Join IDNLearn.com and start exploring the answers to your most pressing questions. Whether it's a simple query or a complex problem, our community has the answers you need.
Sagot :
To graph the function [tex]\( f(x) = -\sqrt[3]{x+2} - 4 \)[/tex] and find points that can be used to represent this function accurately, we first identify the transformations applied to the parent function [tex]\( f(x) = \sqrt[3]{x} \)[/tex].
### Step-by-Step Transformations:
1. Horizontal Shift: The term [tex]\( x + 2 \)[/tex] inside the cube root indicates a horizontal shift. The graph of the function is shifted 2 units to the left.
2. Reflection Across the X-axis: The presence of the negative sign in front of the cube root function [tex]\( -\sqrt[3]{x+2} \)[/tex] indicates a reflection across the x-axis.
3. Vertical Shift: The term [tex]\(-4\)[/tex] outside of the cube root indicates a vertical shift downward by 4 units.
### Applying these transformations to specific points:
To illustrate these transformations, we will use some key points from the parent function [tex]\( g(x) = \sqrt[3]{x} \)[/tex]. Consider the following x-values: [tex]\(-8, -1, 0, 1, 8\)[/tex]. These values are chosen because they are convenient for computing cube roots expressible in simple forms (like integers or easily recognizable decimal values).
### Calculating the transformed points:
1. Begin with the x-values: [tex]\( \{ - 8, - 1, 0, 1, 8 \} \)[/tex].
2. Apply the horizontal shift (subtract 2 from each x-value):
- [tex]\( x = -8 \)[/tex] becomes [tex]\( x' = -8 - 2 = -10 \)[/tex]
- [tex]\( x = -1 \)[/tex] becomes [tex]\( x' = -1 - 2 = -3 \)[/tex]
- [tex]\( x = 0 \)[/tex] becomes [tex]\( x' = 0 - 2 = -2 \)[/tex]
- [tex]\( x = 1 \)[/tex] becomes [tex]\( x' = 1 - 2 = -1 \)[/tex]
- [tex]\( x = 8 \)[/tex] becomes [tex]\( x' = 8 - 2 = 6 \)[/tex]
3. Compute the cube root of the transformed x-values:
- [tex]\( \sqrt[3]{-10} \approx -2.154 \)[/tex]
- [tex]\( \sqrt[3]{-3} \approx -1.4422 \)[/tex]
- [tex]\( \sqrt[3]{-2} \approx -1.2599 \)[/tex]
- [tex]\( \sqrt[3]{-1} = -1 \)[/tex]
- [tex]\( \sqrt[3]{6} \approx 1.817 \)[/tex]
4. Apply the reflection and vertical shift:
- For [tex]\( x' = -10 \)[/tex]: [tex]\( y = -\sqrt[3]{-10} - 4 \approx -(-2.154) - 4 = 2.154 - 4 = -1.846 \)[/tex]
- For [tex]\( x' = -3 \)[/tex]: [tex]\( y = -\sqrt[3]{-3} - 4 \approx -(-1.4422) - 4 = 1.4422 - 4 = -2.5578 \)[/tex]
- For [tex]\( x' = -2 \)[/tex]: [tex]\( y = -\sqrt[3]{-2} - 4 \approx -(-1.2599) - 4 = 1.2599 - 4 = -2.7401 \)[/tex]
- For [tex]\( x' = -1 \)[/tex]: [tex]\( y = -\sqrt[3]{-1} - 4 = -(-1) - 4 = 1 - 4 = -3 \)[/tex]
- For [tex]\( x' = 6 \)[/tex]: [tex]\( y = -\sqrt[3]{6} - 4 \approx -1.817 - 4 = -5.817 \)[/tex]
Based on these calculations, the points that can be used to graph the function [tex]\( f(x) = -\sqrt[3]{x+2} - 4 \)[/tex] accurately are:
[tex]\[ \begin{align*} (-8, & (-5.077217345015942-1.865795172362064j)), \\ (-1, & (-4.721124785153704-1.2490247664834064j)), \\ (0, & (-4.629960524947437-1.0911236359717214j)), \\ (1, & (-4.5-0.8660254037844386j)), \\ (8, & -5.81712059283214) \\ \end{align*} \][/tex]
These points result from the accurate transformations applied to the parent function to achieve the desired function [tex]\( f(x) = -\sqrt[3]{x+2}-4 \)[/tex].
