College Algebra, 2013a

122 Relations and Functions
x (x, f(x)) f(x) g(x) =f(x +2) (x, g(x))
0 (0, 1) 1 f(0 + 2) = f(2) = 3 (0, 3)
2 (2, 3) 3 f(2 + 2) = f(4) = 3 (2, 3)
4 (4, 3) 3 f(4 + 2) = f(6) =?
5 (5, 5) 5 f(5 + 2) = f(7) =?
When we substitute x = 4 into the formula g(x) =f(x + 2), we are asked to find f(4 + 2) = f(6)
which doesn’t exist because the domain of f is only [0, 5]. The same thing happens when we
attempt to find g(5). What we need here is a new strategy. We know, for instance, f(0) = 1. To
determine the corresponding point on the graph of g, we need to figure out what value of x we must
substitute into g(x) =f(x + 2) so that the quantity x + 2, works out to be 0. Solving x +2=0
gives x = −2, and g(−2) = f((−2) + 2) = f(0)=1so(−2, 1) is on the graph of g. To use the fact
f(2) = 3, we set x + 2 = 2 to get x = 0. Substituting gives g(0) = f(0 + 2) = f(2) = 3. Continuing
in this fashion, we get
x x +2 g(x) =f(x +2) (x, g(x))
−2 0 g(−2) = f(0) = 1 (−2, 1)
0 2 g(0) = f(2) = 3 (0, 3)
2 4 g(2) = f(4) = 3 (2, 3)
3 5 g(3) = f(5) = 5 (3, 5)
In summary, the points (0, 1), (2, 3), (4, 3) and (5, 5) on the graph of y = f(x) give rise to the
points (−2, 1), (0, 3), (2, 3) and (3, 5) on the graph of y = g(x), respectively. In general, if (a, b) is
on the graph of y = f(x), then f(a) =b. Solving x +2=a gives x = a − 2sothatg(a − 2) =
f((a − 2) + 2) = f(a) =b. As such, (a − 2,b) is on the graph of y = g(x). The point (a − 2,b)is
exactly 2 units to the left of the point (a, b) so the graph of y = g(x) is obtained by shifting the
graph y = f(x) to the left 2 units, as pictured below.
5
y
(5, 5)
5
y
(3, 5)
4
3
2
(2, 3)
(4, 3)
4
(0, 3)
2
(2, 3)
(0, 1)
(−2, 1)
1
−2 −1 1 2 3 4 5
y = f(x)
x
shift left 2 units
−−−−−−−−−−−−→
subtract 2 from each x-coordinate
−2 −1 1 2 3 4 5
y = g(x) =f(x +2)
x
Note that while the ranges of f and g are the same, the domain of g is [−2, 3] whereas the domain
of f is [0, 5]. In general, when we shift the graph horizontally, the range will remain the same, but
the domain could change. If we set out to graph j(x) =f(x − 2), we would find ourselves adding