Answer :
To understand which graph represents the function [tex]\( f(x) = x^2 + 2x - 1 \)[/tex], let's analyze some important characteristics of this quadratic function.
### 1. Identify the Type of Function:
The function [tex]\( f(x) = x^2 + 2x - 1 \)[/tex] is a quadratic function. Its general form is [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -1 \)[/tex].
### 2. Find the Vertex:
The vertex of a parabola given by a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula for the x-coordinate of the vertex, [tex]\( x = -\frac{b}{2a} \)[/tex].
For this function:
- [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex]
- [tex]\( x = -\frac{2}{2 \times 1} = -1 \)[/tex]
Now, substitute [tex]\( x = -1 \)[/tex] into [tex]\( f(x) \)[/tex] to find the y-coordinate of the vertex:
[tex]\[ f(-1) = (-1)^2 + 2(-1) - 1 = 1 - 2 - 1 = -2 \][/tex]
So, the vertex of the parabola is at [tex]\( (-1, -2) \)[/tex].
### 3. Identify the Y-intercept:
The y-intercept of a graph is the point where the graph crosses the y-axis, which occurs at [tex]\( x = 0 \)[/tex].
For this function:
[tex]\[ f(0) = 0^2 + 2(0) - 1 = -1 \][/tex]
This means the y-intercept is at [tex]\( (0, -1) \)[/tex].
### 4. Sketching the Graph:
- The parabola opens upwards because the coefficient of [tex]\( x^2 \)[/tex] (which is 1) is positive.
- The vertex of the parabola is at [tex]\( (-1, -2) \)[/tex].
- The parabola crosses the y-axis at [tex]\( (0, -1) \)[/tex].
### Conclusion:
A graph that has a vertex at [tex]\( (-1, -2) \)[/tex], passes through the point [tex]\( (0, -1) \)[/tex], and opens upwards would represent the function [tex]\( f(x) = x^2 + 2x - 1 \)[/tex].
In multiple-choice settings, look for a graph with these key features to determine the correct representation of the function.
### 1. Identify the Type of Function:
The function [tex]\( f(x) = x^2 + 2x - 1 \)[/tex] is a quadratic function. Its general form is [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -1 \)[/tex].
### 2. Find the Vertex:
The vertex of a parabola given by a quadratic function [tex]\( ax^2 + bx + c \)[/tex] can be found using the formula for the x-coordinate of the vertex, [tex]\( x = -\frac{b}{2a} \)[/tex].
For this function:
- [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex]
- [tex]\( x = -\frac{2}{2 \times 1} = -1 \)[/tex]
Now, substitute [tex]\( x = -1 \)[/tex] into [tex]\( f(x) \)[/tex] to find the y-coordinate of the vertex:
[tex]\[ f(-1) = (-1)^2 + 2(-1) - 1 = 1 - 2 - 1 = -2 \][/tex]
So, the vertex of the parabola is at [tex]\( (-1, -2) \)[/tex].
### 3. Identify the Y-intercept:
The y-intercept of a graph is the point where the graph crosses the y-axis, which occurs at [tex]\( x = 0 \)[/tex].
For this function:
[tex]\[ f(0) = 0^2 + 2(0) - 1 = -1 \][/tex]
This means the y-intercept is at [tex]\( (0, -1) \)[/tex].
### 4. Sketching the Graph:
- The parabola opens upwards because the coefficient of [tex]\( x^2 \)[/tex] (which is 1) is positive.
- The vertex of the parabola is at [tex]\( (-1, -2) \)[/tex].
- The parabola crosses the y-axis at [tex]\( (0, -1) \)[/tex].
### Conclusion:
A graph that has a vertex at [tex]\( (-1, -2) \)[/tex], passes through the point [tex]\( (0, -1) \)[/tex], and opens upwards would represent the function [tex]\( f(x) = x^2 + 2x - 1 \)[/tex].
In multiple-choice settings, look for a graph with these key features to determine the correct representation of the function.
