Answer :
Alright, let's graph the function [tex]\( f(x) = 3|x-5| + 1 \)[/tex] step-by-step.
### Step 1: Understand the Function
The given function [tex]\( f(x) = 3|x-5| + 1 \)[/tex] incorporates an absolute value term and a linear transformation. Absolute value functions tend to create 'V' shapes in their graph.
### Step 2: Break Down the Function Components
1. Absolute Value: [tex]\( |x-5| \)[/tex].
2. Scaling: [tex]\( 3|x-5| \)[/tex] scales the absolute value by a factor of 3.
3. Translation: Adding 1, [tex]\( 3|x-5| + 1 \)[/tex], shifts the whole graph upwards by 1 unit.
### Step 3: Identify Key Points and Behavior
- The vertex of the absolute value function [tex]\( |x-5| \)[/tex] occurs at [tex]\( x = 5 \)[/tex].
- This vertex is shifted to [tex]\( y = 1 \)[/tex] due to the [tex]\( +1 \)[/tex] term.
- For [tex]\( x \neq 5 \)[/tex], the slope of the lines before and after the vertex will be affected by the scaling factor 3.
Let's calculate some key points around the vertex [tex]\( x = 5 \)[/tex]:
- For [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 3|5-5| + 1 = 3 \cdot 0 + 1 = 1 \][/tex]
So, [tex]\( (5, 1) \)[/tex] is a point on the graph.
- For [tex]\( x < 5 \)[/tex], take [tex]\( x = 4 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[ f(4) = 3|4-5| + 1 = 3 \cdot | -1 | + 1 = 3 \cdot 1 + 1 = 4 \][/tex]
[tex]\[ f(3) = 3|3-5| + 1 = 3 \cdot | -2 | + 1 = 3 \cdot 2 + 1 = 7 \][/tex]
Thus, points [tex]\( (4, 4) \)[/tex] and [tex]\( (3, 7) \)[/tex] lie on the graph.
- For [tex]\( x > 5 \)[/tex], take [tex]\( x = 6 \)[/tex] and [tex]\( x = 7 \)[/tex]:
[tex]\[ f(6) = 3|6-5| + 1 = 3 \cdot 1 + 1 = 4 \][/tex]
[tex]\[ f(7) = 3|7-5| + 1 = 3 \cdot 2 + 1 = 7 \][/tex]
Thus, points [tex]\( (6, 4) \)[/tex] and [tex]\( (7, 7) \)[/tex] lie on the graph.
### Step 4: Sketch the Graph
1. Plot the vertex at [tex]\( (5, 1) \)[/tex].
2. Draw lines with a slope of 3 on both sides of the vertex:
- For [tex]\( x < 5 \)[/tex], the slope is positive indicating an increase in values.
- For [tex]\( x > 5 \)[/tex], the slope is negative indicating an increase in values away from the vertex because of the absolute value.
### Step 5: Connect the Points
- Connect the points [tex]\( (3, 7) \)[/tex], [tex]\( (4, 4) \)[/tex], and [tex]\( (5, 1) \)[/tex] to form the decreasing slope.
- Connect the points [tex]\( (5, 1) \)[/tex], [tex]\( (6, 4) \)[/tex], and [tex]\( (7, 7) \)[/tex] to form the increasing slope.
### Final Graph
- The graph forms a 'V'-shape opening upwards with the vertex at [tex]\( (5, 1) \)[/tex], and slopes of ±3 on either side of the vertex.
By plotting the identified points and connecting them appropriately, you get a graph of [tex]\( f(x) = 3|x-5| + 1 \)[/tex]. The function symmetrically increases away from the vertex due to the absolute value nature and the scaling factor.
### Step 1: Understand the Function
The given function [tex]\( f(x) = 3|x-5| + 1 \)[/tex] incorporates an absolute value term and a linear transformation. Absolute value functions tend to create 'V' shapes in their graph.
### Step 2: Break Down the Function Components
1. Absolute Value: [tex]\( |x-5| \)[/tex].
2. Scaling: [tex]\( 3|x-5| \)[/tex] scales the absolute value by a factor of 3.
3. Translation: Adding 1, [tex]\( 3|x-5| + 1 \)[/tex], shifts the whole graph upwards by 1 unit.
### Step 3: Identify Key Points and Behavior
- The vertex of the absolute value function [tex]\( |x-5| \)[/tex] occurs at [tex]\( x = 5 \)[/tex].
- This vertex is shifted to [tex]\( y = 1 \)[/tex] due to the [tex]\( +1 \)[/tex] term.
- For [tex]\( x \neq 5 \)[/tex], the slope of the lines before and after the vertex will be affected by the scaling factor 3.
Let's calculate some key points around the vertex [tex]\( x = 5 \)[/tex]:
- For [tex]\( x = 5 \)[/tex]:
[tex]\[ f(5) = 3|5-5| + 1 = 3 \cdot 0 + 1 = 1 \][/tex]
So, [tex]\( (5, 1) \)[/tex] is a point on the graph.
- For [tex]\( x < 5 \)[/tex], take [tex]\( x = 4 \)[/tex] and [tex]\( x = 3 \)[/tex]:
[tex]\[ f(4) = 3|4-5| + 1 = 3 \cdot | -1 | + 1 = 3 \cdot 1 + 1 = 4 \][/tex]
[tex]\[ f(3) = 3|3-5| + 1 = 3 \cdot | -2 | + 1 = 3 \cdot 2 + 1 = 7 \][/tex]
Thus, points [tex]\( (4, 4) \)[/tex] and [tex]\( (3, 7) \)[/tex] lie on the graph.
- For [tex]\( x > 5 \)[/tex], take [tex]\( x = 6 \)[/tex] and [tex]\( x = 7 \)[/tex]:
[tex]\[ f(6) = 3|6-5| + 1 = 3 \cdot 1 + 1 = 4 \][/tex]
[tex]\[ f(7) = 3|7-5| + 1 = 3 \cdot 2 + 1 = 7 \][/tex]
Thus, points [tex]\( (6, 4) \)[/tex] and [tex]\( (7, 7) \)[/tex] lie on the graph.
### Step 4: Sketch the Graph
1. Plot the vertex at [tex]\( (5, 1) \)[/tex].
2. Draw lines with a slope of 3 on both sides of the vertex:
- For [tex]\( x < 5 \)[/tex], the slope is positive indicating an increase in values.
- For [tex]\( x > 5 \)[/tex], the slope is negative indicating an increase in values away from the vertex because of the absolute value.
### Step 5: Connect the Points
- Connect the points [tex]\( (3, 7) \)[/tex], [tex]\( (4, 4) \)[/tex], and [tex]\( (5, 1) \)[/tex] to form the decreasing slope.
- Connect the points [tex]\( (5, 1) \)[/tex], [tex]\( (6, 4) \)[/tex], and [tex]\( (7, 7) \)[/tex] to form the increasing slope.
### Final Graph
- The graph forms a 'V'-shape opening upwards with the vertex at [tex]\( (5, 1) \)[/tex], and slopes of ±3 on either side of the vertex.
By plotting the identified points and connecting them appropriately, you get a graph of [tex]\( f(x) = 3|x-5| + 1 \)[/tex]. The function symmetrically increases away from the vertex due to the absolute value nature and the scaling factor.
