To find the points of intersection, we should graph the two functions in the same coordinate plane. To graph the quadratic function, we need to calculate some points that fall on the graph.
|c|c|c|
[-0.8em]
x & (x-2)^2-3 & y [0.2em]
[-0.8em]
-1 & ( -1-2)^2-3 & 6 [0.2em]
[-0.8em]
0 & ( 0-2)^2-3 & 1 [0.2em]
[-0.8em]
1 & ( 1-2)^2-3 & -2 [0.2em]
[-0.8em]
2 & ( 2-2)^2-3 & -3 [0.2em]
[-0.8em]
3 & ( 3-2)^2-3 & -2 [0.2em]
[-0.8em]
4 & ( 4-2)^2-3 & 1 [0.2em]
The values from the table can now be plotted as points and connected with a curve. Since the function is a quadratic, the graph will have the shape of a parabola.
Next, we will graph y=1. This graph is a horizontal line through the y-coordinate 1. Let's graph it and mark the x-values where the line intersects the parabola.
The graphs intersect at x=0 and x=4 which is the solutions to the equation. We can verify this by solving the equation algebraically.
