Chapter 1 Review
(page 1)
1.1 Power Functions
1. State the degree and the leading
coefficient of each polynomial function.
a) f(x) = 2x3 + 3x – 1
b) g(x) = 5x – 6
c) h(x) = x3 – 2x2 – 5x4 + 3
d) p(x) = –3x5 + 2x3 – x – 1
e) r(x) = 21 – 2x + 4x2 – 6x3
2. For each graph
i) state whether the corresponding function
has even degree or odd degree
ii) state whether the leading coefficient is
positive or negative
iii) state the domain and range
iv) identify any symmetry
v) describe the end behaviour
a) Window: x  [–4, 4], y  [–25, 25], Yscl = 5
b) Window: x  [–4, 4], y  [–10, 10],
3. Complete the table. Write each function
in the appropriate row of column 2. Give
reasons for your choices.
1
y = 3x7, y = – x3, y = –0.25x6, y = 2x4
2
End
Behaviour
Extends from
quadrant 3 to
quadrant 1
Extends from
quadrant 2 to
quadrant 4
Extends from
quadrant 2 to
quadrant 1
Extends from
quadrant 3 to
quadrant 4
Function
Reasons
1.2 Characteristics of Polynomial
Function
4. Match each graph of a polynomial
function with the corresponding equation.
Justify your choice.
i) y = 3x2 + 4x – 2x4 + 5
ii) y = –x5 + 3x4 + 7x3 – 15x2 – 18x
a) Window: x  [–5, 5], y  [–50, 50], Yscl = 5
b) Window: x  [–5, 5], y  [–5, 15]
5. For each polynomial function in question 4
a) determine which finite differences are
constant
b) find the value of the constant finite
differences
6. State the degree of the polynomial function
that corresponds to each constant finite
difference and determine the value of the
leading coefficient of each.
a) third differences = – 4
b) first differences = 6
c) sixth differences = –720
d) fourth differences = 96
e) second differences = –12
7. The table represents a polynomial
function.
x
−3
−2
−1
0
1
2
3
4
y
168
0
−40
−24
0
8
0
0
Use finite differences to determine
a) the degree
b) the sign of the leading coefficient
c) the value of the leading coefficient
(page 2)
1.3 Equations and Graphs of Polynomial
Functions
8. Use each graph of a polynomial function
to determine
i) the x-intercept(s) and the factors of the
function
ii) the least possible degree and sign of
the leading coefficient
iii) the interval(s) where the function is
positive and the interval(s) where the
function is negative
a) Window: x  [–4, 6], y  [–5, 15]
b) Window: x  [–4, 4], y  [–10, 60], Yscl = 5
9. Sketch a graph of each polynomial
function.
a) y = 2x(x + 3)(x – 4)
b) y = –3(x – 2)(x + 4)(x2 – 1)
2
10. The zeros of a function are –4, – , and
5
3. Determine an equation for the
function if it has y-intercept 8.
11. Determine algebraically, if each
polynomial function has line symmetry
about the y-axis, point symmetry about
the origin, or neither. Graph the
functions to verify your answer.
a) f(x) = 3x5 – 2x3 + x
b) g(x) = 2x4 + 3x3 – 2x – 6
c) h(x) = 2x6 – 5x4 + x2 + 4
1.4 Transformations
12. i) Describe the transformations that must
be applied to the graph of each power
function, f(x), to obtain the
transformed function. Then, write the
corresponding equation.
ii) State the domain and range of each
transformed function.
a) f(x) = x4, y = –2f(x – 1) + 4
1
b) f(x) = x3, y = f(2x + 6) – 5
3
13. Write an equation for the function that is
the result of each set of transformations.
a) f(x) = x5 is stretched vertically by a
factor of 5, compressed horizontally
1
by a factor of
and translated 2 units
4
to the left and 1 unit down.
b) f(x) = x6 is compressed vertically by a
1
factor of , reflected in the y-axis and
2
translated 4 units to the right and 3
units up.
1.5 Slopes of Secants and Average Rate of
Change
14. The population of a small town, p, is
modelled by the function
p(t) = 10 050 + 225t – 20t2,
where t is the time in years from now.
a) Determine the average rate of change
of the population from
i) year 0 to year 5
ii) year 5 to year 8
iii) year 8 to year 10
b) Interpret your answers in part a).
c) Graph the function to verify your
interpretation in part b).
1.6 Slopes of Tangents and Instantaneous
Rate of Change
15. After being built, a car must be painted.
The revenue, R, in dollars, when x cars
are painted can be modelled by the
function R(x) = 1000x – 0.01x2.
a) Determine the average rate of change
of revenue when painting
20 to 50 cars.
b) Estimate the instantaneous rate of
change of revenue after painting
50 cars.
c) Interpret the results found in parts a)
and b).
Chapter 1 Review ANSWERS
1. a) degree: 3; leading coefficient: 2
b) degree: 1; leading coefficient: 5
c) degree: 4; leading coefficient: –5
d) degree: 5; leading coefficient: –3
e) degree: 3; leading coefficient: –6
2. a) i) even degree
ii) negative
iii) { x  }; { y  , y  15 }
iv) no line or point symmetry
v) quadrant 3 to 4
b) i) odd degree
ii) positive
iii) { x  }; { y  }
iv) point symmetry about (0, 0)
v) quadrant 3 to 1
b) i) –3, –2, 3; (x + 3)(x + 2)(x – 3)2
ii) 4, positive
iii) positive x < −3, –2 < x < 3, x > 3;
negative –3 < x < –2
1
10. y =  (x + 4)(5x + 2)(x – 3)
3
11. a) point symmetry about the origin
b) neither
c) line symmetry about the y-axis
12. a) i) a vertical stretch by a factor of 2, a
reflection in the x-axis and translations
1 unit to the right and 4 units up;
y = −2(x − 1)4 + 4
ii) {x  } , { y  , y  4}
b) i) a vertical compression by a factor of
1
, a horizontal compression by a
3
1
factor of
and translations 3 units to
2
the left and
1
5 units down; y  (2 x  6)3  5
3
ii) {x  } , { y  }
13. a) y = 5[4(x + 2)]5 – 1
b) y = 0.5[–(x − 4)]6 + 3
14. a) i) 125 ii) –35 iii) –135
b) The population grew in the first 5
years towards its maximum; then it
3. y = 3x7: quadrant 3 to 1, odd degree with
1
positive leading coefficient; y =  x3:
2
quadrant 2 to 4, odd degree with negative
leading coefficient; y = 2x4: quadrant 2 to
1, even degree with positive leading
coefficient; y = –0.25x6: quadrant 3 to 4,
even degree with negative leading
coefficient
4. i) and b); ii) and a)
5. a) i) fourth ii) fifth
b) i) – 48 ii) –120
2
6. a) 3, 
b) 1, 6
c) 6, –1
3
d) 4, 4
e) 2, –6
7. a) fourth
b) positive c) 1
8. a) i) –1, 0, 4; x(x + 1)(x − 4)
ii) 3, negative
iii) positive x < −1, 0 < x < 4;
negative −1 < x < 0, x > 4
began to decrease in the next 3 years
and decreased considerably in the
following 2 years.
c) Window: x  [0, 10], y  [0, 13 000],
Yscl = 1000
15. a) $999.30
b) $999.00
c) The slope of the secant line from
x = 20 to x = 50 approaches the slope
of the value of the tangent line at
x = 50, i.e. the average rate of change
from x = 20 to x = 50 is quite close in
value to the instantaneous rate of
change at x = 50.