- The graph of y = (x - h)2 has a vertex at (h,0) that is translated horizontally by h units from the origin. - If h is positive, the translation is to the right, and if h is negative, the translation is to the left. - Other quadratic graphs can be obtained by translating the basic graph of y = x2.

1. Horizontal Translations
Consider what happens when we change the x2 part of the quadratic function
Sketch the following quadratic functions:
y = (x ­ 2)2
Let's compare the tables of values for y = x2 and y = (x ­ 2)2 .
For y = (x ­ 2)2 the value of x that will give the minimum value y = 0 is x = 2.
x y = x2 x y = (x ­ 2)2
0 0 2 0
1 1 Adjust the x coordinates so 3 1
­1 1 that the y coordinates are
the same. This can be done
1 1
2 4 by adding 2 to each of the x 4 4
­2 4 coordinates. The x 0 4
coordinates are shifted or
3 9 translated 2 units to the 5 9
­3 9 right. ­1 9
This has the effect of translating the graph horizontally 2 units to the right.
y = (x ­ 2)2
(2,0)
Sketch the graph of y = (x + 3)2 .
y = (x + 3)2
The vertex has been
translated horizontally 3 units
to the left to (­3,0)
(­3,0)
Summary:
• The graph of y = (x ­ h)2 has a vertex at (h,0).
• The vertex has been translated horizontally by h units.
• If h is positive as in y = (x ­ 2)2, the vertex is translated to the right.
• If h is negative as in y = (x + 3)2 or y = (x ­ (­3))2, the vertex is translated
to the left.
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2. Questions:
1. Write the equation of a quadratic function with a vertex at (­10,0) and a
range of [0, ∞).
2. A parabola has a vertex on the positive x­axis and curves down.
Find a possible equation for this parabola.
3. Compare the graphs of y = x2 and y = 2(x + 5)2.
List at least 2 characteristics of these functions that are the same, and 2
that are different.
4. Sketch the graph of y = (x ­ 3)2 + 4
2

3. Solutions:
1. A possible equation is y = (x + 10)2.
2. A possible equation is y = ­3(x ­ 2)2.
3. The graphs of y = x2 and y = 2(x + 5)2 have the same range, y ≥ 0, and the
same domain.
The coordinate of the vertex for y = 2(x + 5)2 is at (­5,0) and it is narrower
than y = x2.
y = 2(x + 5)2
y = x2
4. The graph of y = (x ­ 3)2 + 4 has a vertex that is shifted right 3 units and up
4 units. The vertex is at (3,4).
y = (x ­ 3)2 + 4
y = x2 (3,4)
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