Answer :
Certainly! Let's identify the graph for the given equation:
[tex]\[ y = 20 \left(\frac{1}{4}\right)^x \][/tex]
This equation represents an exponential decay function. Let's break down the process step-by-step to understand how the function behaves and plot its graph.
### Step-by-Step Solution:
1. Understanding the Equation: The given equation is in the form [tex]\( y = a b^x \)[/tex], where [tex]\( a = 20 \)[/tex] is the coefficient indicating the initial value of [tex]\( y \)[/tex], and [tex]\( b = \frac{1}{4} \)[/tex] is the base, which suggests exponential decay (since [tex]\( 0 < b < 1 \)[/tex]).
2. Identify Key Features of the Exponential Function:
- Initial Value: When [tex]\( x = 0 \)[/tex], [tex]\( y = a \cdot b^0 = 20 \cdot 1 = 20 \)[/tex].
- Decay Factor: Since the base [tex]\( \frac{1}{4} \)[/tex] is less than 1, the function will decrease as [tex]\( x \)[/tex] increases.
3. Generate Points to Plot:
- Start with some [tex]\( x \)[/tex] values and calculate corresponding [tex]\( y \)[/tex] values.
- Since we need a smooth curve, let's consider generating [tex]\( x \)[/tex] values ranging from -2 to 2 with a reasonable step.
4. Plot Points:
From the data:
- For [tex]\( x = -2 \)[/tex], [tex]\( y \approx 320 \)[/tex]
- For [tex]\( x = -1 \)[/tex], [tex]\( y \approx 80 \)[/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 20 \)[/tex]
- For [tex]\( x = 1 \)[/tex], [tex]\( y \approx 5 \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y \approx 1.25 \)[/tex]
5. Examine the Shape of the Graph:
- The graph starts high at [tex]\( x = -2 \)[/tex] and decreases rapidly.
- As [tex]\( x \)[/tex] approaches 0, the function value converges to 20.
- Beyond [tex]\( x = 0 \)[/tex], the function continues to decay towards 0 but never actually reaches 0.
6. Key Observations about the Graph:
- Y-Intercept: The graph intersects the y-axis at [tex]\( (0, 20) \)[/tex].
- Asymptote: The horizontal asymptote is [tex]\( y = 0 \)[/tex], meaning the function value approaches 0 but never actually equals 0 as [tex]\( x \)[/tex] increases to positive infinity.
- Shape: The graph is a downward curve from left to right.
7. Sketch the Graph:
To sketch the graph, plot the points and draw a smooth curve through them considering the rapid decay:
[tex]\[ \begin{align*} x & : -2, -1, 0, 1, 2 \\ y & : 320, 80, 20, 5, 1.25 \\ \end{align*} \][/tex]
Here is a simplified version of how the points and curve would look:
```
y
|
350 |
300 |
250 |
200 |
150 |
100 |
50 |
20 |
0 |---------------------------
-2 -1 0 1 2 x
```
In conclusion, the graph of the equation [tex]\( y = 20 \left(\frac{1}{4}\right)^x \)[/tex] decays rapidly from a high value and approaches zero as [tex]\( x \)[/tex] increases. The y-intercept occurs at [tex]\( (0, 20) \)[/tex] and the graph never touches the x-axis, showing an exponential decay behavior.
[tex]\[ y = 20 \left(\frac{1}{4}\right)^x \][/tex]
This equation represents an exponential decay function. Let's break down the process step-by-step to understand how the function behaves and plot its graph.
### Step-by-Step Solution:
1. Understanding the Equation: The given equation is in the form [tex]\( y = a b^x \)[/tex], where [tex]\( a = 20 \)[/tex] is the coefficient indicating the initial value of [tex]\( y \)[/tex], and [tex]\( b = \frac{1}{4} \)[/tex] is the base, which suggests exponential decay (since [tex]\( 0 < b < 1 \)[/tex]).
2. Identify Key Features of the Exponential Function:
- Initial Value: When [tex]\( x = 0 \)[/tex], [tex]\( y = a \cdot b^0 = 20 \cdot 1 = 20 \)[/tex].
- Decay Factor: Since the base [tex]\( \frac{1}{4} \)[/tex] is less than 1, the function will decrease as [tex]\( x \)[/tex] increases.
3. Generate Points to Plot:
- Start with some [tex]\( x \)[/tex] values and calculate corresponding [tex]\( y \)[/tex] values.
- Since we need a smooth curve, let's consider generating [tex]\( x \)[/tex] values ranging from -2 to 2 with a reasonable step.
4. Plot Points:
From the data:
- For [tex]\( x = -2 \)[/tex], [tex]\( y \approx 320 \)[/tex]
- For [tex]\( x = -1 \)[/tex], [tex]\( y \approx 80 \)[/tex]
- For [tex]\( x = 0 \)[/tex], [tex]\( y = 20 \)[/tex]
- For [tex]\( x = 1 \)[/tex], [tex]\( y \approx 5 \)[/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\( y \approx 1.25 \)[/tex]
5. Examine the Shape of the Graph:
- The graph starts high at [tex]\( x = -2 \)[/tex] and decreases rapidly.
- As [tex]\( x \)[/tex] approaches 0, the function value converges to 20.
- Beyond [tex]\( x = 0 \)[/tex], the function continues to decay towards 0 but never actually reaches 0.
6. Key Observations about the Graph:
- Y-Intercept: The graph intersects the y-axis at [tex]\( (0, 20) \)[/tex].
- Asymptote: The horizontal asymptote is [tex]\( y = 0 \)[/tex], meaning the function value approaches 0 but never actually equals 0 as [tex]\( x \)[/tex] increases to positive infinity.
- Shape: The graph is a downward curve from left to right.
7. Sketch the Graph:
To sketch the graph, plot the points and draw a smooth curve through them considering the rapid decay:
[tex]\[ \begin{align*} x & : -2, -1, 0, 1, 2 \\ y & : 320, 80, 20, 5, 1.25 \\ \end{align*} \][/tex]
Here is a simplified version of how the points and curve would look:
```
y
|
350 |
300 |
250 |
200 |
150 |
100 |
50 |
20 |
0 |---------------------------
-2 -1 0 1 2 x
```
In conclusion, the graph of the equation [tex]\( y = 20 \left(\frac{1}{4}\right)^x \)[/tex] decays rapidly from a high value and approaches zero as [tex]\( x \)[/tex] increases. The y-intercept occurs at [tex]\( (0, 20) \)[/tex] and the graph never touches the x-axis, showing an exponential decay behavior.
