Answer :
Certainly! To graph the equation [tex]\( (x + 1)^2 - 2 = \frac{2}{x} \)[/tex] as two separate functions, let’s follow these steps:
### Step 1: Define the Functions
We will define two functions [tex]\( g1(x) \)[/tex] and [tex]\( g2(x) \)[/tex] such that:
[tex]\[ g1(x) = (x + 1)^2 - 2 \][/tex]
[tex]\[ g2(x) = \frac{2}{x} \][/tex]
### Step 2: Understand the Functions
#### 1. Analyzing [tex]\( g1(x) \)[/tex]:
[tex]\[ g1(x) = (x + 1)^2 - 2 \][/tex]
This is a quadratic function shifted horizontally by -1 and vertically by -2.
#### 2. Analyzing [tex]\( g2(x) \)[/tex]:
[tex]\[ g2(x) = \frac{2}{x} \][/tex]
This is a reciprocal function, which has vertical asymptotes at [tex]\( x = 0 \)[/tex].
### Step 3: Graph Each Function
#### For [tex]\( g1(x) \)[/tex]:
1. Vertex Form of [tex]\( g1(x) \)[/tex]:
[tex]\[ g1(x) = (x + 1)^2 - 2 \][/tex]
- The vertex of this parabola is (-1, -2).
- Opens upwards because the coefficient of [tex]\( (x+1)^2 \)[/tex] is positive.
2. Plot Key Points:
- Vertex: [tex]\((-1, -2)\)[/tex]
- Points: Calculate a few points around the vertex for [tex]\( x = -2, 0, 1, -3 \)[/tex], ...
[tex]\[ \begin{array}{cccc} x & -2 & -3 & 0 & 1 \\ g1(-2) & (-2 + 1)^2 - 2 = (-1)^2 - 2 = -1 & (-3 + 1)^2 - 2 = (-2)^2 - 2 = 2 & (0 + 1)^2 - 2 = 1 - 2 = -1 & (1 + 1)^2 - 2 = 4 - 2 = 2 \\ \end{array} \][/tex]
#### For [tex]\( g2(x) \)[/tex]:
1. Behavior of [tex]\( g2(x) \)[/tex]:
[tex]\[ g2(x) = \frac{2}{x} \][/tex]
2. Vertical Asymptote:
- At [tex]\( x = 0 \)[/tex], the function approaches infinity, indicating a vertical asymptote.
3. Key Points and Symmetry:
- For positive values of [tex]\( x \)[/tex], [tex]\( g2(x) \)[/tex] is positive.
- For negative values of [tex]\( x \)[/tex], [tex]\( g2(x) \)[/tex] is negative.
Calculate a few points:
[tex]\[ \begin{array}{cccc} x & -2 & -1 & 1 & 2 \\ g2(x) & \frac{2}{-2} = -1 & \frac{2}{-1} = -2 & \frac{2}{1} = 2 & \frac{2}{2} = 1 \\ \end{array} \][/tex]
### Step 4: Plot the Graph
1. Draw [tex]\( g1(x) \)[/tex]:
- Plot points such as (-2, -1), (0, -1), (1, 2).
- Draw a smooth curve through these points, opening upwards from the vertex at (-1, -2).
2. Draw [tex]\( g2(x) \)[/tex]:
- Plot points like (-2, -1), (-1, -2), (1, 2), (2, 1).
- Draw hyperbolic curves on both sides of the vertical asymptote at [tex]\( x = 0 \)[/tex].
### Step 5: Intersection Points
The solutions to the equation were found to be [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 1 \)[/tex]. These are the points where [tex]\( g1(x) \)[/tex] and [tex]\( g2(x) \)[/tex] intersect.
- Verify:
- At [tex]\( x = -2 \)[/tex]:
- [tex]\( g1(-2) = (-2 + 1)^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1 \)[/tex]
- [tex]\( g2(-2) = \frac{2}{-2} = -1 \)[/tex]
- At [tex]\( x = -1 \)[/tex]:
- [tex]\( g1(-1) = (-1 + 1)^2 - 2 = 0 - 2 = -2 \)[/tex]
- [tex]\( g2(-1) = \frac{2}{-1} = -2 \)[/tex]
- At [tex]\( x = 1 \)[/tex]:
- [tex]\( g1(1) = (1 + 1)^2 - 2 = 4 - 2 = 2 \)[/tex]
- [tex]\( g2(1) = \frac{2}{1} = 2 \)[/tex]
These points confirm the intersections.
### Final Graph
- Curve of [tex]\( g1(x) \)[/tex] with a vertex at (-1, -2) and parabolic shape.
- Hyperbolic curves of [tex]\( g2(x) \)[/tex] with a vertical asymptote at [tex]\( x = 0 \)[/tex].
- Intersection points at [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 1 \)[/tex].