### Step-by-Step Transformations:
1. Horizontal Shift: The term [tex]\( x + 2 \)[/tex] inside the cube root indicates a horizontal shift. The graph of the function is shifted 2 units to the left.
2. Reflection Across the X-axis: The presence of the negative sign in front of the cube root function [tex]\( -\sqrt[3]{x+2} \)[/tex] indicates a reflection across the x-axis.
3. Vertical Shift: The term [tex]\(-4\)[/tex] outside of the cube root indicates a vertical shift downward by 4 units.
### Applying these transformations to specific points:
To illustrate these transformations, we will use some key points from the parent function [tex]\( g(x) = \sqrt[3]{x} \)[/tex]. Consider the following x-values: [tex]\(-8, -1, 0, 1, 8\)[/tex]. These values are chosen because they are convenient for computing cube roots expressible in simple forms (like integers or easily recognizable decimal values).
### Calculating the transformed points:
1. Begin with the x-values: [tex]\( \{ - 8, - 1, 0, 1, 8 \} \)[/tex].
2. Apply the horizontal shift (subtract 2 from each x-value):
- [tex]\( x = -8 \)[/tex] becomes [tex]\( x' = -8 - 2 = -10 \)[/tex]
- [tex]\( x = -1 \)[/tex] becomes [tex]\( x' = -1 - 2 = -3 \)[/tex]
- [tex]\( x = 0 \)[/tex] becomes [tex]\( x' = 0 - 2 = -2 \)[/tex]
- [tex]\( x = 1 \)[/tex] becomes [tex]\( x' = 1 - 2 = -1 \)[/tex]
- [tex]\( x = 8 \)[/tex] becomes [tex]\( x' = 8 - 2 = 6 \)[/tex]
3. Compute the cube root of the transformed x-values:
- [tex]\( \sqrt[3]{-10} \approx -2.154 \)[/tex]
- [tex]\( \sqrt[3]{-3} \approx -1.4422 \)[/tex]
- [tex]\( \sqrt[3]{-2} \approx -1.2599 \)[/tex]
- [tex]\( \sqrt[3]{-1} = -1 \)[/tex]
- [tex]\( \sqrt[3]{6} \approx 1.817 \)[/tex]
4. Apply the reflection and vertical shift:
- For [tex]\( x' = -10 \)[/tex]: [tex]\( y = -\sqrt[3]{-10} - 4 \approx -(-2.154) - 4 = 2.154 - 4 = -1.846 \)[/tex]
- For [tex]\( x' = -3 \)[/tex]: [tex]\( y = -\sqrt[3]{-3} - 4 \approx -(-1.4422) - 4 = 1.4422 - 4 = -2.5578 \)[/tex]
- For [tex]\( x' = -2 \)[/tex]: [tex]\( y = -\sqrt[3]{-2} - 4 \approx -(-1.2599) - 4 = 1.2599 - 4 = -2.7401 \)[/tex]
- For [tex]\( x' = -1 \)[/tex]: [tex]\( y = -\sqrt[3]{-1} - 4 = -(-1) - 4 = 1 - 4 = -3 \)[/tex]
- For [tex]\( x' = 6 \)[/tex]: [tex]\( y = -\sqrt[3]{6} - 4 \approx -1.817 - 4 = -5.817 \)[/tex]
Based on these calculations, the points that can be used to graph the function [tex]\( f(x) = -\sqrt[3]{x+2} - 4 \)[/tex] accurately are:
[tex]\[ \begin{align*} (-8, & (-5.077217345015942-1.865795172362064j)), \\ (-1, & (-4.721124785153704-1.2490247664834064j)), \\ (0, & (-4.629960524947437-1.0911236359717214j)), \\ (1, & (-4.5-0.8660254037844386j)), \\ (8, & -5.81712059283214) \\ \end{align*} \][/tex]
These points result from the accurate transformations applied to the parent function to achieve the desired function [tex]\( f(x) = -\sqrt[3]{x+2}-4 \)[/tex].
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Your search for answers ends at IDNLearn.com. Thank you for visiting, and we hope to assist you again soon.
