Answer :
### Solution to Question 5
#### a) Two values of [tex]\( x \)[/tex] which would generate the same [tex]\( y \)[/tex] values for both options.
Nontobeko correctly interpreted the function as [tex]\( y = 2x + 1 \)[/tex], whereas Amogelang misunderstood it as a constant function [tex]\( y = 3 \)[/tex] (since 2* + 1 simplifies to 3).
To find the values of [tex]\( x \)[/tex] for which both interpretations yield the same [tex]\( y \)[/tex], we need to solve the equation:
[tex]\[ 2x + 1 = 3 \][/tex]
Subtract 1 from both sides:
[tex]\[ 2x = 2 \][/tex]
Divide by 2:
[tex]\[ x = 1 \][/tex]
Thus, the two values of [tex]\( x \)[/tex] which generate the same [tex]\( y \)[/tex] values are:
[tex]\[ x = 1 \][/tex]
[tex]\[ x = 1 \][/tex]
#### b) Sketch the graphs of [tex]\( y = 2x + 1 \)[/tex] and [tex]\( y = 3 \)[/tex] on the same set of axes.
1. Graph of [tex]\( y = 2x + 1 \)[/tex]:
- X-intercept: Set [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = 2x + 1 \][/tex]
[tex]\[ 2x = -1 \][/tex]
[tex]\[ x = -0.5 \][/tex]
So, the x-intercept is [tex]\((-0.5, 0)\)[/tex].
- Y-intercept: Set [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 2(0) + 1 = 1 \][/tex]
So, the y-intercept is [tex]\((0, 1)\)[/tex].
2. Graph of [tex]\( y = 3 \)[/tex]:
- This is a horizontal line that intercepts the y-axis at [tex]\( y = 3 \)[/tex].
Graphical representation:
Here's a visual aid without actually drawing graphs:
- The line [tex]\( y = 2x + 1 \)[/tex] crosses the x-axis at [tex]\( (-0.5, 0) \)[/tex] and the y-axis at [tex]\( (0, 1) \)[/tex].
- The horizontal line [tex]\( y = 3 \)[/tex] crosses the y-axis at [tex]\( (0, 3) \)[/tex].
- Mark the intersection where both [tex]\( y \)[/tex]-values are equal, [tex]\( x = 1 \)[/tex], yielding [tex]\( (1, 3) \)[/tex].
##### c) Draw the graph of the inverse of [tex]\( y = 2x + 1 \)[/tex] on the same axes.
To find the inverse of the function [tex]\( y = 2x + 1 \)[/tex], we interchange [tex]\( x \)[/tex] and [tex]\( y \)[/tex], and solve for [tex]\( y \)[/tex]:
[tex]\[ x = 2y + 1 \][/tex]
Subtract 1 from both sides:
[tex]\[ x - 1 = 2y \][/tex]
Divide by 2:
[tex]\[ y = \frac{x - 1}{2} \][/tex]
Thus, the inverse function is:
[tex]\[ y = \frac{x - 1}{2} \][/tex]
### Summary of Graphs:
- The original graph [tex]\( y = 2x + 1 \)[/tex].
- The horizontal line [tex]\( y = 3 \)[/tex].
- The inverse graph [tex]\( y = \frac{x - 1}{2} \)[/tex].
### Intercepts:
- For [tex]\( y = 2x + 1 \)[/tex]:
- X-intercept: [tex]\( (-0.5, 0) \)[/tex]
- Y-intercept: [tex]\( (0, 1) \)[/tex]
- For [tex]\( y = 3 \)[/tex]:
- X-intercept: none.
- Y-intercept: [tex]\( (0, 3) \)[/tex]
- For [tex]\( y = \frac{x - 1}{2} \)[/tex]:
- X-intercept: [tex]\( (1, 0) \)[/tex]
- Y-intercept: [tex]\( (0, -0.5) \)[/tex]
These insights complete the steps required to sketch the graphs and indicate results as requested.
