tex]\[ \text{For } x = 7, \quad f(7) = -\left(\frac{1}{5}\right)^{7-5} = -\left(\frac{1}{5}\right)^2 = -\frac{1}{25} \][/tex]
5. Sketch the graph:
- Draw the horizontal asymptote [tex]\( y = 0 \)[/tex].
- Plot the calculated points: [tex]\( (4, -5) \)[/tex], [tex]\( (5, -1) \)[/tex], [tex]\( (6, -0.2) \)[/tex], [tex]\( (3, -25) \)[/tex], and [tex]\( (7, -0.04) \)[/tex].
- Connect these points smoothly, remembering that as [tex]\( x \)[/tex] increases, [tex]\( f(x) \)[/tex] gets closer to 0 from below, and as [tex]\( x \)[/tex] decreases, [tex]\( f(x) \)[/tex] becomes very large negatively.
The graph of [tex]\( f(x) = -\left(\frac{1}{5}\right)^{x-5} \)[/tex] will show a rapid decrease as [tex]\( x \)[/tex] moves left from [tex]\( x = 5 \)[/tex], flattening as it moves right from [tex]\( x = 5 \)[/tex], approaching the horizontal asymptote [tex]\( y = 0 \)[/tex].
1. Understand the function form:
- Notice that [tex]\(\left(\frac{1}{5}\right)^{x-5}\)[/tex] is an exponential function with base [tex]\(\frac{1}{5}\)[/tex] and it is shifted 5 units to the right.
- The negative sign in front of the exponential changes the function from positive to negative.
2. Identify key points:
- When [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = -\left(\frac{1}{5}\right)^{5-5} = -\left(\frac{1}{5}\right)^0 = -1 \][/tex]
- When [tex]\( x > 5 \)[/tex], since the base [tex]\( \frac{1}{5} \)[/tex] is less than 1, [tex]\(\left(\frac{1}{5}\right)^{x-5}\)[/tex] will get smaller as [tex]\( x \)[/tex] increases. This means [tex]\( f(x) \)[/tex] will approach 0 from the negative side.
- When [tex]\( x < 5 \)[/tex], [tex]\(\left(\frac{1}{5}\right)^{x-5}\)[/tex] becomes larger (because raising a fraction to a negative power yields a value greater than 1, for example [tex]\( \left(\frac{1}{5}\right)^{-1} = 5 \)[/tex]). Thus, [tex]\( f(x) \)[/tex] will become more negative as [tex]\( x \)[/tex] decreases further from 5.
3. Asymptotic behavior:
- As [tex]\( x \to \infty \)[/tex], [tex]\( \left(\frac{1}{5}\right)^{x-5} \to 0 \)[/tex], therefore [tex]\( f(x) \to 0 \)[/tex]. This indicates a horizontal asymptote at [tex]\( y = 0 \)[/tex].
- As [tex]\( x \to -\infty \)[/tex], [tex]\( \left(\frac{1}{5}\right)^{x-5} \to \infty \)[/tex], so [tex]\( f(x) \to -\infty \)[/tex].
4. Plotting points and drawing the graph:
- Plot the point [tex]\( (5, -1) \)[/tex].
- Choose a few more points around [tex]\( x = 5 \)[/tex] to see how the function behaves. For instance:
[tex]\[ \text{For } x = 4, \quad f(4) = -\left(\frac{1}{5}\right)^{4-5} = -\left(\frac{1}{5}\right)^{-1} = -5 \][/tex]
[tex]\[ \text{For } x = 6, \quad f(6) = -\left(\frac{1}{5}\right)^{6-5} = -\left(\frac{1}{5}\right)^1 = -\frac{1}{5} \][/tex]
[tex]\[ \text{For } x = 3, \quad f(3) = -\left(\frac{1}{5}\right)^{3-5} = -\left(\frac{1}{5}\right)^{-2} = -25 \][/tex]
[tex]\[ \text{For } x = 7, \quad f(7) = -\left(\frac{1}{5}\right)^{7-5} = -\left(\frac{1}{5}\right)^2 = -\frac{1}{25} \][/tex]
5. Sketch the graph:
- Draw the horizontal asymptote [tex]\( y = 0 \)[/tex].
- Plot the calculated points: [tex]\( (4, -5) \)[/tex], [tex]\( (5, -1) \)[/tex], [tex]\( (6, -0.2) \)[/tex], [tex]\( (3, -25) \)[/tex], and [tex]\( (7, -0.04) \)[/tex].
- Connect these points smoothly, remembering that as [tex]\( x \)[/tex] increases, [tex]\( f(x) \)[/tex] gets closer to 0 from below, and as [tex]\( x \)[/tex] decreases, [tex]\( f(x) \)[/tex] becomes very large negatively.
The graph of [tex]\( f(x) = -\left(\frac{1}{5}\right)^{x-5} \)[/tex] will show a rapid decrease as [tex]\( x \)[/tex] moves left from [tex]\( x = 5 \)[/tex], flattening as it moves right from [tex]\( x = 5 \)[/tex], approaching the horizontal asymptote [tex]\( y = 0 \)[/tex].
