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Sagot :
### (a) Graph [tex]\( f(x) = |x-7| - 3 \)[/tex] Using Transformations
To graph [tex]\( f(x) = |x-7| - 3 \)[/tex], let's break down the function into basic transformations of the absolute value function.
1. Start with the basic absolute value function [tex]\( y = |x| \)[/tex].
2. Horizontal shift:
- The term [tex]\( |x-7| \)[/tex] represents a horizontal shift of the absolute value function 7 units to the right.
So, [tex]\( y = |x-7| \)[/tex].
3. Vertical shift:
- The term [tex]\( -3 \)[/tex] at the end of the function represents a vertical shift 3 units downward.
Combining these transformations, we get: [tex]\( y = |x-7| - 3 \)[/tex].
Now, summarizing the transformations:
- Shift the vertex of the absolute value function right to [tex]\( (7, 0) \)[/tex]
- Shift the entire function down by 3 units to [tex]\( (7, -3) \)[/tex]
Thus, the function [tex]\( f(x) \)[/tex] is V-shaped with the vertex at (7, -3), opening upwards.
### (b) Finding the Area Below the [tex]\( x \)[/tex]-axis
To determine the area of the region bounded by [tex]\( f(x) \)[/tex] and the [tex]\( x \)[/tex]-axis that lies below the [tex]\( x \)[/tex]-axis, we first need to find the points where [tex]\( f(x) \)[/tex] crosses the [tex]\( x \)[/tex]-axis.
Finding the x-intercepts:
[tex]\[ |x-7| - 3 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ |x-7| = 3 \][/tex]
This gives two solutions:
[tex]\[ x - 7 = 3 \quad \text{or} \quad x - 7 = -3 \][/tex]
[tex]\[ x = 10 \quad \text{or} \quad x = 4 \][/tex]
Thus, [tex]\( f(x) \)[/tex] crosses the [tex]\( x \)[/tex]-axis at [tex]\( x = 4 \)[/tex] and [tex]\( x = 10 \)[/tex].
Determining the area under the curve:
The function [tex]\( f(x) \)[/tex] is below the [tex]\( x \)[/tex]-axis for [tex]\( 4 \leq x \leq 10 \)[/tex].
Write the integral for the area under the [tex]\( x \)[/tex]-axis:
[tex]\[ A = \int_{4}^{10} [|x-7| - 3] \, dx \][/tex]
Since the integral of an absolute value can be split at the vertex point where the absolute value functions change, we split the integral into two parts:
1. Evaluate from [tex]\( x = 4 \)[/tex] to [tex]\( x = 7 \)[/tex]:
- Here, [tex]\( |x-7| = 7-x \)[/tex]
- Therefore, the integral is:
[tex]\[ \int_4^7 (7 - x - 3) \, dx = \int_4^7 (4 - x) \, dx \][/tex]
2. Evaluate from [tex]\( x = 7 \)[/tex] to [tex]\( x = 10 \)[/tex]:
- Here, [tex]\( |x-7| = x - 7 \)[/tex]
- Therefore, the integral is:
[tex]\[ \int_7^{10} (x - 7 - 3) \, dx = \int_7^{10} (x - 10) \, dx \][/tex]
Calculate the integrals:
1. From [tex]\( x = 4 \)[/tex] to [tex]\( x = 7 \)[/tex]:
[tex]\[ \int_4^7 (4 - x) \, dx \][/tex]
[tex]\[ = [4x - \frac{x^2}{2}]_4^7 \][/tex]
[tex]\[ = (4 \times 7 - \frac{7^2}{2}) - (4 \times 4 - \frac{4^2}{2}) \][/tex]
[tex]\[ = (28 - \frac{49}{2}) - (16 - \frac{16}{2}) \][/tex]
[tex]\[ = (28 - 24.5) - (16 - 8) \][/tex]
[tex]\[ = 3.5 - 8 \][/tex]
[tex]\[ = -4.5 \][/tex]
2. From [tex]\( x = 7 \)[/tex] to [tex]\( x = 10 \)[/tex]:
[tex]\[ \int_7^{10} (x - 10) \, dx \][/tex]
[tex]\[ = [\frac{x^2}{2} - 10x]_7^{10} \][/tex]
[tex]\[ = (\frac{10^2}{2} - 10 \times 10) - (\frac{7^2}{2} - 10 \times 7) \][/tex]
[tex]\[ = (50 - 100) - (\frac{49}{2} - 70) \][/tex]
[tex]\[ = -50 - (-45.5) \][/tex]
[tex]\[ = -50 + 45.5 \][/tex]
[tex]\[ = -4.5 \][/tex]
Thus, the area below the [tex]\( x \)[/tex]-axis from [tex]\( 4 \leq x \leq 10 \)[/tex] is:
[tex]\[ -4.5 + -4.5 = -9 \][/tex]
Finally, the area is:
[tex]\[ \boxed{9 \, \text{square units}} \][/tex]
Note: The positive sign for the area because the area itself is always a positive quantity.
