Core Connections Integrated II, 2015
CC
Core Connections Integrated II, 2015 View details
1. Section 9.1
Continue to next subchapter

Exercise 7 Page 485

Practice makes perfect
a Let's begin by calculating a few values of the function in the given domain.
x (x-2)^2 Simplify y
-1 ( -1-2)^2 (-3)^2 9
0 ( 0-2)^2 (-2)^2 4
1 ( 1-2)^2 (-1)^2 1
2 ( 2-2)^2 0^2 0
3 ( 3-2)^2 1^2 1
4 ( 4-2)^2 2^2 4
5 ( 5-2)^2 3^2 9

By plotting these points in a coordinate plane, we will outline the shape of the graph.

We can now draw the function's graph. Since it is a quadratic function, we know that the graph will take the shape of a parabola.

b To compare y=(x-2)^2 with its parent function, y=x^2, we will put them in the same coordinate plane.

Examining the diagram, we notice that y=(x-2)^2 is a horizontal shift of its parent function. To determine how much it has shifted, we must find the horizontal distance between two corresponding points, such as the parabola's vertex.

As we can see, when subtracting a constant from the input of a function, it shifts the graph to the right by the same number of units as the constant.