- d.Statement ais true because, by definition, a
positive slope ofmmeans that the graph of the
line rises vertically m units for every positive
unit increase in x. For example the graph of
the function f(x) = x^1 2 intersects both Quad-
rants I and II, illustrating the truth of state-
mentb. (See the following graph.) Statement c
is true because the vertex is the point at which
the maximum or minimum value of a quadratic
function occurs. These graphs resemble the let-
ter U or an upside down U, so that the graph is
indeed decreasing on one side of the vertex and
increasing on the other side of the vertex.
- b.The x-values of the points of intersection of
the graphs off(x) = –4xand g(x) = 2xmust
satisfy the equation–4x= 2x, which is solved
as follows:
–4x= 2x
(–4x)^2 = (2x)^2
16 x^2 = 4x
16 x^2 – 4x= 0
4 x(4x– 1) = 0
x= 0,^14
- b.The x-values of the points of intersection of
the graphs off(x) = xand g(x) = 3xmust
satisfy the equation x= 3x, which is
solved as follows:
x= 3x
x– 3x= 0
–2x= 0
x= 0
Set 47 (Page 112)
- d.We apply the general principle that the graph
ofy= g(x+ h) + kis obtained by shifting the
graph ofy= g(x) right (resp. left) hunits ifh
0 (resp.h0), and up (resp.) down kunits
ifk0 (resp.k0). Here, observe that f(x)
= (x+ 2)^3 – 3 = g(x+ 2) – 3, so the correct
choice is d.
- d.In order for the turning point of the parabola
to be in Quadrant II, its x-coordinate must be
negative and its y-coordinate must be positive.
Note that the turning point ofy= –(x+ 2)^2 + 1
is (–2, 1), so that the correct choice is d.
- b.The turning point ofy= (x– 2)^2 – 2 is (2, –2),
while the turning point ofy= x^2 is (0,0). There-
fore, we would shift the graph ofy= x^2 to the
right 2 units and down 2 units.
- c.The graph ofy= g(x+ h) + kis obtained by
shifting the graph ofy= g(x) right (resp. left)
hunits ifh0 (resp.h0), and up (resp.
down) kunits ifk0 (resp.k0). Here,
observe that f(x) = (x– 4)^3 + 1 = g(x– 4) + 1,
so the correct choice is c.
- d.The graph ofy= g(x+ h) + kis obtained by
shifting the graph ofy= g(x) right (resp. left)
hunits ifh0 (resp.h0), and up (resp.
down) kunits ifk0 (resp.k0). Here,
observe that f(x) = (x– 2)^2 – 4 = g(x– 2) – 4,
so the correct choice is d.
–10 –8 –6 –4 –2 2 4 6 8 10
–2
–4
–6
–8
–10
10
8
6
4
2
ANSWERS & EXPLANATIONS–
