420 McGRAW-HILL’S SAT
2.C The function h(x) is equivalent to the function
g(x) after it has been reflected over the x-axis, ver-
tically stretched by a factor of 3, and shifted down-
ward one unit. After these transformations, the
vertex of the parabola will be at (1, 5), so the
maximum value isy= 5.
3.D Call the original point (a, b). Its reflection over
the line y= xis (b, a). (Draw a graph to see.) The
reflection of this point over the x-axis is (b, a),
and the reflection of this point over the y-axis is
(b, a). If the final point is (3, 4), then the origi-
nal point was (4, 3).Concept Review 3
- y= f(x5). Although the 5 seems to suggest a
shift to the left (because when we subtract 5 from
a number, we move five units to the left on the
number line), this change actually shifts the graph
to the right. To see why, look back at the first two
examples in Lesson 3, and pay particular attention
to how the changed equations produce the individ-
ual points on the graph and how these points com-
pare with the points on the original graph. It also
may help to pick a simple function, such as y= x^2 ,
graph it by hand (by choosing values for x, calcu-
lating the corresponding values for y, and plotting
the ordered pairs), and then graph y= (x5)^2 in
the same way to see how the graphs compare. - y= x^2 + 5. Since the point (x, y) is the reflection
of (x, y) over the x-axis, reflecting any function
over the x-axis simply means multiplying yby 1.
This “negates” every term in the function. - It is the original graph after it has been “flipped”
vertically and “stretched” vertically. The graph of
y= 4 f(x) is a “vertically stretched” version of
y= f(x) that also has been reflected over the x-axis.
Every point on y= 4 f(x) is four times farther from
the x-axis as its corresponding point on y= f(x) and
is also on the opposite side of the x-axis. - Overall shape and maximum and minimum val-
ues. The graph of y= f(x+ 15) is simply the graph
of y= f(x) shifted to the left 15 units. It maintains
the shape of the original graph and has the same
maximum and minimum values. (That is, if the
greatest value of yon y= f(x) is 10, then the great-
est value of yon y= f(x+ 15) is also 10.)
5. Zeroes, vertical lines of symmetry, and xcoordi-
nates of maximum and minimum values. The
graph of y= 6f(x) is simply the graph of y= f(x)
that has been “vertically stretched” by a factor of
6. Imagine drawing the graph of y= f(x) on a rub-
ber sheet and then attaching sticks across the top
and bottom of the sheet and pulling the sheet until
it’s six times as tall as it was originally. The
stretched graph looks like y= 6f(x). Although most
of the points move because of this stretch, the
ones on the x-axis do not. These points, called the
zeroesbecause their ycoordinates are 0, remain
the same, as do any vertical lines of symmetry and
the xcoordinates of any maximum or minimum
points.
6. The graph of h(x) = ax^2 + bx+ c, since it is qua-
dratic, looks like a parabola. If ais negative, the
parabola is “open down.” (Remember that y= x^2
is the graph of the “open up” parabola y= x^2 after
it has been “flipped” over the x-axis.) Also, notice
that cis the “y-intercept” of the graph, since h(0) =
a(0)^2 + b(0) + c= c.Therefore, the graph is an up-
side-down parabola with a positive y-intercept.
The only choice that qualifies is (B).
7. The point on the d-axis represents the starting
depth of the water. If the tank begins with twice as
much water, the starting point, or “d-intercept,”
must be twice that of the original graph. Also, if
the water drains out at twice the rate, the line
must be twice as steep. Since twice as much water
drains out at twice the rate, the tank should empty
in the same amount of time it took the original
tank to drain. The only graph that depicts this sit-
uation is (A).
Answer Key 3: Transformations
SAT Practice 3
1.D The transformation in (D) is a shift of the orig-
inal function upward two units. This creates a tri-
angular region above the x-axis with a greater
height and base than those of the original graph,
and therefore creates a greater overall area. The
transformation in (A) will create a triangle with
area^1 / 2 A,the transformations in (B) and (C) are
horizontal shifts, and so will not change the area.
The downward shift in (E) will reduce the height
and base, and therefore the total area.