By plotting these correctly, you can visualize the solutions and behavior of the two functions in relation to each other.
### Step 1: Define the Functions
We will define two functions [tex]\( g1(x) \)[/tex] and [tex]\( g2(x) \)[/tex] such that:
[tex]\[ g1(x) = (x + 1)^2 - 2 \][/tex]
[tex]\[ g2(x) = \frac{2}{x} \][/tex]
### Step 2: Understand the Functions
#### 1. Analyzing [tex]\( g1(x) \)[/tex]:
[tex]\[ g1(x) = (x + 1)^2 - 2 \][/tex]
This is a quadratic function shifted horizontally by -1 and vertically by -2.
#### 2. Analyzing [tex]\( g2(x) \)[/tex]:
[tex]\[ g2(x) = \frac{2}{x} \][/tex]
This is a reciprocal function, which has vertical asymptotes at [tex]\( x = 0 \)[/tex].
### Step 3: Graph Each Function
#### For [tex]\( g1(x) \)[/tex]:
1. Vertex Form of [tex]\( g1(x) \)[/tex]:
[tex]\[ g1(x) = (x + 1)^2 - 2 \][/tex]
- The vertex of this parabola is (-1, -2).
- Opens upwards because the coefficient of [tex]\( (x+1)^2 \)[/tex] is positive.
2. Plot Key Points:
- Vertex: [tex]\((-1, -2)\)[/tex]
- Points: Calculate a few points around the vertex for [tex]\( x = -2, 0, 1, -3 \)[/tex], ...
[tex]\[ \begin{array}{cccc} x & -2 & -3 & 0 & 1 \\ g1(-2) & (-2 + 1)^2 - 2 = (-1)^2 - 2 = -1 & (-3 + 1)^2 - 2 = (-2)^2 - 2 = 2 & (0 + 1)^2 - 2 = 1 - 2 = -1 & (1 + 1)^2 - 2 = 4 - 2 = 2 \\ \end{array} \][/tex]
#### For [tex]\( g2(x) \)[/tex]:
1. Behavior of [tex]\( g2(x) \)[/tex]:
[tex]\[ g2(x) = \frac{2}{x} \][/tex]
2. Vertical Asymptote:
- At [tex]\( x = 0 \)[/tex], the function approaches infinity, indicating a vertical asymptote.
3. Key Points and Symmetry:
- For positive values of [tex]\( x \)[/tex], [tex]\( g2(x) \)[/tex] is positive.
- For negative values of [tex]\( x \)[/tex], [tex]\( g2(x) \)[/tex] is negative.
Calculate a few points:
[tex]\[ \begin{array}{cccc} x & -2 & -1 & 1 & 2 \\ g2(x) & \frac{2}{-2} = -1 & \frac{2}{-1} = -2 & \frac{2}{1} = 2 & \frac{2}{2} = 1 \\ \end{array} \][/tex]
### Step 4: Plot the Graph
1. Draw [tex]\( g1(x) \)[/tex]:
- Plot points such as (-2, -1), (0, -1), (1, 2).
- Draw a smooth curve through these points, opening upwards from the vertex at (-1, -2).
2. Draw [tex]\( g2(x) \)[/tex]:
- Plot points like (-2, -1), (-1, -2), (1, 2), (2, 1).
- Draw hyperbolic curves on both sides of the vertical asymptote at [tex]\( x = 0 \)[/tex].
### Step 5: Intersection Points
The solutions to the equation were found to be [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 1 \)[/tex]. These are the points where [tex]\( g1(x) \)[/tex] and [tex]\( g2(x) \)[/tex] intersect.
- Verify:
- At [tex]\( x = -2 \)[/tex]:
- [tex]\( g1(-2) = (-2 + 1)^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1 \)[/tex]
- [tex]\( g2(-2) = \frac{2}{-2} = -1 \)[/tex]
- At [tex]\( x = -1 \)[/tex]:
- [tex]\( g1(-1) = (-1 + 1)^2 - 2 = 0 - 2 = -2 \)[/tex]
- [tex]\( g2(-1) = \frac{2}{-1} = -2 \)[/tex]
- At [tex]\( x = 1 \)[/tex]:
- [tex]\( g1(1) = (1 + 1)^2 - 2 = 4 - 2 = 2 \)[/tex]
- [tex]\( g2(1) = \frac{2}{1} = 2 \)[/tex]
These points confirm the intersections.
### Final Graph
- Curve of [tex]\( g1(x) \)[/tex] with a vertex at (-1, -2) and parabolic shape.
- Hyperbolic curves of [tex]\( g2(x) \)[/tex] with a vertical asymptote at [tex]\( x = 0 \)[/tex].
- Intersection points at [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 1 \)[/tex].
By plotting these correctly, you can visualize the solutions and behavior of the two functions in relation to each other.
