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Quadratic Function Example 3

The question is how to move the graph of y = (1/2)x2 to get the graph of y = (1/2)(x - 3)2 - 2?

The function of y = (1/2)x2 can be written to y = (1/2)(x - 0)2 - 0, so its vertex is (0, 0), axis of symmetry is x = 0 which is y-axis, and the graph is open upward because of the coefficient of x square term, a > 0.

The vertex of y = (1/2)(x - 3)2 - 2 is (3, -2), the axis of symmetry is x = 3, and the graph is open upward because of the coefficient of the x square term, a > 0. Because both functions have the same coefficient of the x square term a, so we can move the graph of y = (1/2)x2 to get the graph of y = (1/2)(x - 3)2 - 2.

Note: If the quadratic function is written in the form y = a(x - h)2 + k, then the vertex is (h, k), the axis of symmetry is x = h.

The graph of the quadratic function y = ax2 + bx + x, where the coefficient a can not be zero, is the set of points (x, y) that satisfy the quadratic function y = ax2 + bx + c.

To draw the graph of y = (1/2)x2, because the coefficient a is larger than zero, so the graph is open upward. So x = 0 is the minimum value of the quadratic function. We let x = 0 be the middle value of the set of x.

x ... -2 -1 0 1 2 ...
y ... 2 1/2 0 1/2 2 ...

So we move the graph of y = (1/2)x2 right 3 units to get the graph of y = (x - 3)2, then move the graph down 2 units to get the graph of y = (1/2)(x - 3)2 - 2.