To graph the function [tex]\( y = -2(x+2)^2 - 1 \)[/tex] using the given table of values, we will select and plot at least five points from the table. The chosen points are as follows:
1. [tex]\((-10, -129)\)[/tex]
2. [tex]\((-8, -73)\)[/tex]
3. [tex]\((-4, -9)\)[/tex]
4. [tex]\((0, -9)\)[/tex]
5. [tex]\((4, -73)\)[/tex]
Now, let's proceed with plotting these points:
1. Point (-10, -129):
- This point is located 10 units to the left of the y-axis and 129 units down from the x-axis.
2. Point (-8, -73):
- This point is located 8 units to the left of the y-axis and 73 units down from the x-axis.
3. Point (-4, -9):
- This point is located 4 units to the left of the y-axis and 9 units down from the x-axis.
4. Point (0, -9):
- This point is located on the y-axis at 9 units down from the x-axis.
5. Point (4, -73):
- This point is located 4 units to the right of the y-axis and 73 units down from the x-axis.
By plotting these points on the graph, we get a visual representation of the function [tex]\( y = -2(x+2)^2 - 1 \)[/tex]. The resulting plot should illustrate that the function is a downward-opening parabola with a vertex at [tex]\((-2, -1)\)[/tex].
The process entails carefully placing each point based on the given coordinates and then joining the points to visualize the curve.
#### Let's summarize the coordinates to visualize them more clearly:
- Point 1: [tex]\((-10, -129)\)[/tex]
- Point 2: [tex]\((-8, -73)\)[/tex]
- Point 3: [tex]\((-4, -9)\)[/tex]
- Point 4: [tex]\((0, -9)\)[/tex]
- Point 5: [tex]\((4, -73)\)[/tex]
When these points are graphed, the resulting parabola reflects the nature of the quadratic function, specifically how it opens downwards due to the negative coefficient in front of the [tex]\((x+2)^2\)[/tex] term.
