Precalculus: Enhanced with Graphing Utilities, 9th Edition — Michael Sullivan
,TABLE OF CONTENTS
Precalculus: Enhanced with Graphing Utilities, 9th Edition — Michael Sullivan
Complete Comprehensive Test Bank | Chapters 1–14
Chapter Major Content Area High-Yield Coverage
Graphing utilities, coordinate geometry, intercepts, symmetry,
1 Graphs
equations, lines, and circles
Functions, domains and ranges, graph properties, piecewise
2 Functions and Their Graphs
functions, transformations, and mathematical models
Linear models, regression, quadratic functions, quadratic
3 Linear and Quadratic Functions
modeling, and inequalities
Polynomial and Rational Polynomial graphs, zeros, complex roots, Fundamental Theorem
4
Functions of Algebra, rational functions, and inequalities
Exponential and Logarithmic Composite and inverse functions, exponential and logarithmic
5
Functions models, equations, growth, decay, finance, and logistic models
Angles, unit-circle trigonometry, trig properties, graphs,
6 Trigonometric Functions
transformations, phase shifts, and sinusoidal models
Inverse trig functions, equations, identities, sum/difference,
7 Analytic Trigonometry
double-angle, half-angle, and product/sum formulas
Applications of Trigonometric Right-triangle trigonometry, Law of Sines, Law of Cosines, triangle
8
Functions area, harmonic and damped motion
Polar coordinates and graphs, complex numbers, De Moivre’s
9 Polar Coordinates; Vectors
Theorem, vectors, dot product, 3D vectors, and cross product
Conic sections, parabolas, ellipses, hyperbolas, rotated conics,
10 Analytic Geometry
polar conics, and parametric equations
Linear systems, matrices, determinants, matrix algebra, partial
Systems of Equations and
11 fractions, nonlinear systems, inequalities, and linear
Inequalities
programming
Sequences; Induction; The Sequences, arithmetic and geometric series, infinite geometric
12
Binomial Theorem series, mathematical induction, and binomial expansions
,Chapter Major Content Area High-Yield Coverage
Counting principles, permutations, combinations, probability
13 Counting and Probability
rules, conditional probability, and independent events
A Preview of Calculus: The Limit, Limits, continuity, tangent slopes, derivatives, area problems,
14
Derivative, and Integral Riemann sums, and definite integrals
Test Bank Highlights
Complete Chapters 1–14 • 1,400 Exam-Style Questions • Detailed Step-by-Step Solutions • Advanced
Mathematical Reasoning • Graph & Data Interpretation • Multi-Step Problems • Exam Strategy Guidance •
Comprehensive Precalculus Review
Content Progression
Algebra & Graphs → Functions & Modeling → Polynomial/Exponential Functions → Trigonometry → Polar
Coordinates & Vectors → Analytic Geometry → Systems & Matrices → Sequences & Probability →
Introductory Calculus
This version is more compact, professional, and suitable for the opening Table of Contents pages of the test
bank.
CHAPTER 1 — GRAPHS
Precalculus Enhanced with Graphing Utilities, 9th Edition — Michael Sullivan
Question 1. The graph of (y=(x-2)^2-9) is displayed with (Y_{\min}=0) and (Y_{\max}=12); which viewing-
window adjustment is necessary to display the vertex?
A. Increase (Y_{\max}) above (12)
B. Increase (X_{\max}) only
C. Decrease (X_{\min}) only
D. Decrease (Y_{\min}) below (-9)
Correct Answer: D
Detailed Solution/Rationale: The equation is in vertex form (y=(x-h)^2+k), so the vertex is ((2,-9)). Since
(Y_{\min}=0), the vertex is below the displayed region. The lower vertical bound must therefore be decreased
below (-9).
Why the other options are less appropriate: A extends the window upward, while B and C alter only the
horizontal range.
Graphing Utility Insight: A graphing window must contain the coordinates of significant points such as vertices
and intercepts.
Exam Strategy: Identify the vertex algebraically before adjusting the viewing window.
,Question 2. Which method most efficiently determines the exact distance between ((-5,3)) and ((7,-2))?