To graph [tex]\( f(x) = |x-7| - 3 \)[/tex], let's break down the function into basic transformations of the absolute value function.
1. Start with the basic absolute value function [tex]\( y = |x| \)[/tex].
2. Horizontal shift:
- The term [tex]\( |x-7| \)[/tex] represents a horizontal shift of the absolute value function 7 units to the right.
So, [tex]\( y = |x-7| \)[/tex].
3. Vertical shift:
- The term [tex]\( -3 \)[/tex] at the end of the function represents a vertical shift 3 units downward.
Combining these transformations, we get: [tex]\( y = |x-7| - 3 \)[/tex].
Now, summarizing the transformations:
- Shift the vertex of the absolute value function right to [tex]\( (7, 0) \)[/tex]
- Shift the entire function down by 3 units to [tex]\( (7, -3) \)[/tex]
Thus, the function [tex]\( f(x) \)[/tex] is V-shaped with the vertex at (7, -3), opening upwards.
### (b) Finding the Area Below the [tex]\( x \)[/tex]-axis
To determine the area of the region bounded by [tex]\( f(x) \)[/tex] and the [tex]\( x \)[/tex]-axis that lies below the [tex]\( x \)[/tex]-axis, we first need to find the points where [tex]\( f(x) \)[/tex] crosses the [tex]\( x \)[/tex]-axis.
Finding the x-intercepts:
[tex]\[ |x-7| - 3 = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ |x-7| = 3 \][/tex]
This gives two solutions:
[tex]\[ x - 7 = 3 \quad \text{or} \quad x - 7 = -3 \][/tex]
[tex]\[ x = 10 \quad \text{or} \quad x = 4 \][/tex]
Thus, [tex]\( f(x) \)[/tex] crosses the [tex]\( x \)[/tex]-axis at [tex]\( x = 4 \)[/tex] and [tex]\( x = 10 \)[/tex].
Determining the area under the curve:
The function [tex]\( f(x) \)[/tex] is below the [tex]\( x \)[/tex]-axis for [tex]\( 4 \leq x \leq 10 \)[/tex].
Write the integral for the area under the [tex]\( x \)[/tex]-axis:
[tex]\[ A = \int_{4}^{10} [|x-7| - 3] \, dx \][/tex]
Since the integral of an absolute value can be split at the vertex point where the absolute value functions change, we split the integral into two parts:
1. Evaluate from [tex]\( x = 4 \)[/tex] to [tex]\( x = 7 \)[/tex]:
- Here, [tex]\( |x-7| = 7-x \)[/tex]
- Therefore, the integral is:
[tex]\[ \int_4^7 (7 - x - 3) \, dx = \int_4^7 (4 - x) \, dx \][/tex]
2. Evaluate from [tex]\( x = 7 \)[/tex] to [tex]\( x = 10 \)[/tex]:
- Here, [tex]\( |x-7| = x - 7 \)[/tex]
- Therefore, the integral is:
[tex]\[ \int_7^{10} (x - 7 - 3) \, dx = \int_7^{10} (x - 10) \, dx \][/tex]
Calculate the integrals:
1. From [tex]\( x = 4 \)[/tex] to [tex]\( x = 7 \)[/tex]:
[tex]\[ \int_4^7 (4 - x) \, dx \][/tex]
[tex]\[ = [4x - \frac{x^2}{2}]_4^7 \][/tex]
[tex]\[ = (4 \times 7 - \frac{7^2}{2}) - (4 \times 4 - \frac{4^2}{2}) \][/tex]
[tex]\[ = (28 - \frac{49}{2}) - (16 - \frac{16}{2}) \][/tex]
[tex]\[ = (28 - 24.5) - (16 - 8) \][/tex]
[tex]\[ = 3.5 - 8 \][/tex]
[tex]\[ = -4.5 \][/tex]
2. From [tex]\( x = 7 \)[/tex] to [tex]\( x = 10 \)[/tex]:
[tex]\[ \int_7^{10} (x - 10) \, dx \][/tex]
[tex]\[ = [\frac{x^2}{2} - 10x]_7^{10} \][/tex]
[tex]\[ = (\frac{10^2}{2} - 10 \times 10) - (\frac{7^2}{2} - 10 \times 7) \][/tex]
[tex]\[ = (50 - 100) - (\frac{49}{2} - 70) \][/tex]
[tex]\[ = -50 - (-45.5) \][/tex]
[tex]\[ = -50 + 45.5 \][/tex]
[tex]\[ = -4.5 \][/tex]
Thus, the area below the [tex]\( x \)[/tex]-axis from [tex]\( 4 \leq x \leq 10 \)[/tex] is:
[tex]\[ -4.5 + -4.5 = -9 \][/tex]
Finally, the area is:
[tex]\[ \boxed{9 \, \text{square units}} \][/tex]
Note: The positive sign for the area because the area itself is always a positive quantity.
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