From the above observations, we can say that:
When |k| units are added to y, the graph shifts upward by k units.
When |k| units are subtracted from y, the graph shifts downwards by k units.
The phenomenon of the graph shifting upwards or downwards is called vertical translation.
2. Consider the graphs of:
a. y = x2
b. y = (x – 2)2
c. y = (x + 3)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 = (x – 2)2 shifts two units rightwards and the graph of y = (x + 3)2 shifts three units leftwards relative to y = x2. The following table explains why this happens:
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). In the given equation, b is -4 and c is 5. Therefore the point of the vertex is found at (-b, c) = (4, 5). The following graph proves the aforementioned statement:
4a. Express x2 – 10x + 25 in the form (x – h)2
h = 5
b. Express x2 – 10x + 32 in the form (x – h)2 + g
h = 5, g = 7
c. Repeat this procedure with some examples of your own.
d. Describe a method of writing the quadratic expression x2 + bx + c in the form (x – h)2 + g
The quadratic expression
can be written in the form
by using the completing square method.
Following steps are involved in this method:
Notice that the value of h in the example is positive. To make the two equations agree,
h = -2
5. Describe the shape and the position of the graph of y = (x – h)2 + g
Since the value of x is positive, the graph would be ‘U’ shaped. The vertex of the graph would be (h, g). Consider the following example:
As the coefficient of x2 is positive (and thus the value of a in
), the graph is of ‘U’ shape. Because the vertex of a parabola is given by (-h, g), the vertex for the above example is (3,-34). Following is the proof:
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.