A. Midpoint formula
B. Distance formula
C. Slope formula
D. Intercept formula
Correct Answer: B
Detailed Solution/Rationale: The distance formula directly calculates the length of the segment joining two
coordinate points:
[
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
]
The other formulas determine different geometric quantities.
Why the other options are less appropriate: The midpoint formula finds the segment center; slope measures
rate of change; intercept methods locate axis crossings.
Exam Strategy: Match the requested geometric quantity directly to its defining formula.
Question 3. Find the midpoint of the segment with endpoints ((-6,5)) and ((8,-3)).
A. ((2,1))
B. ((1,-1))
C. ((1,1))
D. ((-1,1))
Correct Answer: C
Detailed Solution/Rationale:
[
M=\left(\frac{-6+8}{2},\frac{5+(-3)}{2}\right)
=\left(\frac22,\frac22\right)
=(1,1).
]
Why the other options are less appropriate: The incorrect choices result from subtraction or sign errors while
averaging coordinates.
Exam Strategy: Average corresponding coordinates independently.
,Question 4. For (y=x^2-7x+12), determine all intercepts.
A. ((3,0),(4,0),(0,12))
B. ((-3,0),(-4,0),(0,12))
C. ((3,0),(4,0),(0,-12))
D. ((-3,0),(4,0),(0,12))
Correct Answer: A
Detailed Solution/Rationale: For the (x)-intercepts,
[
x^2-7x+12=(x-3)(x-4)=0,
]
so (x=3,4). For the (y)-intercept, substitute (x=0), giving (y=12).
Why the other options are less appropriate: They contain sign errors in either the zeros or constant term.
Exam Strategy: Set (y=0) for (x)-intercepts and (x=0) for the (y)-intercept.
Question 5. Which substitution is the most direct test for determining whether the graph of (x=y^2+4) is
symmetric about the (x)-axis?
A. Replace (x) with (-x)
B. Interchange (x) and (y)
C. Replace (y) with (-y)
D. Replace both (x) and (y) with their negatives
Correct Answer: C
Detailed Solution/Rationale: Symmetry about the (x)-axis is tested by replacing (y) with (-y). Here,
[
x=(-y)^2+4=y^2+4,
]
so the equation is unchanged.
Why the other options are less appropriate: Replacing (x) tests (y)-axis symmetry; replacing both variables tests
origin symmetry.
Exam Strategy: (x)-axis: (y\to -y); (y)-axis: (x\to -x); origin: both.
Question 6. Determine the zeros of (x^3-16x=0).
A. (-16,0,16)
B. (-4,0,4)
,C. (-4,4)
D. (0,4,16)
Correct Answer: B
Detailed Solution/Rationale:
[
x^3-16x=x(x^2-16)=x(x-4)(x+4).
]
Thus,
[
x=-4,\quad0,\quad4.
]
Why the other options are less appropriate: A and D confuse coefficients with zeros; C omits the factor (x=0).
Exam Strategy: Factor out the greatest common factor before using difference of squares.
Question 7. The equation (y=x^2-6x+8) is graphed together with the (x)-axis; at which (x)-values do the two
graphs intersect?
A. (x=2,4)
B. (x=-2,-4)
C. (x=1,8)
D. (x=-1,-8)
Correct Answer: A
Detailed Solution/Rationale:
[
x^2-6x+8=0
]
factors as
[
(x-2)(x-4)=0.
]
Thus the intersections occur at (x=2) and (x=4).
Why the other options are less appropriate: The other values do not satisfy the quadratic.
Graphing Utility Insight: Zeros of a function are the (x)-coordinates of its intersections with the (x)-axis.
Exam Strategy: Predict graphical zeros algebraically whenever factorization is simple.
,Question 8. Which approach most efficiently determines the equation of a line through ((3,-1)) with known
slope (m=4)?
A. Distance formula
B. Standard circle form
C. Midpoint formula
D. Point-slope form
Correct Answer: D
Detailed Solution/Rationale: A known point and slope fit directly into
[
y-y_1=m(x-x_1).
]
Thus,
[
y+1=4(x-3).
]
Why the other options are less appropriate: The other formulas concern distance, midpoint, or circles.
