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Investigating transformations of quadratic graphs

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Introduction

Trial Portfolio 1

XYZ

21st October 2012

Portfolio task – trail 1

1. Sketch the graphs of:

a. y = x2 b. y = x2 + 3 c. y = x2 – 2 What do you notice? Can you generalize?

The sketches above are graphs of quadratic functions and therefore parabolas. Although the shapes of all three graphs are the same, the graphs’ positions differ from one another. Taking the other two graphs relative to y = x2, we make observations. The graph of y = x2 + 3 shifts three units upwards and the graph of y = x2 – 2 shifts two units downwards relative to y = x2. The following table explains why this happens:

 x -3 -2 -1 0 1 2 3 y = x2 9 4 1 0 1 4 9 y = x2 + 3 12 7 4 3 4 7 12 y = x2 – 2 7 2 -1 -2 -2 2 7

The table shows that for y = x2 + 3, three is added to the value of y (relative to y = x2). That means when, for example, x = 0, y = 3 ( x = 0, y = 0 for     y = x2). This explains the vertex of y = x2 + 3 moving three units upwards (the y value changes). The same principle applies to y = x2 – 2. In this case, two is subtracted from the value of y (relative to y = x2) and therefore the vertex (and the graph) shifts two units downwards (when x = 0, y = -2).

Middle

-4

-3

-2

-1

0

1

2

3

y = x2

16

9

4

1

0

1

4

9

y = (x – 2)2

36

25

16

9

4

1

0

1

y = (x + 3)2

1

0

1

4

9

16

25

36

For y = (x – 2)2, value is subtracted from x and for y = (x + 3)2, value is added to x (relative to y = x2). Unlike the previous case where the value was added to y, the value is added to x which explains the horizontal movement. For y = x2, when y = 0, x = 0; for y = (x – 2)2, when y = 0, x = 2; for y = (x + 3)2, when y = 0, x = -3. The sketch of the combined graph is as follows: From the above observations, we can say that:

When |k| units are added to x, the graph shifts leftwards by k units

When |k| units are subtracted from x, the graph shifts rightwards by k units.

The phenomenon of the graph shifting leftwards or rightwards is called horizontal translation.

3. Where would you expect the vertex on the graph of y = (x – 4)2 + 5 to be? Explain why.

The vertex would be found at (4, 5)

The equation is given in the form y = a(x – b)2 + c. For this general form of the equation, the vertex can be found by taking opposite value of b (the x-coordinate) and same value of c (the y-coordinate).

Conclusion Do your findings apply to the graphs of other types of functions? Can you generalize?

The vertical and horizontal translations’ theory mentioned above applies to all types of functions. Consider the linear function graph: In the above sketch we observe that when three is added to the original equation y = x, the graph shifts three units upwards and when two is subtracted from the original equation, the graph shifts two units downwards (for y = x, when x = 0, y = 0; for

y = x + 3, when x = 0, y = 3; for y = x – 2, when x = 0, y = -2).

In the above sketch we also observe that when two is subtracted from the original equation y = x, the graph shifts two units rightwards and when three is added to the original equation, the graph shifts three units leftwards (for y = x, when y = 0, x = 0; for y = x + 3, when y = 0, x = -3; for y = x – 2, when y = 0, x = 2). Therefore we can generalize that for y = f(x) + k, the graph shifts upwards if k > 0 by k units and shifts downwards if k < 0 by k units, and for y = f(x + k), the graph shifts leftwards if k > 0 by k units and the graph shifts rightwards if k < 0 by k units.

This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.

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