Answer :
Sure! Let's analyze the quadratic function [tex]\( f(x) = (x - 2)^2 + 1 \)[/tex] step by step.
### 1) Vertex (and plot it)
The function [tex]\( f(x) = (x - 2)^2 + 1 \)[/tex] is in vertex form [tex]\( f(x) = a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
Here, [tex]\( h = 2 \)[/tex] and [tex]\( k = 1 \)[/tex], thus the vertex is [tex]\( (2, 1) \)[/tex].
Plotting the vertex:
You would plot the point (2, 1) on the xy-plane.
### 2) Direction of graph
The coefficient of the quadratic term [tex]\((x - 2)^2\)[/tex] is positive (implicitly 1 in this case), so the parabola opens upwards.
### 3) Domain
For any quadratic function, the domain is the set of all real numbers because you can substitute any real number for [tex]\( x \)[/tex].
Domain: [tex]\(\mathbb{R} \)[/tex] (all real numbers)
### 4) Range
Since the parabola opens upwards and the vertex [tex]\((2, 1)\)[/tex] is its lowest point, the function's output values (f(x)) starts from [tex]\(1\)[/tex] and increases to [tex]\(+\infty\)[/tex].
Range: [tex]\([1, \infty)\)[/tex]
### Graphing from Vertex Form
To graph the function [tex]\( f(x) = (x - 2)^2 + 1 \)[/tex]:
- Start by plotting the vertex [tex]\( (2, 1) \)[/tex].
- Next, because the parabola opens upwards, plot additional points on both sides of the vertex to get the shape.
Example points:
- When [tex]\( x = 1 \)[/tex]: [tex]\( f(1) = (1 - 2)^2 + 1 = 1^2 + 1 = 2 \)[/tex]
- When [tex]\( x = 3 \)[/tex]: [tex]\( f(3) = (3 - 2)^2 + 1 = 1^2 + 1 = 2 \)[/tex]
- When [tex]\( x = 0 \)[/tex]: [tex]\( f(0) = (0 - 2)^2 + 1 = 4 + 1 = 5 \)[/tex]
- When [tex]\( x = 4 \)[/tex]: [tex]\( f(4) = (4 - 2)^2 + 1 = 4 + 1 = 5 \)[/tex]
Connecting these points while ensuring the parabola remains symmetric and opens upwards will give you the accurate shape of the graph.
### Summary:
1. Vertex: [tex]\( (2, 1) \)[/tex]
2. Direction of graph: Upwards
3. Domain: [tex]\(\mathbb{R} \)[/tex] (all real numbers)
4. Range: [tex]\([1, \infty)\)[/tex]
You can confirm this by graphing it on a graphing calculator or using graphing software.
### 1) Vertex (and plot it)
The function [tex]\( f(x) = (x - 2)^2 + 1 \)[/tex] is in vertex form [tex]\( f(x) = a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex of the parabola.
Here, [tex]\( h = 2 \)[/tex] and [tex]\( k = 1 \)[/tex], thus the vertex is [tex]\( (2, 1) \)[/tex].
Plotting the vertex:
You would plot the point (2, 1) on the xy-plane.
### 2) Direction of graph
The coefficient of the quadratic term [tex]\((x - 2)^2\)[/tex] is positive (implicitly 1 in this case), so the parabola opens upwards.
### 3) Domain
For any quadratic function, the domain is the set of all real numbers because you can substitute any real number for [tex]\( x \)[/tex].
Domain: [tex]\(\mathbb{R} \)[/tex] (all real numbers)
### 4) Range
Since the parabola opens upwards and the vertex [tex]\((2, 1)\)[/tex] is its lowest point, the function's output values (f(x)) starts from [tex]\(1\)[/tex] and increases to [tex]\(+\infty\)[/tex].
Range: [tex]\([1, \infty)\)[/tex]
### Graphing from Vertex Form
To graph the function [tex]\( f(x) = (x - 2)^2 + 1 \)[/tex]:
- Start by plotting the vertex [tex]\( (2, 1) \)[/tex].
- Next, because the parabola opens upwards, plot additional points on both sides of the vertex to get the shape.
Example points:
- When [tex]\( x = 1 \)[/tex]: [tex]\( f(1) = (1 - 2)^2 + 1 = 1^2 + 1 = 2 \)[/tex]
- When [tex]\( x = 3 \)[/tex]: [tex]\( f(3) = (3 - 2)^2 + 1 = 1^2 + 1 = 2 \)[/tex]
- When [tex]\( x = 0 \)[/tex]: [tex]\( f(0) = (0 - 2)^2 + 1 = 4 + 1 = 5 \)[/tex]
- When [tex]\( x = 4 \)[/tex]: [tex]\( f(4) = (4 - 2)^2 + 1 = 4 + 1 = 5 \)[/tex]
Connecting these points while ensuring the parabola remains symmetric and opens upwards will give you the accurate shape of the graph.
### Summary:
1. Vertex: [tex]\( (2, 1) \)[/tex]
2. Direction of graph: Upwards
3. Domain: [tex]\(\mathbb{R} \)[/tex] (all real numbers)
4. Range: [tex]\([1, \infty)\)[/tex]
You can confirm this by graphing it on a graphing calculator or using graphing software.