Solution Key to Homework #5: College Algebra
P.187
14. Determine symmetry with respect to the x-axis, y-axis, or origin, if any exists, and
graph. y = 2 x
a) With respect to the x-axis: Replace y by –y in the equation:
y = 2 x → (1)
− y = 2 x ⇒ y = −2 x → (2)
The equation (1) is not equivalent to the equation (2) and so it is not symmetric
with respect to x-axis.
b) With respect to the y-axis: Replace x by –x in the equation:
y = 2 x → (1)
y = 2(− x) ⇒ y = −2 x → (2)
The equation (1) is not equivalent to the equation (2) and so it is not symmetric
with respect to y-axis.
c) With respect to the origin: Replace x by –x and y by –y in the equation:
y = 2 x → (1)
− y = 2(− x) ⇒ − y = −2 x ⇒ y = 2 x → (2)
The equation (1) is equivalent to the equation (2) and so it is symmetric with
respect to the origin.
38. Determine symmetry with respect to the x-axis, y-axis, or origin, if any exists, and
graph. x 2 + 9 y 2 = 9 .
a) With respect to the x-axis: Replace y by –y in the equation:
x 2 + 9 y 2 = 9 → (1)
x 2 + 9( − y ) 2 = 9 ⇒ x 2 + 9 y 2 = 9 → ( 2)
The equation (1) is equivalent to the equation (2) and so it is symmetric with
respect to x-axis.
b) With respect to the y-axis: Replace x by –x in the equation:
x 2 + 9 y 2 = 9 → (1)
( − x ) 2 + 9 y 2 = 9 ⇒ x 2 + 9 y 2 = 9 → ( 2)
The equation (1) is equivalent to the equation (2) and so it is symmetric with
respect to y-axis.
College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 5
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c) With respect to the origin: Replace x by –x and y by –y in the equation:
x 2 + 9 y 2 = 9 → (1)
( − x ) 2 + 9( − y ) 2 = 9 ⇒ x 2 + 9 y 2 = 9 → ( 2)
The equation (1) is equivalent to the equation (2) and so it is symmetric with
respect to the origin.
86. If a graph is symmetric with respect to the y-axis and to the origin, must it be
symmetry with respect to the x-axis ? Explain your answer.
Yes, it is symmetric with respect to the x-axis as well.
Note: A graph is said to be symmetric with respect to the x-axis if, for every point (x, y)
on the graph, the point (x, -y) is also on the graph.
Answer:
For every (x, y) on the graph, by the symmetry with respect to the y-axis,
the point (-x, y) is also on the graph. Since the point (-x,y) is on the graph,
by the symmetry with respect to the origin, (-(-x), -y) = (x, -y) is also on
the graph.
College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 5
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P. 237
6.
Domain: (−∞,−3) ∪ (−3, ∞)
Range: (−∞,−2) ∪ [2, ∞)
x-intercepts: None
y-intercept = 2
Intervals over which f is increasing: (−∞,−3) ∪ [3, ∞)
Intervals over which f is decreasing: None
Intervals over which f is constant: (−3,3]
Any points of discontinuity: at x = -3.
10. Describe the graph of a continuous function f over the interval [-5,5]. Sketch the
graph of a function that is consistent with the given information.
The function f is increasing on [-5,-2], constant on [-2,2], and increasing on [2,5].
22. Graph, finding the axis, vertex, maximum or minimum, and range.
f ( x) = −( x − 2) 2 + 4 ⇒ the vertex is (2, 4), the axis is the line x = 2,
the maximum is 4 when x = 2, the range is (−∞,4).
College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 5
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30. Graph, finding the axis, vertex, intervals over which f is increasing, and intervals over
which f is decreasing.
f ( x) = − x 2 − 10 x − 24 = −( x 2 + 10 x) − 24 = −( x 2 + 10 x + 25 − 25) − 24
= −( x + 5) 2 + 25 − 24 = −( x + 5) 2 + 1 ⇒ the vertex : (−5,1), the axis is the line x = −5,
f is increasing on (−∞,−5] and decreasing on [−5, ∞)
The following ones are from your textbook:
P. 263
14.The graph of an equation is given. Indicate whether the graph is symmetric with
respect to the x-axis, the y-axis, or the origin.
Answer: It is symmetric with respect to the y-axis only.
16. The graph of an equation is given. Indicate whether the graph is symmetric with
respect to the x-axis, the y-axis, or the origin.
Answer: It is symmetric with respect to the x-axis, the y-axis, and the origin.
30. Determine symmetry with respect to the x-axis, y-axis, or origin, if any exists, and
graph. y = x 4 − 1 .
College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 5
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a) With respect to the x-axis: Replace y by –y in the equation:
y = x 4 − 1 → (1)
− y = x 4 − 1 ⇒ y = − x 4 + 1 → ( 2)
The equation (1) is not equivalent to the equation (2) and so it is not symmetric
with respect to the x-axis.
b) With respect to the y-axis: Replace x by –x in the equation:
y = x 4 − 1 → (1)
y = ( − x ) 4 − 1 ⇒ y = x 4 − 1 → ( 2)
The equation (1) is equivalent to the equation (2) and so it is symmetric with
respect to the y-axis.
c) With respect to the origin: Replace x by –x and y by –y in the equation:
y = x 4 − 1 → (1)
− y = ( − x ) 4 − 1 ⇒ − y = x 4 − 1 ⇒ y = − x 4 + 1 → ( 2)
The equation (1) is not equivalent to the equation (2) and so it is not symmetric
with respect to the origin.
P. 274
22. At what number(s), if any, does f have a local maximum?
What are the local maximum?
At x = 0, f has the local maximum 2.
At what number(s), if any, does f have a local minimum?
What are the local minimum?
At x = -1 and 1, f has the local minimum 0.
P. 274
30.
Given f ( x) = x − 2 x 2 .
a) Find the average rate of change of f from 1 to x.
f ( x) − f (1) x − 2 x 2 − (−1) x − 2 x 2 + 1 − 2 x 2 + x + 1
=
=
=
x −1
x −1
x −1
x −1
2
− (2 x − x − 1) − (2 x + 1)( x − 1)
=
=
= −(2 x + 1), if x ≠ 1
x −1
x −1
Note : f ( x) = x − 2 x 2 ⇒ f (1) = 1 − 2(1) 2 = 1 − 2 = −1
b) Use the result from (a) to compute the average rate of change form x = 1 to x=2.
College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 5
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f ( x) − f (1)
= −(2 x + 1), if x ≠ 1
x −1
f (2) − f (1)
= −(2(2) + 1) = −5
2 −1
c) Find an equation of the secant line containing (1, f(1)) and (1, f(2)).
Note that the average rate of change in (b) is the slope of the secant line.
The secant line passes through (1, f(1))=(1, -1) and it has the slope m =-5.
y − (−1) = −5( x − 1)
y + 1 = −5 x + 5
y = −5 x + 5 − 1
y = −5 x + 4
d) Graph f and the secant line on the same viewing window.
College Algebra by Chinyoung Bergbauer at NHMCCD: Solhw 5
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