Answer :
To find the graph of the function [tex]\( f(x) = x^2 + 2x + 3 \)[/tex], follow these steps:
### Step 1: Identify the Function Type
The given function [tex]\( f(x) = x^2 + 2x + 3 \)[/tex] is a quadratic function. Quadratic functions generally graph as parabolas.
### Step 2: Finding the Vertex
The vertex of a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For the given function:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = 2 \][/tex]
So,
[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 [tex]\((-1, 2)\)[/tex].
### Step 3: Finding the Y-intercept
The y-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 + 2 \cdot 0 + 3 = 3 \][/tex]
So, the y-intercept is [tex]\( (0, 3) \)[/tex].
### Step 4: Finding Additional Points
To have a more detailed graph, let's find several more points by substituting some x-values into the function:
For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = (-2)^2 + 2(-2) + 3 = 4 - 4 + 3 = 3 \][/tex]
So, another point is [tex]\((-2, 3)\)[/tex].
For [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = (-3)^2 + 2(-3) + 3 = 9 - 6 + 3 = 6 \][/tex]
So, another point is [tex]\((-3, 6)\)[/tex].
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = (1)^2 + 2(1) + 3 = 1 + 2 + 3 = 6 \][/tex]
So, another point is [tex]\((1, 6)\)[/tex].
For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = (2)^2 + 2(2) + 3 = 4 + 4 + 3 = 11 \][/tex]
So, another point is [tex]\((2, 11)\)[/tex].
### Step 5: Plotting the Points
Using the points we have identified:
- Vertex: [tex]\((-1, 2)\)[/tex]
- Y-intercept: [tex]\( (0, 3) \)[/tex]
- Additional points: [tex]\((-2, 3), (-3, 6), (1, 6), (2, 11)\)[/tex]
### Step 6: Drawing the Parabola
Using the points, draw the parabola. The graph is symmetric with respect to the vertical line that passes through the vertex [tex]\(x = -1\)[/tex]. The curve opens upwards since the coefficient of [tex]\(x^2\)[/tex] is positive.
Given the numerical result, here are some sample points to plot:
[tex]\[ \begin{array}{c} \begin{array}{c|c} \text{x values} & \text{y values} \\ \hline -10 & 83 \\ -9.94987469 & 82.1002569079340 \\ -9.89974937 & 81.2055389099315 \\ -9.84962406 & 80.3158460059924 \\ -9.79949875 & 79.4311781961169 \\ \end{array} \end{array} \][/tex]
... and so on, which span from [tex]\( x = -10 \)[/tex] to [tex]\( x = 10 \)[/tex].
The graphical representation of the points mentioned above, maintaining the values consistently along x-coordinates from [tex]\(-10\)[/tex] to [tex]\(10\)[/tex] shows the parabolic shape. The exact numerical implementation provided results in:
[tex]\[ \begin{align*} & \{-10.0, 83.0\}, \{-9.94987469, 82.1002569079340\}, \{-9.89974937, 81.2055389099315\}, \{-9.84962406, 80.3158460059924\}, \{-9.79949875, 79.4311781961169\}, \\ & \{-2.0, 3.0\}, \{-1.0, 2.0\}, \{0.0, 3.0\}, \{1.0, 6.0\}, \{2.0, 11.0\}, \{10.0, 123.0\} \end{align*} \][/tex]
This gives the complete set of both x and y values to be plotted, ensuring the proper parabolic curve for [tex]\( f(x) = x^2 + 2x + 3 \)[/tex]. Ultimately, visualizing these values will depict a parabola opening upwards with the described features.
### Step 1: Identify the Function Type
The given function [tex]\( f(x) = x^2 + 2x + 3 \)[/tex] is a quadratic function. Quadratic functions generally graph as parabolas.
### Step 2: Finding the Vertex
The vertex of a quadratic function [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
For the given function:
[tex]\[ a = 1 \][/tex]
[tex]\[ b = 2 \][/tex]
So,
[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 [tex]\((-1, 2)\)[/tex].
### Step 3: Finding the Y-intercept
The y-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = 0^2 + 2 \cdot 0 + 3 = 3 \][/tex]
So, the y-intercept is [tex]\( (0, 3) \)[/tex].
### Step 4: Finding Additional Points
To have a more detailed graph, let's find several more points by substituting some x-values into the function:
For [tex]\( x = -2 \)[/tex]:
[tex]\[ f(-2) = (-2)^2 + 2(-2) + 3 = 4 - 4 + 3 = 3 \][/tex]
So, another point is [tex]\((-2, 3)\)[/tex].
For [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = (-3)^2 + 2(-3) + 3 = 9 - 6 + 3 = 6 \][/tex]
So, another point is [tex]\((-3, 6)\)[/tex].
For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = (1)^2 + 2(1) + 3 = 1 + 2 + 3 = 6 \][/tex]
So, another point is [tex]\((1, 6)\)[/tex].
For [tex]\( x = 2 \)[/tex]:
[tex]\[ f(2) = (2)^2 + 2(2) + 3 = 4 + 4 + 3 = 11 \][/tex]
So, another point is [tex]\((2, 11)\)[/tex].
### Step 5: Plotting the Points
Using the points we have identified:
- Vertex: [tex]\((-1, 2)\)[/tex]
- Y-intercept: [tex]\( (0, 3) \)[/tex]
- Additional points: [tex]\((-2, 3), (-3, 6), (1, 6), (2, 11)\)[/tex]
### Step 6: Drawing the Parabola
Using the points, draw the parabola. The graph is symmetric with respect to the vertical line that passes through the vertex [tex]\(x = -1\)[/tex]. The curve opens upwards since the coefficient of [tex]\(x^2\)[/tex] is positive.
Given the numerical result, here are some sample points to plot:
[tex]\[ \begin{array}{c} \begin{array}{c|c} \text{x values} & \text{y values} \\ \hline -10 & 83 \\ -9.94987469 & 82.1002569079340 \\ -9.89974937 & 81.2055389099315 \\ -9.84962406 & 80.3158460059924 \\ -9.79949875 & 79.4311781961169 \\ \end{array} \end{array} \][/tex]
... and so on, which span from [tex]\( x = -10 \)[/tex] to [tex]\( x = 10 \)[/tex].
The graphical representation of the points mentioned above, maintaining the values consistently along x-coordinates from [tex]\(-10\)[/tex] to [tex]\(10\)[/tex] shows the parabolic shape. The exact numerical implementation provided results in:
[tex]\[ \begin{align*} & \{-10.0, 83.0\}, \{-9.94987469, 82.1002569079340\}, \{-9.89974937, 81.2055389099315\}, \{-9.84962406, 80.3158460059924\}, \{-9.79949875, 79.4311781961169\}, \\ & \{-2.0, 3.0\}, \{-1.0, 2.0\}, \{0.0, 3.0\}, \{1.0, 6.0\}, \{2.0, 11.0\}, \{10.0, 123.0\} \end{align*} \][/tex]
This gives the complete set of both x and y values to be plotted, ensuring the proper parabolic curve for [tex]\( f(x) = x^2 + 2x + 3 \)[/tex]. Ultimately, visualizing these values will depict a parabola opening upwards with the described features.