Exam Strategy: Point + slope implies point-slope form.
Question 9. Determine the equation of the line through ((-2,7)) and ((4,-5)).
A. (y=2x+3)
B. (y=-2x+3)
C. (y=-2x-3)
D. (y=2x-3)
Correct Answer: B
Detailed Solution/Rationale:
[
m=\frac{-5-7}{4-(-2)}=-2.
]
Then
[
7=-2(-2)+b
]
,gives (b=3). Therefore,
[
y=-2x+3.
]
Why the other options are less appropriate: They contain slope or intercept sign errors.
Exam Strategy: Verify the final line equation using both given points.
Question 10. A line is required to pass through ((6,1)) and be perpendicular to (y=-\frac13x+5); determine its
equation.
A. (y=-3x+19)
B. (y=\frac13x-1)
C. (y=3x-17)
D. (y=-\frac13x+3)
Correct Answer: C
Detailed Solution/Rationale: The given slope is (-\frac13). Its perpendicular slope is (3). Thus,
[
y-1=3(x-6),
]
so
[
y=3x-17.
]
Why the other options are less appropriate: A uses the wrong sign; B and D do not use the negative reciprocal.
Exam Strategy: Perpendicular slopes satisfy (m_1m_2=-1).
Question 11. Which equation represents a circle with center ((3,-4)) and radius (6)?
A. ((x-3)^2+(y+4)^2=36)
B. ((x+3)^2+(y-4)^2=36)
C. ((x-3)^2+(y+4)^2=6)
D. ((x+3)^2+(y-4)^2=6)
Correct Answer: A
Detailed Solution/Rationale: Standard circle form is
,[
(x-h)^2+(y-k)^2=r^2.
]
Substituting (h=3), (k=-4), and (r=6) gives
[
(x-3)^2+(y+4)^2=36.
]
Why the other options are less appropriate: B reverses the center signs; C and D use (r) rather than (r^2).
Exam Strategy: Center signs appear reversed inside the squared binomials.
Question 12. The endpoints of a diameter are ((-4,-2)) and ((6,4)); determine the center of the circle.
A. ((2,2))
B. ((-1,1))
C. ((5,3))
D. ((1,1))
Correct Answer: D
Detailed Solution/Rationale: The center is the midpoint:
[
\left(\frac{-4+6}{2},\frac{-2+4}{2}\right)
=(1,1).
]
Why the other options are less appropriate: They do not equal the coordinate-wise average of the diameter
endpoints.
Exam Strategy: The midpoint of any diameter is the circle's center.
Question 13. The graph of (y=(x+12)^2+4) is displayed with (-10\le x\le10); which modification is required to
include the vertex?
A. Increase (Y_{\max}) only
B. Decrease (X_{\min}) to a value at or below (-12)
C. Increase (X_{\max}) above (12) only
D. Decrease (Y_{\min}) below (-12)
Correct Answer: B
Detailed Solution/Rationale: The vertex is
, [
(-12,4).
]
Its (x)-coordinate lies outside the current interval, so (X_{\min}) must be decreased.
Why the other options are less appropriate: They do not include the required horizontal coordinate (-12).
Graphing Utility Insight: A horizontal translation changes the critical (x)-coordinate of a graph.
Exam Strategy: Rewrite (x+12) mentally as (x-(-12)).
Question 14. Determine the exact distance between ((-1,2)) and ((5,5)).
A. (3\sqrt5)
B. (5\sqrt3)
C. (\sqrt{15})
D. (9)
Correct Answer: A
Detailed Solution/Rationale:
[
d=\sqrt{6^2+3^2}
=\sqrt{45}
=3\sqrt5.
]
Why the other options are less appropriate: They result from incorrect addition or radical simplification.
Exam Strategy: Simplify radicals completely.
Question 15. Determine the midpoint of (\left(-\frac32,4\right)) and (\left(\frac52,-2\right)).
