**Unformatted text preview: **e multiply the output by a, we multiply
the y -coordinates on the graph of g by a, so the point (0, 0) remains (0, 0) and remains the vertex.
Next, we deﬁne g2 (x) = g1 (x − h) = a(x − h)2 . This induces a horizontal shift right or left h units3
moves the vertex, in either case, to (h, 0). Finally, f (x) = g2 (x) + k = a(x − h)2 + k which eﬀects
a vertical shift up or down k units4 resulting in the vertex moving from (h, 0) to (h, k ).
To verify the vertex formula for a quadratic function in general form, we complete the square to
convert the general form into the standard form.5
f (x) = ax2 + bx + c
b
= a x2 + x + c
a
b
b2
= a x2 + x + 2 + c − a
a
4a
2
b
4ac − b2
= a x+
+
2a
4a b2
4a2 complete the square
factor; get a common denominator Comparing this last expression with the standard form, we identify (x − h) as x + 2ba so that
b
2
2
h = − . Instead of memorizing the value k = 4ac−b , we see that f − 2ba = 4ac−b . As such,
4a
4a
2a
2 Just a scaling if a > 0. If a < 0, there is a reﬂection involved.
Right if h > 0, left if h &l...

View
Full
Document