Homework 1 Solutions
1. (a) Describe the effect of the transformations cf (x), f (x + c), and
f (x) + c on the graph of f (x).
If c > 1, then the transformation cf (x) stretches the graph vertically by a factor of c. If 1 > c > 0, then cf (x) compresses
the graph vertically by a factor of c. The vertical scaling often
described as a narrowing (c > 1) or widening (1 > c > 0) of the
graph. If c < 0, then cf (x) reflects the graph in the x-axis and
also scales the graph according to whether |c| > 1 or 1 > |c| > 0.
If c > 0, then the transformation f (x + c) translates the graph
horizontally to the left by c units. If c < 0, then the graph is
translated to the right by |c| units.
If c > 0, then the transformation f (x) + c translates the graph
vertically upward by c units. If c < 0, then the graph is translated
downward by |c| units.
(b) Suppose that g(x) is a function whose graph is a parabola with
vertex (−1, 1), all of the points of the graph lie above the x-axis,
and the graph appears to be more narrow than that of y = x2 .
Write a suitable expression for g(x) and justify your answer by
graphing the curve y = g(x).
One such solution is g(x) = 2(x + 1)2 + 1.
Note that in general, the vertex of a quadratic function q(x) =
ax2 + bx + c, a 6= 0, can be found by completing the square.
2. In this problem you will investigate a cosine wave. Suppose that a
spring is attached to a horizontal surface so that the spring hangs
downward towards the ground. A block is then attached to the spring.
After some time the system is in equilibrium as in the figure below.
Suppose that the block is then displaced vertically upwards by A =
0.01 meters (1 cm) and then released. In an ideal situation (ignoring
friction and other outside factors), the displacement of the block from
1
the height of the equilibrium position is described by a cosine wave.
Let y be the displacement at time t (measured in seconds). Thus,
y = A cos (ωt),
where ω (the lowercase Greek letter omega) depends on how long it
takes the block to complete one cycle (i.e. travel downwards and back
up again to the initial position).
(a) If the block takes 0.25 seconds to complete one cycle, what is the
value of ω? What if it takes 0.5 seconds to complete one cycle?
Since the period (the time to complete one cycle) T is related to
the frequency ω by the equation
T=
2π
,
ω
we can solve for ω and use the given information. Thus, if T =
0.25 seconds, then ω = 2π/0.25 = 8π inverse seconds (called
Hertz and written Hz). Similarly, if T = 0.5, then ω = 4π.
(b) Graph the curve y = A cos (ωt) where A = 0.01, ω = 0.25, and
0 ≤ t ≤ T (where T is the time to complete one cycle). Label
the exact values of coordinates of the points where the curve
intersects the t-axis.
The graph of y = 0.01 cos (0.25t) is that of a cosine wave that is
scaled vertically by a factor of 0.01 and which has frequency 0.25
and, hence, period equal to 2π/0.25 = 8π. You will see that if we
graph the complete cycle which occurs for 0 ≤ t ≤ 8π, then the
x-intercepts are (0, 2π) and (0, 6π).
3. Solve exercise 32 in Appendix C.3. Your solution should begin with
a clear statement of the problem. Explain how you decided that you
had found all of the solutions to the equation in the interval [0, 2π).
Solve for θ, where 0 ≤ θ < 2π, in the equation
tan2 θ = 3.
To solve this, take the square root. We then obtain two equations:
√
√
tan θ = 3 and
tan θ = − 3.
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Using the unit circle as a guide (and a 30-60-90 triangle), you find that
there are four solutions:
π 2π 4π 5π
θ ∈ { , , , }.
3 3 3 3
One way to see that you have found all the solutions is to just use the
unit circle. Another way is to look
√ at the graph
√ of y = tan x. It is
clear that the horizontal lines y = 3 and y = − 3 both intersect the
graph twice in the interval [0, 2π).
4. A textbook problem: please solve exercise 42 in Appendix C.3.
amplitude = 1.5, period = 4π; you can compute the period by noticing that the distance between consecutive x-intercepts is 2π, whereas
ordinarily for y = cos x, the distance is π. Therefore, this wave has
double the period of y = cos x.
5. Another textbook problem: please solve exercise 64 in Appendix C.3.
There are several solutions. Here are two:
1. a = 1/2, b = 2, c = π 2. a = −1/2, b = 2, c = 0.
The difficult part is to compute the value of c. If you would like to see
another example of how to compute this value, please stop by during
either my office hours or one of the LA’s office hours.
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