A. ((1,2))
B. (\left(\frac12,2\right))
C. (\left(\frac12,1\right))
D. ((2,1))
Correct Answer: C
Detailed Solution/Rationale:
[
x_M=\frac{-\frac32+\frac52}{2}
,TABLE OF CONTENTS
Precalculus: Enhanced with Graphing Utilities, 9th Edition — Michael Sullivan
Complete Comprehensive Test Bank | Chapters 1–14
Chapter Major Content Area High-Yield Coverage
Graphing utilities, coordinate geometry, intercepts, symmetry,
1 Graphs
equations, lines, and circles
Functions, domains and ranges, graph properties, piecewise
2 Functions and Their Graphs
functions, transformations, and mathematical models
Linear models, regression, quadratic functions, quadratic
3 Linear and Quadratic Functions
modeling, and inequalities
Polynomial and Rational Polynomial graphs, zeros, complex roots, Fundamental Theorem
4
Functions of Algebra, rational functions, and inequalities
Exponential and Logarithmic Composite and inverse functions, exponential and logarithmic
5
Functions models, equations, growth, decay, finance, and logistic models
Angles, unit-circle trigonometry, trig properties, graphs,
6 Trigonometric Functions
transformations, phase shifts, and sinusoidal models
Inverse trig functions, equations, identities, sum/difference,
7 Analytic Trigonometry
double-angle, half-angle, and product/sum formulas
Applications of Trigonometric Right-triangle trigonometry, Law of Sines, Law of Cosines, triangle
8
Functions area, harmonic and damped motion
Polar coordinates and graphs, complex numbers, De Moivre’s
9 Polar Coordinates; Vectors
Theorem, vectors, dot product, 3D vectors, and cross product
Conic sections, parabolas, ellipses, hyperbolas, rotated conics,
10 Analytic Geometry
polar conics, and parametric equations
Linear systems, matrices, determinants, matrix algebra, partial
Systems of Equations and
11 fractions, nonlinear systems, inequalities, and linear
Inequalities
programming
Sequences; Induction; The Sequences, arithmetic and geometric series, infinite geometric
12
Binomial Theorem series, mathematical induction, and binomial expansions
,Chapter Major Content Area High-Yield Coverage
Counting principles, permutations, combinations, probability
13 Counting and Probability
rules, conditional probability, and independent events
A Preview of Calculus: The Limit, Limits, continuity, tangent slopes, derivatives, area problems,
14
Derivative, and Integral Riemann sums, and definite integrals
Test Bank Highlights
Complete Chapters 1–14 • 1,400 Exam-Style Questions • Detailed Step-by-Step Solutions • Advanced
Mathematical Reasoning • Graph & Data Interpretation • Multi-Step Problems • Exam Strategy Guidance •
Comprehensive Precalculus Review
Content Progression
Algebra & Graphs → Functions & Modeling → Polynomial/Exponential Functions → Trigonometry → Polar
Coordinates & Vectors → Analytic Geometry → Systems & Matrices → Sequences & Probability →
Introductory Calculus
This version is more compact, professional, and suitable for the opening Table of Contents pages of the test
bank.
CHAPTER 1 — GRAPHS
Precalculus Enhanced with Graphing Utilities, 9th Edition — Michael Sullivan
Question 1. The graph of (y=(x-2)^2-9) is displayed with (Y_{\min}=0) and (Y_{\max}=12); which viewing-
window adjustment is necessary to display the vertex?
A. Increase (Y_{\max}) above (12)
B. Increase (X_{\max}) only
C. Decrease (X_{\min}) only
D. Decrease (Y_{\min}) below (-9)
Correct Answer: D
Detailed Solution/Rationale: The equation is in vertex form (y=(x-h)^2+k), so the vertex is ((2,-9)). Since
(Y_{\min}=0), the vertex is below the displayed region. The lower vertical bound must therefore be decreased
below (-9).
Why the other options are less appropriate: A extends the window upward, while B and C alter only the
horizontal range.
Graphing Utility Insight: A graphing window must contain the coordinates of significant points such as vertices
and intercepts.
Exam Strategy: Identify the vertex algebraically before adjusting the viewing window.
,Question 2. Which method most efficiently determines the exact distance between ((-5,3)) and ((7,-2))?
