Discover the answers to your questions at Westonci.ca, where experts share their knowledge and insights with you. Get quick and reliable solutions to your questions from a community of seasoned experts on our user-friendly platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.
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.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.
