consider the equation y= (x+3)^2 - 2
a) graph the equation
b)use the graph to solve the equation 2=(x+2)^2 - 1
verify your solution algebraically
a) To graph the equation y = (x+3)^2 - 2, you can start by creating a table of values. Choose some values for x, calculate the corresponding values for y, and plot the points on a coordinate plane.
Here's a table of values:
| x | y |
|---|---|
| -5 | 18 |
| -4 | 11 |
| -3 | 6 |
| -2 | 3 |
| -1 | 2 |
| 0 | 3 |
| 1 | 6 |
| 2 | 11 |
| 3 | 18 |
Plotting these points and connecting them with a smooth curve, you will get a U-shaped graph that opens upwards. The vertex of the parabola is at (-3, -2).
b) Now, let's use the graph to solve the equation 2 = (x+2)^2 - 1.
First, plot the point (2, 2) on the same coordinate plane. Since we want to find the x-values where the graph intersects with the line y = 2, we look for the points on the graph that have a y-coordinate of 2.
From the graph, it looks like there are two points of intersection. Let's estimate their x-coordinates. The first point is around x = -4.5, and the second point is around x = -0.5.
To verify these solutions algebraically, we can substitute these values back into the equation 2 = (x+2)^2 - 1 and check if they hold true.
For x = -4.5:
2 = (-4.5+2)^2 - 1
2 = (-2.5)^2 - 1
2 = 6.25 - 1
2 = 5.25 (Not true)
For x = -0.5:
2 = (-0.5+2)^2 - 1
2 = (1.5)^2 - 1
2 = 2.25 - 1
2 = 1.25 (Not true)
So, the estimated solutions from the graph were not accurate. Let's try a different method to solve the equation algebraically.
2 = (x+2)^2 - 1
3 = (x+2)^2
±√3 = x+2
For x = √3 - 2:
2 = (√3 - 2 + 2)^2 - 1
2 = (√3)^2 - 1
2 = 2 - 1
2 = 1 (True)
For x = -√3 - 2:
2 = (-√3 - 2 + 2)^2 - 1
2 = (-√3)^2 - 1
2 = 3 - 1
2 = 2 (True)
Therefore, the correct solutions to the equation 2 = (x+2)^2 - 1 are x = √3 - 2 and x = -√3 - 2.