A. Midpoint formula
B. Distance formula
C. Slope formula
D. Intercept formula
Correct Answer: B
Detailed Solution/Rationale: The distance formula directly calculates the length of the segment joining two
coordinate points:
[
d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
]
The other formulas determine different geometric quantities.
Why the other options are less appropriate: The midpoint formula finds the segment center; slope measures
rate of change; intercept methods locate axis crossings.
Exam Strategy: Match the requested geometric quantity directly to its defining formula.
Question 3. Find the midpoint of the segment with endpoints ((-6,5)) and ((8,-3)).
A. ((2,1))
B. ((1,-1))
C. ((1,1))
D. ((-1,1))
Correct Answer: C
Detailed Solution/Rationale:
[
M=\left(\frac{-6+8}{2},\frac{5+(-3)}{2}\right)
=\left(\frac22,\frac22\right)
=(1,1).
]
Why the other options are less appropriate: The incorrect choices result from subtraction or sign errors while
averaging coordinates.
Exam Strategy: Average corresponding coordinates independently.
,Question 4. For (y=x^2-7x+12), determine all intercepts.
A. ((3,0),(4,0),(0,12))
B. ((-3,0),(-4,0),(0,12))
C. ((3,0),(4,0),(0,-12))
D. ((-3,0),(4,0),(0,12))
Correct Answer: A
Detailed Solution/Rationale: For the (x)-intercepts,
[
x^2-7x+12=(x-3)(x-4)=0,
]
so (x=3,4). For the (y)-intercept, substitute (x=0), giving (y=12).
Why the other options are less appropriate: They contain sign errors in either the zeros or constant term.
Exam Strategy: Set (y=0) for (x)-intercepts and (x=0) for the (y)-intercept.
Question 5. Which substitution is the most direct test for determining whether the graph of (x=y^2+4) is
symmetric about the (x)-axis?
A. Replace (x) with (-x)
B. Interchange (x) and (y)
C. Replace (y) with (-y)
D. Replace both (x) and (y) with their negatives
Correct Answer: C
Detailed Solution/Rationale: Symmetry about the (x)-axis is tested by replacing (y) with (-y). Here,
[
x=(-y)^2+4=y^2+4,
]
so the equation is unchanged.
Why the other options are less appropriate: Replacing (x) tests (y)-axis symmetry; replacing both variables tests
origin symmetry.
Exam Strategy: (x)-axis: (y\to -y); (y)-axis: (x\to -x); origin: both.
Question 6. Determine the zeros of (x^3-16x=0).
A. (-16,0,16)
B. (-4,0,4)
,C. (-4,4)
D. (0,4,16)
Correct Answer: B
Detailed Solution/Rationale:
[
x^3-16x=x(x^2-16)=x(x-4)(x+4).
]
Thus,
[
x=-4,\quad0,\quad4.
]
Why the other options are less appropriate: A and D confuse coefficients with zeros; C omits the factor (x=0).
Exam Strategy: Factor out the greatest common factor before using difference of squares.
Question 7. The equation (y=x^2-6x+8) is graphed together with the (x)-axis; at which (x)-values do the two
graphs intersect?
A. (x=2,4)
B. (x=-2,-4)
C. (x=1,8)
D. (x=-1,-8)
Correct Answer: A
Detailed Solution/Rationale:
[
x^2-6x+8=0
]
factors as
[
(x-2)(x-4)=0.
]
Thus the intersections occur at (x=2) and (x=4).
Why the other options are less appropriate: The other values do not satisfy the quadratic.
Graphing Utility Insight: Zeros of a function are the (x)-coordinates of its intersections with the (x)-axis.
Exam Strategy: Predict graphical zeros algebraically whenever factorization is simple.
,Question 8. Which approach most efficiently determines the equation of a line through ((3,-1)) with known
slope (m=4)?
A. Distance formula
B. Standard circle form
C. Midpoint formula
D. Point-slope form
Correct Answer: D
Detailed Solution/Rationale: A known point and slope fit directly into
[
y-y_1=m(x-x_1).
]
Thus,
[
y+1=4(x-3).
]
Why the other options are less appropriate: The other formulas concern distance, midpoint, or circles.
