161
Precalculus 1
Review 2
1. Consider the quadratic function y  3x 2  5x  2 .
a. Use the completion of squares to write the function in the standard
form y  a ( x  xv ) 2  yv .
Solution. First we will factor out the coefficient by x 2 .
5
2
3x 2  5 x  2  3( x 2  x  ) .
3
3
Next we will complete the square inside the parentheses using the formula
2
2
B B

x  Bx   x      .
2 2

2
2
B
5
25
B
  , and   
.
2
6
 2  36
5
In our case, B   ,
3
2
5
5  25

Therefore x 2  x   x    , and
3
6  36

2
5
2 
5  25 2 
5  25 24
x  x x    x   

3
3 
6  36 3 
6)  36 36
2
2
2
5
1

x  
6  36

2
5
1

Finally, 3x  5 x  2  3  x   
and this is the standard form of our quadratic
6  12

function.
2
b. Find the coordinates of the vertex, the x-intercepts (if any), and the y-intercept.
Solution. The coordinates of the vertex can be found immediately from the standard form
5
1
of the function. Namely, xv  and yv  .
6
12
The x-intercepts we can find by solving the quadratic equation 3x 2  5 x  2  0 .
We can solve it by factoring, or using the quadratic formula, or the standard form we
have found. If we make the last choice we write
2
2
5
1
5
1


3  x     0;  3  x     ;
6  12
6
12


2
5
1
5
1
1
5 1

 ; x  .
x   ; x 
6  36
6
36
6
6 6

Therefore the x-intercepts are
2
and 1 .
3
Finally, the y-intercept equals to y (0)  2 .
c. Graph the function.
5 1 
Solution. To graph the function we can plot the vertex  ,  , the x-intercepts
 6 12 
2 
 , 0  and (1,0) , and the y-intercept (0, 2) . To plot a good graph you are advised to
3 
plot a few more points on the graph. For example we can compute the values of the
1 3
 1 1   3 5 
, and 2 and plot the points  ,  ,  ,  , (2, 4) .
function at the points x  ,
2 2
2 4  2 4 
2. A projectile is fired vertically up from the top of a 120 ft high tower with initial
velocity 48 ft/sec.
a. When will the projectile reach its maximum height? What is this maximum height?
Solution. If an object is fired from initial height h0 ft with initial velocity v0 ft/sec its
height at moment t is h(t )  16t 2  v0t  h0 . In our case it will be h(t )  16t 2  48t  120 .
The height h is a quadratic function of t with the coefficients a  16, b  48, and c  120 .
The function h takes its greatest value at the vertex tv  
b
48

 1.5 . This greatest
2a
32
b2
482
 120 
 156 . Therefore the projectile will reach the maximum
4a
64
height after 1.5 sec. and this maximum height is 156 ft.
value is hv  c 
b. When will the projectile hit the ground?
Solution. When the projectile hits the ground we have h(t )  0 . So, to answer the
question we have to solve the equation 16t 2  48t  120  0 . After dividing both parts by
-8 we get 2t 2  6t  15  0 . Let us solve it using the quadratic formula
(6)  (6) 2  4  2  (15) 6  156

2 2
4
6  156
 4.62 sec.
Only the positive solution makes sense whence t 
4
t
3. Compute and write as a complex number in the standard form
(1  i )(2  i )
.
(3  i )(4  i)
Solution. Let us first perform multiplications in the numerator and the denominator of
(1  i)(2  i)
2  i  2i  i 2
our fraction.
. Recall that i 2  1 whence the last fraction

2
(3  i)(4  i) 12  3i  4i  i
3i
is equal to
. Division is multiplication by the reciprocal and the reciprocal to the
11  7i
a  bi
complex number a  bi is 2
. In particular the reciprocal to 11  7i is
a  b2
11  7i 11  7i

. Now we have to perform the
112  7 2
170
(3  i)(11  7i) 33  21i  11i  7i 2 40  10i 4 1


  i . This is our
multiplication
170
170
170
17 17
answer.
4. Find all the solutions of the polynomial equation 2 x 4  3x3  6 x 2  13x  6  0 and factor
the polynomial completely.
Solution. We will use the rational root test. The numerators of possible rational roots can
be only factors of 6: 1, 2, 3, 6 . The denominators are positive factors of 2:1, 2.
1
3
Therefore the complete list of possible rational roots is  ,  1,  ,  2,  3,  6 . Let us
2
2
check these numbers using synthetic division. We will start with 1/2.
1/2
2
-3
1
-2
2
-6
-1
-7
13
-7/2
19/2
-6
19/4
-5/4
We see that 1/2 is not a solution of our equation. Next we try -1/2.
-1/2
2
-3
-1
-4
2
-6
2
-4
13
2
15
-6
-15/2
-27/2
-6
-1
-7
13
-7
6
-6
-6
0
-1/2 is not a solution either. Next we try 1.
1
2
-3
2
-1
2
1 is a solution from the result of synthetic division we have the following factorization.
2 x 4  3x3  6 x 2  13x  6  ( x  1)(2 x3  x 2  7 x  6) .
Now we have to solve the equation 2 x3  x 2  7 x  6  0 . The list of possible rational
3
roots has shrunk to  1,  ,  2,  3,  6 . We will try 1 once again.
2
1
2
2
-1
2
1
-7
1
-6
6
-6
0
So 1 is the solution of our new equation and also a multiple solution of our original
equation of multiplicity at least 2. Also we get the following factorization.
2 x4  3x3  6 x 2  13x  6  ( x  1)(2 x3  x 2  7 x  6)  ( x  1)2 (2 x 2  x  6) .
It remains to solve the quadratic equation 2 x 2  x  6  0 . The quadratic formula gives us
1  (1) 2  4  2  (6) 1  49 1  7
x


. Therefore we have two more
2 2
4
4
solutions x  2 and x  6/ 4  3/ 2 . Notice that both of them are, of course, on our list of
possible rational roots. So all the solutions of our equation are
1, 1, 2, 3 / 2
And the complete factorization is
2 x 4  3x3  6 x 2  13x  6  ( x  1)2 ( x  2)(2 x  3) .
5. Solve the inequality 2 x 4  3x3  6 x 2  13x  6  0 .
Solution. The solution of the previous problem tells us that the left part of the inequality
is 0 at points -2, 1, 3/2. The sign of the polynomial can change only at these points. Let us
look at the following table
Interval
Sign
(-∞, -2)
+
(-2, 1)
-
(1, 3/2)
-
(3/2, ∞)
+
The signs in the second row can be explained as follows. The sign to the left of -2
and to the right of 3/2 is positive because the leading term 2x4 is always non-negative.
The point -2 is a simple root of our polynomial so the sign at -2 changes from + to –. The
point 1 is a root of even multiplicity (two) and the sign does not change. Finally, 3/2 is
again a simple root and the sign at 3/2 changes from – to +.
Another way to see it is to pick up a point inside each interval and to look at the
sign of the polynomial at that point. For example, the value at 0 is -6 and therefore the
sign of the polynomial on the interval (-2, 1) is negative.
Now we see from the table that the solution of the inequality is the union of two
open intervals (2,1)  (1,3 / 2) .
Notice that if the inequality were not strict, 2 x 4  3x3  6 x 2  13x  6  0 , the
solution would be the closed interval [-2, 3/2].
6. Graph the polynomial P( x)  0.2(2 x 4  3x3  6 x 2  13x  6) .
Solution. The polynomial P is proportional to the polynomial from the previous problem.
The factor 0.2 is so called scaling factor; we use it to prevent the graph from going to far
down or up and to see better its features.
We see from the table in the previous problem what the x-intercepts are and what
is the sign of P. So we know where the graph goes below and above the x-axis. We can
compute additionally a few values of P. P(0)  6, P(1)  20, P(2)  4 .
7. Solve the equation 5x  6  3  x  3 .
Solution. Let us square both parts of our equation.

5x  6
  3 
2

2
x3 .
Applying to the right part the formula (a  b)2  a 2  2ab  b2 we get
5 x  6  32  2  3  x  3 

x3

2
 9  6 x  3  x  3  12  x  6 x  3 .
From here,
5x  6  x  12  6 x  3 or 4 x  6  6 x  3 .
After dividing both parts by 2 we get
2x  3  3 x  3 .
Let us again square both parts.


2
(2 x  3) 2  3 x  3 .
Or,
4 x 2  12 x  9  9( x  3) .
After moving all the terms to the left we have a quadratic equation for x.
4 x 2  21x  18  0 .
The quadratic formula tells us that
(21)  (21) 2  4  4  (18) 21  729 21  27
x


.
2 4
8
8
So either x  6 or x  6/ 8  3/ 4 . We must check these solutions by plugging them into
original equation because when we square both parts of such an equation we can get so
called false solutions. If we plug in x  6 then the left part is 36  6 and the right part
is 3  9  6 , so 6 is a solution of our equation. If we plug in x  3/ 4 then the left part
15
9 3
3
9
3
6 
 but the right part is 3    3  3 
 3  . Finally we see
4
4 2
4
4
2
that the equation has only one solution x  6 .
is

8. Find all the solutions of the equation x 6  63x3  64  0 .
Solution. We have here an equation of the quadratic type. Indeed, the left part can be
written as  x3   63x3  64 . Because u 2  63u  64  (u  1)(u  64) we can write our
2
equation as ( x3  1)( x3  64)  0 . To factor the left part completely let us recall the
formulas a3  b3  (a  b)(a 2  ab  b2 ) and a3  b3  (a  b)(a 2  ab  b2 ) . Applying these
formulas we see that x3  1  ( x  1)( x 2  x  1)
and x3  64  x3  43  ( x  4)( x 2  4 x  16) . Our equation can be now written
as ( x  1)( x  4)( x 2  x  1)( x 2  4 x  16)  0 .
We can see at once two real solutions: x  1 and x  4 . To find the remaining
four solutions we have to solve two quadratic equations: x 2  x  1  0
and x 2  4 x  16  0 .
Applying the quadratic formula to the first of these equations we obtain
 
1  12  4 11 1  3 1  3 i
.


2
2
2
Similarly solving the second equation we get
x
 
4  42  4 116 4  48 4  16  3 4  4 3 i



 22 3 i .
2
2
2
2
Thus we have found all six solutions of our polynomial equation of degree six – two real
solutions and two pairs of conjugate complex solutions.
x
 
9. Find all the solutions of the equation | x 2  3x  1| 5 x  3 .
Solution. We have to consider two cases. The first one is x 2  3x  1  5 x  3 . In this case
(8)  (8) 2  4 1 (2) 8  72 8  6 2


 4  3 2 . Both
2
2
2
solutions satisfy the original equation because for both of them the expression 5x  3 will
be positive.
x 2  8 x  2  0 and x 
In the second case we have x 2  3x  1  5 x  3 whence x 2  2 x  4  0 . This
quadratic equation has only complex solutions (its discriminant b 2  4ac equals to -12)
and therefore our original equation has only two solutions 4  3 2 .
10. Consider the rational function R( x) 
x3  x
.
x2 1
a. Find the x and y intercepts and the vertical asymptotes (if any).
Solution. To find the x intercepts we have to solve the equation R ( x )  0 . A fraction is 0
if and only if its numerator is 0, so we have to solve the equation x3  x  0 . Because
x3  x  x( x 2  1) this equation has only one real solution x  0 . So the only x-intercept is
0 and the y intercept is 0 as well.
To find the vertical asymptotes we consider the points where the
denominator, x 2  1 , is 0. These points are -1 and 1 and therefore we have two vertical
asymptotes x  1 and x  1 .
b. Find the intervals where R ( x )  0 and where R ( x )  0 .
x( x 2  1)
. The function can change sign only at the
( x  1)( x  1)
points -1, 0, and 1. The sign of the function is given by the following table
Solution. We can write R( x) 
Interval
Sign
(-∞, -1)
-
(-1,0)
+
(0,1)
-
(1, ∞)
+
To explain the table notice that if x < -1 then each of the factors x  1, x, x  1 is negative
whence R ( x ) is negative. Each factor is simple (of multiplicity one) therefore at each of
the points -1, 0, 1 the function changes sign.
c. Is the function R ( x ) even or odd or neither? How does it affect its graph?
Solution. The function is odd. Indeed,
(  x )3  (  x)  x 3  x
x3  x
R(  x) 



  R( x) . Therefore the graph of R ( x ) is
(  x) 2  1
x2 1
x2 1
symmetric about the origin.
d. Does the graph of R ( x ) have a horizontal or a slant asymptote? If it does what is an
equation of such an asymptote?
Solution. The graph has an oblique asymptote because the degree of the numerator is 3
and the degree of the denominator is 2 and the difference of degrees is one.
To find an equation of the oblique asymptote we will divide the numerator by the
denominator using the long division.
x
2
3
x 1 x  x
 x3  x
_______________
2x
The quotient of division is x and therefore the equation of the slant asymptote is y  x .
e. Graph the function and its oblique asymptote in the same coordinate system.
Solution. Far left the graph is approaching the oblique asymptote. Then it will go along
the vertical asymptote, x  1 , down, because the sign is negative. Between 0 and 1 the
function is positive and the graph goes along the same vertical asymptote and intersects
the x-axis at 0. The right part of the graph is symmetric to the left part about the origin.