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To identify the graph of the function [tex]\( f(x) = x^2 + 2x + 3 \)[/tex], we need to understand the characteristics and shape of this quadratic function. Let's break it down step by step:
1. General Shape and Vertex:
- The function [tex]\( f(x) = x^2 + 2x + 3 \)[/tex] is a quadratic function, which means its graph is a parabola.
- Since the coefficient of [tex]\( x^2 \)[/tex] (which is 1) is positive, the parabola opens upwards.
2. Finding the Vertex:
- The vertex of a quadratic function in the form [tex]\( ax^2 + bx + c \)[/tex] can be found using the vertex formula: [tex]\[ x = -\frac{b}{2a} \][/tex]
- In this function, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = 3 \)[/tex]. Plugging in these values:
[tex]\[ x = -\frac{2}{2 \cdot 1} = -1 \][/tex]
- To find the y-coordinate of the vertex, substitute [tex]\( x = -1 \)[/tex] back into the function:
[tex]\[ f(-1) = (-1)^2 + 2(-1) + 3 = 1 - 2 + 3 = 2 \][/tex]
- Therefore, the vertex of the parabola is at [tex]\( (-1, 2) \)[/tex].
3. Y-Intercept:
- The y-intercept occurs where [tex]\( x = 0 \)[/tex].
- Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = 0^2 + 2(0) + 3 = 3 \][/tex]
- Thus, the y-intercept is [tex]\( (0, 3) \)[/tex].
4. Axis of Symmetry:
- The axis of symmetry for the parabola is the vertical line that passes through the vertex. Therefore, the axis of symmetry is [tex]\( x = -1 \)[/tex].
5. Plotting Points:
- Let's consider several values of [tex]\( x \)[/tex] and calculate corresponding [tex]\( y \)[/tex]-values to plot additional points and draw the graph accurately:
- For [tex]\( x = -2 \)[/tex], [tex]\[ f(-2) = (-2)^2 + 2(-2) + 3 = 4 - 4 + 3 = 3 \][/tex]
- For [tex]\( x = 1 \)[/tex], [tex]\[ f(1) = (1)^2 + 2(1) + 3 = 1 + 2 + 3 = 6 \][/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\[ f(2) = (2)^2 + 2(2) + 3 = 4 + 4 + 3 = 11 \][/tex]
These calculations provide the following points: [tex]\( (-2, 3) \)[/tex], [tex]\( (-1, 2) \)[/tex], [tex]\( (0, 3) \)[/tex], [tex]\( (1, 6) \)[/tex], and [tex]\( (2, 11) \)[/tex].
6. Confirming the graph:
- The resulting points and the structure of the function match a characteristic upward-opening parabola.
Summarizing the graph:
- The vertex is [tex]\( (-1, 2) \)[/tex].
- The y-intercept is [tex]\( (0, 3) \)[/tex].
- The parabola is symmetric about the line [tex]\( x = -1 \)[/tex].
- The general shape is an upward-opening parabola, becoming broader as [tex]\( |x| \)[/tex] increases.
By plotting these critical points and drawing a smooth curve through them, you would get the correct graph of the function [tex]\( f(x) = x^2 + 2x + 3 \)[/tex].
1. General Shape and Vertex:
- The function [tex]\( f(x) = x^2 + 2x + 3 \)[/tex] is a quadratic function, which means its graph is a parabola.
- Since the coefficient of [tex]\( x^2 \)[/tex] (which is 1) is positive, the parabola opens upwards.
2. Finding the Vertex:
- The vertex of a quadratic function in the form [tex]\( ax^2 + bx + c \)[/tex] can be found using the vertex formula: [tex]\[ x = -\frac{b}{2a} \][/tex]
- In this function, [tex]\( a = 1 \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = 3 \)[/tex]. Plugging in these values:
[tex]\[ x = -\frac{2}{2 \cdot 1} = -1 \][/tex]
- To find the y-coordinate of the vertex, substitute [tex]\( x = -1 \)[/tex] back into the function:
[tex]\[ f(-1) = (-1)^2 + 2(-1) + 3 = 1 - 2 + 3 = 2 \][/tex]
- Therefore, the vertex of the parabola is at [tex]\( (-1, 2) \)[/tex].
3. Y-Intercept:
- The y-intercept occurs where [tex]\( x = 0 \)[/tex].
- Substitute [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = 0^2 + 2(0) + 3 = 3 \][/tex]
- Thus, the y-intercept is [tex]\( (0, 3) \)[/tex].
4. Axis of Symmetry:
- The axis of symmetry for the parabola is the vertical line that passes through the vertex. Therefore, the axis of symmetry is [tex]\( x = -1 \)[/tex].
5. Plotting Points:
- Let's consider several values of [tex]\( x \)[/tex] and calculate corresponding [tex]\( y \)[/tex]-values to plot additional points and draw the graph accurately:
- For [tex]\( x = -2 \)[/tex], [tex]\[ f(-2) = (-2)^2 + 2(-2) + 3 = 4 - 4 + 3 = 3 \][/tex]
- For [tex]\( x = 1 \)[/tex], [tex]\[ f(1) = (1)^2 + 2(1) + 3 = 1 + 2 + 3 = 6 \][/tex]
- For [tex]\( x = 2 \)[/tex], [tex]\[ f(2) = (2)^2 + 2(2) + 3 = 4 + 4 + 3 = 11 \][/tex]
These calculations provide the following points: [tex]\( (-2, 3) \)[/tex], [tex]\( (-1, 2) \)[/tex], [tex]\( (0, 3) \)[/tex], [tex]\( (1, 6) \)[/tex], and [tex]\( (2, 11) \)[/tex].
6. Confirming the graph:
- The resulting points and the structure of the function match a characteristic upward-opening parabola.
Summarizing the graph:
- The vertex is [tex]\( (-1, 2) \)[/tex].
- The y-intercept is [tex]\( (0, 3) \)[/tex].
- The parabola is symmetric about the line [tex]\( x = -1 \)[/tex].
- The general shape is an upward-opening parabola, becoming broader as [tex]\( |x| \)[/tex] increases.
By plotting these critical points and drawing a smooth curve through them, you would get the correct graph of the function [tex]\( f(x) = x^2 + 2x + 3 \)[/tex].
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