Exam Strategy: Point + slope implies point-slope form.
Question 9. Determine the equation of the line through ((-2,7)) and ((4,-5)).
A. (y=2x+3)
B. (y=-2x+3)
C. (y=-2x-3)
D. (y=2x-3)
Correct Answer: B
Detailed Solution/Rationale:
[
m=\frac{-5-7}{4-(-2)}=-2.
]
Then
[
7=-2(-2)+b
]
,gives (b=3). Therefore,
[
y=-2x+3.
]
Why the other options are less appropriate: They contain slope or intercept sign errors.
Exam Strategy: Verify the final line equation using both given points.
Question 10. A line is required to pass through ((6,1)) and be perpendicular to (y=-\frac13x+5); determine its
equation.
A. (y=-3x+19)
B. (y=\frac13x-1)
C. (y=3x-17)
D. (y=-\frac13x+3)
Correct Answer: C
Detailed Solution/Rationale: The given slope is (-\frac13). Its perpendicular slope is (3). Thus,
[
y-1=3(x-6),
]
so
[
y=3x-17.
]
Why the other options are less appropriate: A uses the wrong sign; B and D do not use the negative reciprocal.
Exam Strategy: Perpendicular slopes satisfy (m_1m_2=-1).
Question 11. Which equation represents a circle with center ((3,-4)) and radius (6)?
A. ((x-3)^2+(y+4)^2=36)
B. ((x+3)^2+(y-4)^2=36)
C. ((x-3)^2+(y+4)^2=6)
D. ((x+3)^2+(y-4)^2=6)
Correct Answer: A
Detailed Solution/Rationale: Standard circle form is
,[
(x-h)^2+(y-k)^2=r^2.
]
Substituting (h=3), (k=-4), and (r=6) gives
[
(x-3)^2+(y+4)^2=36.
]
Why the other options are less appropriate: B reverses the center signs; C and D use (r) rather than (r^2).
Exam Strategy: Center signs appear reversed inside the squared binomials.
Question 12. The endpoints of a diameter are ((-4,-2)) and ((6,4)); determine the center of the circle.
A. ((2,2))
B. ((-1,1))
C. ((5,3))
D. ((1,1))
Correct Answer: D
Detailed Solution/Rationale: The center is the midpoint:
[
\left(\frac{-4+6}{2},\frac{-2+4}{2}\right)
=(1,1).
]
Why the other options are less appropriate: They do not equal the coordinate-wise average of the diameter
endpoints.
Exam Strategy: The midpoint of any diameter is the circle's center.
Question 13. The graph of (y=(x+12)^2+4) is displayed with (-10\le x\le10); which modification is required to
include the vertex?
A. Increase (Y_{\max}) only
B. Decrease (X_{\min}) to a value at or below (-12)
C. Increase (X_{\max}) above (12) only
D. Decrease (Y_{\min}) below (-12)
Correct Answer: B
Detailed Solution/Rationale: The vertex is
, [
(-12,4).
]
Its (x)-coordinate lies outside the current interval, so (X_{\min}) must be decreased.
Why the other options are less appropriate: They do not include the required horizontal coordinate (-12).
Graphing Utility Insight: A horizontal translation changes the critical (x)-coordinate of a graph.
Exam Strategy: Rewrite (x+12) mentally as (x-(-12)).
Question 14. Determine the exact distance between ((-1,2)) and ((5,5)).
A. (3\sqrt5)
B. (5\sqrt3)
C. (\sqrt{15})
D. (9)
Correct Answer: A
Detailed Solution/Rationale:
[
d=\sqrt{6^2+3^2}
=\sqrt{45}
=3\sqrt5.
]
Why the other options are less appropriate: They result from incorrect addition or radical simplification.
Exam Strategy: Simplify radicals completely.
Question 15. Determine the midpoint of (\left(-\frac32,4\right)) and (\left(\frac52,-2\right)).
A. ((1,2))
B. (\left(\frac12,2\right))
C. (\left(\frac12,1\right))
D. ((2,1))
Correct Answer: C
Detailed Solution/Rationale:
[
x_M=\frac{-\frac32+\frac52}{2}
