Honors Pre-Calc - