Download Multiple-Choice Questions 1. D The equation of line is , so the

Multiple-Choice Questions
1.
3
5
D The equation of line l is y   x  , so the equation of any line perpendicular to
2
2
2
l is given by y  x  b . Since the line passes through B , the coordinates of B  (6,1)
3

2
must satisfy this equation. This yields b  3 , so the equation of the line is y  x  3 ,
3


or 2x  3y  9 in standard form.
2.

B

The equation of the circle centered at A passing through B is given by
 x  3
2
  y  2   82 since the center is A  (3,2) and the radius is the distance
2
between A and B , which is
x 2  y 2  6x  4 y  69  0 .

82 . Expanding this equation yields the general form




1
1
 3 , or y  x3 
 3 , has no symmetries since
x 2
x2
replacing x by  x in the equation does not return y or  y .
 3.
D
The graph of y  x 3 

4x
is given by Df  x | x  4, x  1,0,1
x3  x
since 4  x  0 implies that x  4 and x3  x  0 implies that x( x 2  1)  0 , or
x( x  1)( x  1)  0 , or x  0, 1 .


The domain of the function f (x) 
4.
C
5.
B If x  4 is a zero of f , then f (4)  0 . This implies that 2(4)  A  0 , or A  8 .
Now, if x  1 a vertical asymptote of the graph of f , then B  1 .
6.


A The graph of the quadratic function f (x)  x 2  6x  a , where a is some real

coefficient, is a parabola whose vertex is the point  3, f (3)    3,9  a  and whose
vertex is also a global maximum (i.e. the parabola opens down). Therefore, the range of

f is given by R f  y | y  9  a.

7.
D The graph of y  3  x 3  1 x 3  1  7 is also the graph of y  3x 6 shifted vertically 4
units upwards since y  3  x3  1 x3  1  7  3  x 6  1  7  3x 6  4 .
8.
B If a parabola passes through the origin and its line of symmetry is the vertical line
x  4 , then the parabola must be described by an equation of the form
Only the equation
y  a( x  4)2  16a , where a is some real coefficient.
2
(x  4)
y
 8 does not satisfy this form. [You can check that in this equation if
2
0 . Hence, this parabola does not pass through the origin.]
x  0 , then y  16  
9.
B If a polynomial function f has a zero at x  3 of multiplicity 2 , another zero at
x  2 of multiplicity 3, and a third zero at x  1 of multiplicity 5, then the function f
has the form f ( x)  a( x  3)2 ( x  2)3 ( x  1)5 , where a is some real coefficient. Now if
the graph of f intersects the y-axis at 216, then f (0)  216 . This implies that a  3
and so f ( x)  3( x  3)2 ( x  2)3 ( x  1)5 .

10.
D


The graph of R( x) 
horizontal
R( x) 
asymptote,
2 x
2 x
has three vertical asymptotes, one
 x  3  x 3  4 x 2  4 x 
and
no x-intercept.
To check this, note that
( x  2)
1
for x  2 . From this


2
x  x  3 x  2 
x  x  3 x  2 
x  x  3 x  2 
reduced form we see that the graph has 3 vertical asymptote at x  0 , x  3 , and
x  2 . Since the numerator is 1, the graph has no x-intercept. Since R is proper, the
graph has one horizontal asymptote: y  0 (the x-axis).
2
Problem 1
Start with the graph of y 
#1
y
1
, then use the following transformations:
x2
1
1
1
y
. This is a horizontal shift of the graph of y  2 4 units to the
2
2
x
x
 x  4
1
, we see
x2
that these are now changed to, respectively, points (3,1) and (5,1) on the graph of
1
1
y
. Note that the vertical asymptote x  0 on the graph of y  2 is now
2
x
 x  4
right. Using test points ( 1,1) and (1,1) , for example, on the graph of y 
moved to x  4 on the graph of y 
#2
y
1
 x  4
2
 y
1
 x  4
2
1
 x  4
2
.
. This is a reflection of the graph of y 
1
 x  4
about
2
the x-axis. Our test points are now moved to, respectively, points (3, 1) and (5, 1) .
#3
y
1
 x  4
2
 y  3
1
 x  4
2
. This is a vertical shift of the graph of y  
1
 x  4
2
3
units upwards. Our test points are now moved to, respectively, points (3, 2) and (5, 2) .
Note that the horizontal asymptote is now the line y  3 .
The range of the function y  3 
1
 x  4
2
is then given by  y | y  3 , or (,3) .
Problem 2
a)
One possibility is to place the vertex of the arch at (0, 25) and the bases of the two legs
at (60, 0) and ( 60, 0) . Then the equation of the parabola is given by y  ax 2  25 ,
for some positive coefficient a . Using the point (60, 0) , for example, we get
a
25
25
1
x2


.
So
the
equation
of
the
parabolic
arch
is
y


 25 .
602 3600 144
144
[Note: you could also place the vertex at the origin, or come up with a different
configuration. The equation would then change accordingly]
b)
x2
 25 . We then need to solve for x in the equation f ( x)  6 to find
144
out how far a 6-feet tall man would have to walk from the midpoint of the arch legs
Let f ( x)  
before hitting his head on the arch.
approximately 52.3 feet.
This yields a distance of 12 19 feet, or
Problem 3
If f is an odd function, then f ( x)   f ( x) for any x in the domain of f . Since
y  f ( x)   f ( x)  f ( x) , we conclude that the function defined by y  f ( x) is necessarily
even (so its graph is symmetrical with respect to the y-axis) whenever f is odd.
Problem 4
1 2 4 7 14 28
are 1, 2, 4, 7, 14, 28,  ,  ,  ,  ,  , 
3 3 3 3
3
3
according to the Rational Zero Theorem.
a)
The potential zeros of
b)
Factoring by grouping, we have:
f
f ( x)  3x3  7 x 2  12 x  28  x 2  3x  7   4  3x  7    3 x  7   x 2  4  .
Therefore,
irreducible.
7
is the only real zero of f . Note that the quadratic factor
3
x
2
 4  is
Problem 5
a)
 x  1  x 2  x  1
x3  1
Note that R( x)  2
is already in lowest terms.

 x  9  x  3 x  3
DR , the domain of R , is then given by DR  x | x  3  (, 3)  (3,3)  (3, ) .
b)
The vertical asymptotes of R are the vertical lines x  3 and x  3 . GR , the graph of
R , also has the oblique asymptote y  x since
R( x) 
c)
x3  1
9x 1
 2
 x.
2
 x  9 x  9
[Check this result using long division.]
To find the point where GR intersects its non-vertical asymptote, solve for x in the
equation R( x)  x , or
x3  1
 x . We then have x3  1  x  x 2  9   x 3  9 x , which
2
 x  9
1
1 1
gives 1  9x , or x  . The intersection point is then  ,  .
9
9 9
[Check this result graphically!]
Bonus Problem 1
The two centers of the circles are (2, 3) and ( 3, 2) . The equation of the line passing
1
13
through both of these points is then given by y   x  , or x  5 y  13 in standard form.
5
5
Bonus Problem 2
Let C be the point on the line passing through O and A (the line y  2 x ) which is located in
Quadrant I such that OC  OB  10 . Then C 


2, 2 2 . Now, let M be the midpoint of
 2 1 2 2  3 
,
line segment BC . Then M  
 . The line passing through O and M is then
2 
 2
the bisector of angle AOB . This line has slope approximately equal to 14.1.