#### a) Two values of [tex]\( x \)[/tex] which would generate the same [tex]\( y \)[/tex] values for both options.
Nontobeko correctly interpreted the function as [tex]\( y = 2x + 1 \)[/tex], whereas Amogelang misunderstood it as a constant function [tex]\( y = 3 \)[/tex] (since 2* + 1 simplifies to 3).
To find the values of [tex]\( x \)[/tex] for which both interpretations yield the same [tex]\( y \)[/tex], we need to solve the equation:
[tex]\[ 2x + 1 = 3 \][/tex]
Subtract 1 from both sides:
[tex]\[ 2x = 2 \][/tex]
Divide by 2:
[tex]\[ x = 1 \][/tex]
Thus, the two values of [tex]\( x \)[/tex] which generate the same [tex]\( y \)[/tex] values are:
[tex]\[ x = 1 \][/tex]
[tex]\[ x = 1 \][/tex]
#### b) Sketch the graphs of [tex]\( y = 2x + 1 \)[/tex] and [tex]\( y = 3 \)[/tex] on the same set of axes.
1. Graph of [tex]\( y = 2x + 1 \)[/tex]:
- X-intercept: Set [tex]\( y = 0 \)[/tex]:
[tex]\[ 0 = 2x + 1 \][/tex]
[tex]\[ 2x = -1 \][/tex]
[tex]\[ x = -0.5 \][/tex]
So, the x-intercept is [tex]\((-0.5, 0)\)[/tex].
- Y-intercept: Set [tex]\( x = 0 \)[/tex]:
[tex]\[ y = 2(0) + 1 = 1 \][/tex]
So, the y-intercept is [tex]\((0, 1)\)[/tex].
2. Graph of [tex]\( y = 3 \)[/tex]:
- This is a horizontal line that intercepts the y-axis at [tex]\( y = 3 \)[/tex].
Graphical representation:
Here's a visual aid without actually drawing graphs:
- The line [tex]\( y = 2x + 1 \)[/tex] crosses the x-axis at [tex]\( (-0.5, 0) \)[/tex] and the y-axis at [tex]\( (0, 1) \)[/tex].
- The horizontal line [tex]\( y = 3 \)[/tex] crosses the y-axis at [tex]\( (0, 3) \)[/tex].
- Mark the intersection where both [tex]\( y \)[/tex]-values are equal, [tex]\( x = 1 \)[/tex], yielding [tex]\( (1, 3) \)[/tex].
##### c) Draw the graph of the inverse of [tex]\( y = 2x + 1 \)[/tex] on the same axes.
To find the inverse of the function [tex]\( y = 2x + 1 \)[/tex], we interchange [tex]\( x \)[/tex] and [tex]\( y \)[/tex], and solve for [tex]\( y \)[/tex]:
[tex]\[ x = 2y + 1 \][/tex]
Subtract 1 from both sides:
[tex]\[ x - 1 = 2y \][/tex]
Divide by 2:
[tex]\[ y = \frac{x - 1}{2} \][/tex]
Thus, the inverse function is:
[tex]\[ y = \frac{x - 1}{2} \][/tex]
### Summary of Graphs:
- The original graph [tex]\( y = 2x + 1 \)[/tex].
- The horizontal line [tex]\( y = 3 \)[/tex].
- The inverse graph [tex]\( y = \frac{x - 1}{2} \)[/tex].
### Intercepts:
- For [tex]\( y = 2x + 1 \)[/tex]:
- X-intercept: [tex]\( (-0.5, 0) \)[/tex]
- Y-intercept: [tex]\( (0, 1) \)[/tex]
- For [tex]\( y = 3 \)[/tex]:
- X-intercept: none.
- Y-intercept: [tex]\( (0, 3) \)[/tex]
- For [tex]\( y = \frac{x - 1}{2} \)[/tex]:
- X-intercept: [tex]\( (1, 0) \)[/tex]
- Y-intercept: [tex]\( (0, -0.5) \)[/tex]
These insights complete the steps required to sketch the graphs and indicate results as requested.
