Its vertex is the point (−5, 0) and the axis of symmetry is x = −5. Similarly, the graph of y = ( x + 5) 2 is congruent to the basic parabola, but is translated 5 units to the left. The graph of y = ( x − 3) 2 is congruent to the basic parabola, but is translated 3 units to the right. If we put x = 0 we obtain y = 9 and this is called the y- intercept. Thus x = 3 is the equation of the axis symmetry for this graph, which has its vertex at (3, 0). Since the same thing will happen as we move the same number of units to the right or left of 3, we see that the graph of this function is symmetric about the line x = 3. X = 1, which is 2 units to the left of 3. The value x = 5, which is 2 units to the right of 3, produces the same y value as does Since the right-hand side is a square, the y-values are all non-negative and takes the value 0 when x = 3. We begin by considering the equation y = ( x − 3) 2. What happens when we translate the basic parabola to the left or to the right? We can confirm this with a table of values. Similarly, the basic parabola becomes y = x 2 − 9 when translated down 9 units, with vertex (0, 9). The vertex of this parabola is now (0, 9), but it has the same axis of symmetry. The equation of this new parabola is thus y = x 2 + 9. Thus, for example, translating the parabola upwards by 9 units, shifts the general point ( a, a 2) to ( a, a 2 + 9). When we translate the parabola vertically upwards or downwards, the y-value of each point on the basic parabola is increased or decreased. Thus the value a = 2, gives the point (2, 4). Since each point on the basic parabola satisfies the equation y = x 2 we can express the coordinates of the general point by ( a, a 2). It is also possible to reflect in other lines and rotate parabolas, but the corresponding algebra is difficult and is generally studied in later years using matrices. This basic parabola can be reflected in vertical and horizontal lines and translated to produce congruent parabolas.
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