Answer:
by transforming the base function y=x
Step-by-step explanation:
Solution.
1. Since g(x) = (x + 2)2 − 3 = f(x + 2) − 3, Theorem 1.7 instructs us to first subtract 2 from
each of the x-values of the points on y = f(x). This shifts the graph of y = f(x) to the left
2 units and moves (−2, 4) to (−4, 4), (−1, 1) to (−3, 1), (0, 0) to (−2, 0), (1, 1) to (−1, 1) and
(2, 4) to (0, 4). Next, we subtract 3 from each of the y-values of these new points. This moves
the graph down 3 units and moves (−4, 4) to (−4, 1), (−3, 1) to (−3, −2), (−2, 0) to (−2, 3),
(−1, 1) to (−1, −2) and (0, 4) to (0, 1). We connect the dots in parabolic fashion to get
__________________________________________________________
From the graph, we see that the vertex has moved from (0, 0) on the graph of y = f(x)
to (−2, −3) on the graph of y = g(x). This sets [−3, ∞) as the range of g. We see that
the graph of y = g(x) crosses the x-axis twice, so we expect two x-intercepts. To find
these, we set y = g(x) = 0 and solve. Doing so yields the equation (x + 2)2 − 3 = 0, or
(x + 2)2 = 3. Extracting square roots gives x + 2 = ±
√
3, or x = −2 ±
√
3. Our x-intercepts
are (−2 −
√
3, 0) ≈ (−3.73, 0) and (−2 + √
3, 0) ≈ (−0.27, 0). The y-intercept of the graph,
(0, 1) was one of the points we originally plotted, so we are done.
