Answer :
To graph the function [tex]\( f(x) = \frac{1}{2} x^2 + 2 x - 6 \)[/tex]:
1. Recognize the type of function:
The given function is a quadratic function of the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] where [tex]\( a = \frac{1}{2} \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -6 \)[/tex]. Quadratic functions produce parabolas.
2. Determine the direction of the parabola:
Since the coefficient [tex]\( a \)[/tex] is positive ([tex]\( a = \frac{1}{2} \)[/tex]), the parabola opens upwards.
3. Find the vertex of the parabola:
The vertex of a parabola defined by [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in our values:
[tex]\[ x = -\frac{2}{2 \cdot \frac{1}{2}} = -\frac{2}{1} = -2 \][/tex]
Now we find the [tex]\( y \)[/tex]-coordinate of the vertex by plugging [tex]\( x = -2 \)[/tex] back into the function:
[tex]\[ f(-2) = \frac{1}{2}(-2)^2 + 2(-2) - 6 = \frac{1}{2}(4) - 4 - 6 = 2 - 4 - 6 = -8 \][/tex]
So, the vertex is at [tex]\( (-2, -8) \)[/tex].
4. Identify the axis of symmetry:
The axis of symmetry is a vertical line passing through the vertex. Thus, the axis of symmetry is [tex]\( x = -2 \)[/tex].
5. Find the y-intercept:
The y-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{2}(0)^2 + 2(0) - 6 = -6 \][/tex]
So, the y-intercept is [tex]\( (0, -6) \)[/tex].
6. Find additional points:
Calculating function values at selected x-values helps in plotting the parabola accurately. Here are a few points:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = \frac{1}{2}(-3)^2 + 2(-3) - 6 = \frac{1}{2}(9) - 6 - 6 = 4.5 - 6 - 6 = -7.5 \][/tex]
So, [tex]\( (-3, -7.5) \)[/tex].
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = \frac{1}{2}(-1)^2 + 2(-1) - 6 = \frac{1}{2}(1) - 2 - 6 = 0.5 - 2 - 6 = -7.5 \][/tex]
So, [tex]\( (-1, -7.5) \)[/tex].
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{1}{2}(1)^2 + 2(1) - 6 = \frac{1}{2}(1) + 2 - 6 = 0.5 + 2 - 6 = -3.5 \][/tex]
So, [tex]\( (1, -3.5) \)[/tex].
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{1}{2}(3)^2 + 2(3) - 6 = \frac{1}{2}(9) + 6 - 6 = 4.5 + 6 - 6 = 4.5 \][/tex]
So, [tex]\( (3, 4.5) \)[/tex].
7. Plotting the points and drawing the parabola:
Now that we have key points [tex]\((-2, -8)\)[/tex], [tex]\((0, -6)\)[/tex], [tex]\((-3, -7.5)\)[/tex], [tex]\((-1, -7.5)\)[/tex], [tex]\((1, -3.5)\)[/tex], and [tex]\((3, 4.5)\)[/tex], we can plot these on a coordinate plane.
By connecting these points in a smooth, parabolic shape, we get the graph of the function [tex]\( f(x) = \frac{1}{2} x^2 + 2 x - 6 \)[/tex]. The parabola should open upwards and symmetrically about the line [tex]\( x = -2 \)[/tex]. The vertex is the minimum point on the graph, located at [tex]\((-2, -8)\)[/tex].
1. Recognize the type of function:
The given function is a quadratic function of the form [tex]\( f(x) = ax^2 + bx + c \)[/tex] where [tex]\( a = \frac{1}{2} \)[/tex], [tex]\( b = 2 \)[/tex], and [tex]\( c = -6 \)[/tex]. Quadratic functions produce parabolas.
2. Determine the direction of the parabola:
Since the coefficient [tex]\( a \)[/tex] is positive ([tex]\( a = \frac{1}{2} \)[/tex]), the parabola opens upwards.
3. Find the vertex of the parabola:
The vertex of a parabola defined by [tex]\( f(x) = ax^2 + bx + c \)[/tex] can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
Plugging in our values:
[tex]\[ x = -\frac{2}{2 \cdot \frac{1}{2}} = -\frac{2}{1} = -2 \][/tex]
Now we find the [tex]\( y \)[/tex]-coordinate of the vertex by plugging [tex]\( x = -2 \)[/tex] back into the function:
[tex]\[ f(-2) = \frac{1}{2}(-2)^2 + 2(-2) - 6 = \frac{1}{2}(4) - 4 - 6 = 2 - 4 - 6 = -8 \][/tex]
So, the vertex is at [tex]\( (-2, -8) \)[/tex].
4. Identify the axis of symmetry:
The axis of symmetry is a vertical line passing through the vertex. Thus, the axis of symmetry is [tex]\( x = -2 \)[/tex].
5. Find the y-intercept:
The y-intercept occurs when [tex]\( x = 0 \)[/tex]:
[tex]\[ f(0) = \frac{1}{2}(0)^2 + 2(0) - 6 = -6 \][/tex]
So, the y-intercept is [tex]\( (0, -6) \)[/tex].
6. Find additional points:
Calculating function values at selected x-values helps in plotting the parabola accurately. Here are a few points:
- For [tex]\( x = -3 \)[/tex]:
[tex]\[ f(-3) = \frac{1}{2}(-3)^2 + 2(-3) - 6 = \frac{1}{2}(9) - 6 - 6 = 4.5 - 6 - 6 = -7.5 \][/tex]
So, [tex]\( (-3, -7.5) \)[/tex].
- For [tex]\( x = -1 \)[/tex]:
[tex]\[ f(-1) = \frac{1}{2}(-1)^2 + 2(-1) - 6 = \frac{1}{2}(1) - 2 - 6 = 0.5 - 2 - 6 = -7.5 \][/tex]
So, [tex]\( (-1, -7.5) \)[/tex].
- For [tex]\( x = 1 \)[/tex]:
[tex]\[ f(1) = \frac{1}{2}(1)^2 + 2(1) - 6 = \frac{1}{2}(1) + 2 - 6 = 0.5 + 2 - 6 = -3.5 \][/tex]
So, [tex]\( (1, -3.5) \)[/tex].
- For [tex]\( x = 3 \)[/tex]:
[tex]\[ f(3) = \frac{1}{2}(3)^2 + 2(3) - 6 = \frac{1}{2}(9) + 6 - 6 = 4.5 + 6 - 6 = 4.5 \][/tex]
So, [tex]\( (3, 4.5) \)[/tex].
7. Plotting the points and drawing the parabola:
Now that we have key points [tex]\((-2, -8)\)[/tex], [tex]\((0, -6)\)[/tex], [tex]\((-3, -7.5)\)[/tex], [tex]\((-1, -7.5)\)[/tex], [tex]\((1, -3.5)\)[/tex], and [tex]\((3, 4.5)\)[/tex], we can plot these on a coordinate plane.
By connecting these points in a smooth, parabolic shape, we get the graph of the function [tex]\( f(x) = \frac{1}{2} x^2 + 2 x - 6 \)[/tex]. The parabola should open upwards and symmetrically about the line [tex]\( x = -2 \)[/tex]. The vertex is the minimum point on the graph, located at [tex]\((-2, -8)\)[/tex].
