Answer :
To tackle the problem given, we begin by examining the two functions derived from the equation:
[tex]\[ (x + 1)^2 - 2 = \frac{2}{x} \][/tex]
We denote the left-hand side function as [tex]\( f(x) \)[/tex] and the right-hand side function as [tex]\( g(x) \)[/tex]:
[tex]\[ f(x) = (x + 1)^2 - 2 \][/tex]
[tex]\[ g(x) = \frac{2}{x} \][/tex]
1. Graphing the Functions
Let’s analyze each function:
Function [tex]\( f(x) = (x + 1)^2 - 2 \)[/tex]:
This is a quadratic function. Its graph will be a parabola:
- Vertex: The vertex form of a quadratic function is [tex]\( a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex of the parabola. Here, [tex]\( f(x) = (x + 1)^2 - 2 \)[/tex] can be compared to the vertex form where [tex]\( (h, k) = (-1, -2) \)[/tex]. The parabola opens upwards because the coefficient of the squared term (which is 1) is positive.
Function [tex]\( g(x) = \frac{2}{x} \)[/tex]:
This is a rational function. Its graph will display hyperbolic behavior:
- Vertical Asymptote: [tex]\( x = 0 \)[/tex] (since division by zero is undefined)
- Horizontal Asymptote: [tex]\( y = 0 \)[/tex] as [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex]
2. Intersection Points
To find the intersection points, where [tex]\( f(x) = g(x) \)[/tex]:
[tex]\[ (x + 1)^2 - 2 = \frac{2}{x} \][/tex]
Using the result from the solution, the [tex]\( x \)[/tex]-values that satisfy this equation are:
[tex]\[ x = -2, \quad x = -1, \quad x = 1 \][/tex]
3. Interpretation of the Solutions
By graphing the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] and identifying where they intersect, we see the following:
- At [tex]\( x = -2 \)[/tex]:
[tex]\( f(-2) = ((-2) + 1)^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1 \)[/tex]
and
[tex]\( g(-2) = \frac{2}{-2} = -1 \)[/tex]
So, [tex]\( f(-2) = g(-2) \)[/tex].
- At [tex]\( x = -1 \)[/tex]:
[tex]\( f(-1) = ((-1) + 1)^2 - 2 = 0^2 - 2 = -2 \)[/tex]
and
[tex]\( g(-1) = \frac{2}{-1} = -2 \)[/tex]
So, [tex]\( f(-1) = g(-1) \)[/tex].
- At [tex]\( x = 1 \)[/tex]:
[tex]\( f(1) = (1 + 1)^2 - 2 = 2^2 - 2 = 4 - 2 = 2 \)[/tex]
and
[tex]\( g(1) = \frac{2}{1} = 2 \)[/tex]
So, [tex]\( f(1) = g(1) \)[/tex].
Therefore, the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] intersect at points:
[tex]\[ x = -2, \quad x = -1, \quad x = 1 \][/tex]
These intersection points correspond to the solutions obtained previously when solving the equation algebraically. The [tex]\( x \)[/tex]-values of the solutions are [tex]\(-2\)[/tex], [tex]\(-1\)[/tex], and [tex]\(1\)[/tex]. The solutions derived from graphing match exactly with our earlier findings.
In summary:
- The functions [tex]\( f(x) = (x + 1)^2 - 2 \)[/tex] and [tex]\( g(x) = \frac{2}{x} \)[/tex] intersect at [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 1 \)[/tex].
- Therefore, the [tex]\( x \)[/tex]-values of the solutions are [tex]\(-2\)[/tex], [tex]\(-1\)[/tex], and [tex]\(1\)[/tex].
[tex]\[ (x + 1)^2 - 2 = \frac{2}{x} \][/tex]
We denote the left-hand side function as [tex]\( f(x) \)[/tex] and the right-hand side function as [tex]\( g(x) \)[/tex]:
[tex]\[ f(x) = (x + 1)^2 - 2 \][/tex]
[tex]\[ g(x) = \frac{2}{x} \][/tex]
1. Graphing the Functions
Let’s analyze each function:
Function [tex]\( f(x) = (x + 1)^2 - 2 \)[/tex]:
This is a quadratic function. Its graph will be a parabola:
- Vertex: The vertex form of a quadratic function is [tex]\( a(x - h)^2 + k \)[/tex], where [tex]\((h, k)\)[/tex] is the vertex of the parabola. Here, [tex]\( f(x) = (x + 1)^2 - 2 \)[/tex] can be compared to the vertex form where [tex]\( (h, k) = (-1, -2) \)[/tex]. The parabola opens upwards because the coefficient of the squared term (which is 1) is positive.
Function [tex]\( g(x) = \frac{2}{x} \)[/tex]:
This is a rational function. Its graph will display hyperbolic behavior:
- Vertical Asymptote: [tex]\( x = 0 \)[/tex] (since division by zero is undefined)
- Horizontal Asymptote: [tex]\( y = 0 \)[/tex] as [tex]\( x \to \infty \)[/tex] or [tex]\( x \to -\infty \)[/tex]
2. Intersection Points
To find the intersection points, where [tex]\( f(x) = g(x) \)[/tex]:
[tex]\[ (x + 1)^2 - 2 = \frac{2}{x} \][/tex]
Using the result from the solution, the [tex]\( x \)[/tex]-values that satisfy this equation are:
[tex]\[ x = -2, \quad x = -1, \quad x = 1 \][/tex]
3. Interpretation of the Solutions
By graphing the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] and identifying where they intersect, we see the following:
- At [tex]\( x = -2 \)[/tex]:
[tex]\( f(-2) = ((-2) + 1)^2 - 2 = (-1)^2 - 2 = 1 - 2 = -1 \)[/tex]
and
[tex]\( g(-2) = \frac{2}{-2} = -1 \)[/tex]
So, [tex]\( f(-2) = g(-2) \)[/tex].
- At [tex]\( x = -1 \)[/tex]:
[tex]\( f(-1) = ((-1) + 1)^2 - 2 = 0^2 - 2 = -2 \)[/tex]
and
[tex]\( g(-1) = \frac{2}{-1} = -2 \)[/tex]
So, [tex]\( f(-1) = g(-1) \)[/tex].
- At [tex]\( x = 1 \)[/tex]:
[tex]\( f(1) = (1 + 1)^2 - 2 = 2^2 - 2 = 4 - 2 = 2 \)[/tex]
and
[tex]\( g(1) = \frac{2}{1} = 2 \)[/tex]
So, [tex]\( f(1) = g(1) \)[/tex].
Therefore, the functions [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] intersect at points:
[tex]\[ x = -2, \quad x = -1, \quad x = 1 \][/tex]
These intersection points correspond to the solutions obtained previously when solving the equation algebraically. The [tex]\( x \)[/tex]-values of the solutions are [tex]\(-2\)[/tex], [tex]\(-1\)[/tex], and [tex]\(1\)[/tex]. The solutions derived from graphing match exactly with our earlier findings.
In summary:
- The functions [tex]\( f(x) = (x + 1)^2 - 2 \)[/tex] and [tex]\( g(x) = \frac{2}{x} \)[/tex] intersect at [tex]\( x = -2 \)[/tex], [tex]\( x = -1 \)[/tex], and [tex]\( x = 1 \)[/tex].
- Therefore, the [tex]\( x \)[/tex]-values of the solutions are [tex]\(-2\)[/tex], [tex]\(-1\)[/tex], and [tex]\(1\)[/tex].
