To determine the graph of the function [tex]\( f(x) = x^2 + 3x + 2 \)[/tex], we need to follow these steps:
1. Identify the type of function: The function [tex]\( f(x) = x^2 + 3x + 2 \)[/tex] is a quadratic function of the form [tex]\( ax^2 + bx + c \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = 2 \)[/tex]. Quadratic functions graph as parabolas.
2. Determine the direction of the parabola: Since the coefficient of [tex]\( x^2 \)[/tex] (which is [tex]\( a \)[/tex]) is positive ([tex]\( a = 1 \)[/tex]), the parabola opens upwards.
3. Find the vertex: The vertex of a parabola in the form [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]
Plugging in the values of [tex]\( a \)[/tex] and [tex]\( b \)[/tex]:
[tex]\[
x = -\frac{3}{2 \cdot 1} = -\frac{3}{2} = -1.5
\][/tex]
To find the y-coordinate of the vertex, plug [tex]\( x = -1.5 \)[/tex] back into the function:
[tex]\[
f(-1.5) = (-1.5)^2 + 3(-1.5) + 2 = 2.25 - 4.5 + 2 = -0.25
\][/tex]
Thus, the vertex is at [tex]\( (-1.5, -0.25) \)[/tex].
4. Find the y-intercept: The y-intercept occurs when [tex]\( x = 0 \)[/tex]. Plugging [tex]\( x = 0 \)[/tex] into the function:
[tex]\[
f(0) = 0^2 + 3(0) + 2 = 2
\][/tex]
Thus, the y-intercept is at [tex]\( (0, 2) \)[/tex].
5. Find the x-intercepts: The x-intercepts occur where [tex]\( f(x) = 0 \)[/tex]. Solve the quadratic equation:
[tex]\[
x^2 + 3x + 2 = 0
\][/tex]
Factor the quadratic:
[tex]\[
(x + 1)(x + 2) = 0
\][/tex]
Set each factor equal to zero and solve for [tex]\( x \)[/tex]:
[tex]\[
x + 1 = 0 \quad \Rightarrow \quad x = -1
\][/tex]
[tex]\[
x + 2 = 0 \quad \Rightarrow \quad x = -2
\][/tex]
Thus, the x-intercepts are at [tex]\( (-1, 0) \)[/tex] and [tex]\( (-2, 0) \)[/tex].
6. Sketch or describe the graph: Based on the information:
- The parabola opens upwards.
- The vertex is at [tex]\( (-1.5, -0.25) \)[/tex].
- The y-intercept is at [tex]\( (0, 2) \)[/tex].
- The x-intercepts are at [tex]\( (-1, 0) \)[/tex] and [tex]\( (-2, 0) \)[/tex].
The graph of [tex]\( f(x) = x^2 + 3x + 2 \)[/tex] is a parabola that opens upwards, with its vertex at [tex]\( (-1.5, -0.25) \)[/tex]. It crosses the x-axis at [tex]\( (-1, 0) \)[/tex] and [tex]\( (-2, 0) \)[/tex], and it crosses the y-axis at [tex]\( (0, 2) \)[/tex]. The basic shape is a symmetric curve about the vertical line passing through the vertex.
This is how you can arrive at determining the correct graph for the function [tex]\( f(x) = x^2 + 3x + 2 \)[/tex].
