Let's graph the function [tex]\( f(x) = x^2 + 4x + 6 \)[/tex] by first identifying the vertex and then finding a second point on the graph.
### Finding the Vertex
For a quadratic function of the form [tex]\( f(x) = ax^2 + bx + c \)[/tex], the x-coordinate of the vertex can be found using the formula:
[tex]\[ x = -\frac{b}{2a} \][/tex]
In our case, [tex]\( a = 1 \)[/tex] and [tex]\( b = 4 \)[/tex]. Plugging in these values, we get:
[tex]\[ x = -\frac{4}{2 \cdot 1} = -2 \][/tex]
Next, we find the y-coordinate of the vertex by substituting [tex]\( x = -2 \)[/tex] back into the original function:
[tex]\[ f(-2) = (-2)^2 + 4(-2) + 6 = 4 - 8 + 6 = 2 \][/tex]
Thus, the vertex of the function is at [tex]\( (-2, 2) \)[/tex].
### Selecting a Second Point
To graph the function accurately, we need a second point. One common choice is the y-intercept, which occurs when [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the function:
[tex]\[ f(0) = 0^2 + 4 \cdot 0 + 6 = 6 \][/tex]
This gives us the point [tex]\( (0, 6) \)[/tex].
### Plotting the Graph
Now we have two key points to plot: the vertex [tex]\( (-2, 2) \)[/tex] and the y-intercept [tex]\( (0, 6) \)[/tex].
1. Plot the vertex: Click on and plot the point [tex]\( (-2, 2) \)[/tex].
2. Plot the second point: Click on and plot the point [tex]\( (0, 6) \)[/tex].
With these points marked, you can draw a smooth parabolic curve that opens upwards, passing through both points. Remember, because the quadratic term [tex]\( x^2 \)[/tex] is positive, the parabola opens upwards.
### Summary of Points to Plot
- Vertex: [tex]\( (-2, 2) \)[/tex]
- Second Point: [tex]\( (0, 6) \)[/tex]
Graphing these points will help you visualize the quadratic function [tex]\( f(x) = x^2 + 4x + 6 \)[/tex].
