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Sagot :
Let's sketch a graph that satisfies the given characteristics step by step.
### a. Crosses the [tex]$y$[/tex]-axis at [tex]$(0,4)$[/tex]
This means that the value of the function [tex]\( f(x) \)[/tex] when [tex]\( x = 0 \)[/tex] is 4. So, the point (0, 4) is on the graph.
### b. Increases in the interval [tex]$-12 \leq x \leq -2$[/tex]
In this interval, the graph should be rising as [tex]\( x \)[/tex] moves from -12 to -2. One possible linear function that increases in this interval and ensures continuity with the y-axis crossing point is [tex]\( y = x + 16 \)[/tex].
- At [tex]\( x = -12 \)[/tex]:
[tex]\[ y = -12 + 16 = 4 \][/tex]
- At [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -2 + 16 = 14 \][/tex]
So in this interval [tex]$-12 \leq x \leq -2$[/tex], the function linearly increases from the point [tex]$(-12, 4)$[/tex] to the point [tex]$(-2, 14)$[/tex].
### c. Constant in the interval [tex]$-2 \leq x \leq 2$[/tex]
In this interval, the function has a constant value. To match the continuity and y-intercept at (0, 4), the function will be:
[tex]\[ y = 4 \][/tex]
So in the interval [tex]$-2 \leq x \leq 2$[/tex], the function is a horizontal line at [tex]$y = 4$[/tex].
### d. Decreases in the interval [tex]$2 \leq x \leq 12$[/tex]
In this interval, the graph should be falling as [tex]\( x \)[/tex] moves from 2 to 12. One possible linear function that decreases in this interval and maintains continuity is [tex]\( y = -x + 8 \)[/tex]:
- At [tex]\( x = 2 \)[/tex]:
[tex]\[ y = -2 + 8 = 6 \][/tex]
- At [tex]\( x = 12 \)[/tex]:
[tex]\[ y = -12 + 8 = -4 \][/tex]
So in this interval [tex]$2 < x \leq 12$[/tex], the function linearly decreases from the point [tex]$(2, 6)$[/tex] to the point [tex]$(12, -4)$[/tex].
### Summary
Now, let's summarize all the intervals and sketch the graph:
- From [tex]$-12 \leq x < -2$[/tex], [tex]\( f(x) = x + 16 \)[/tex]: increases from the point [tex]$(-12, 4)$[/tex] to [tex]$(-2, 14)$[/tex].
- From [tex]$-2 \leq x \leq 2$[/tex], [tex]\( f(x) = 4 \)[/tex]: constant at [tex]\( y = 4 \)[/tex].
- From [tex]$2 < x \leq 12$[/tex], [tex]\( f(x) = -x + 8 \)[/tex]: decreases from the point [tex]$(2, 6)$[/tex] to [tex]$(12, -4)$[/tex].
Here's a rudimentary sketch of the graph:
1. Plot the point (0, 4) where it crosses the y-axis.
2. Draw a line from [tex]$(-12, 4)$[/tex] to [tex]$(-2, 14)$[/tex] for the increasing interval.
3. Draw a horizontal line from [tex]$(-2, 14)$[/tex] to [tex]$(2, 4)$[/tex] for the constant interval.
4. Draw a line from [tex]$(2, 6)$[/tex] to [tex]$(12, -4)$[/tex] for the decreasing interval.
This piecewise function will give you a graph that meets all the given characteristics.
### a. Crosses the [tex]$y$[/tex]-axis at [tex]$(0,4)$[/tex]
This means that the value of the function [tex]\( f(x) \)[/tex] when [tex]\( x = 0 \)[/tex] is 4. So, the point (0, 4) is on the graph.
### b. Increases in the interval [tex]$-12 \leq x \leq -2$[/tex]
In this interval, the graph should be rising as [tex]\( x \)[/tex] moves from -12 to -2. One possible linear function that increases in this interval and ensures continuity with the y-axis crossing point is [tex]\( y = x + 16 \)[/tex].
- At [tex]\( x = -12 \)[/tex]:
[tex]\[ y = -12 + 16 = 4 \][/tex]
- At [tex]\( x = -2 \)[/tex]:
[tex]\[ y = -2 + 16 = 14 \][/tex]
So in this interval [tex]$-12 \leq x \leq -2$[/tex], the function linearly increases from the point [tex]$(-12, 4)$[/tex] to the point [tex]$(-2, 14)$[/tex].
### c. Constant in the interval [tex]$-2 \leq x \leq 2$[/tex]
In this interval, the function has a constant value. To match the continuity and y-intercept at (0, 4), the function will be:
[tex]\[ y = 4 \][/tex]
So in the interval [tex]$-2 \leq x \leq 2$[/tex], the function is a horizontal line at [tex]$y = 4$[/tex].
### d. Decreases in the interval [tex]$2 \leq x \leq 12$[/tex]
In this interval, the graph should be falling as [tex]\( x \)[/tex] moves from 2 to 12. One possible linear function that decreases in this interval and maintains continuity is [tex]\( y = -x + 8 \)[/tex]:
- At [tex]\( x = 2 \)[/tex]:
[tex]\[ y = -2 + 8 = 6 \][/tex]
- At [tex]\( x = 12 \)[/tex]:
[tex]\[ y = -12 + 8 = -4 \][/tex]
So in this interval [tex]$2 < x \leq 12$[/tex], the function linearly decreases from the point [tex]$(2, 6)$[/tex] to the point [tex]$(12, -4)$[/tex].
### Summary
Now, let's summarize all the intervals and sketch the graph:
- From [tex]$-12 \leq x < -2$[/tex], [tex]\( f(x) = x + 16 \)[/tex]: increases from the point [tex]$(-12, 4)$[/tex] to [tex]$(-2, 14)$[/tex].
- From [tex]$-2 \leq x \leq 2$[/tex], [tex]\( f(x) = 4 \)[/tex]: constant at [tex]\( y = 4 \)[/tex].
- From [tex]$2 < x \leq 12$[/tex], [tex]\( f(x) = -x + 8 \)[/tex]: decreases from the point [tex]$(2, 6)$[/tex] to [tex]$(12, -4)$[/tex].
Here's a rudimentary sketch of the graph:
1. Plot the point (0, 4) where it crosses the y-axis.
2. Draw a line from [tex]$(-12, 4)$[/tex] to [tex]$(-2, 14)$[/tex] for the increasing interval.
3. Draw a horizontal line from [tex]$(-2, 14)$[/tex] to [tex]$(2, 4)$[/tex] for the constant interval.
4. Draw a line from [tex]$(2, 6)$[/tex] to [tex]$(12, -4)$[/tex] for the decreasing interval.
This piecewise function will give you a graph that meets all the given characteristics.
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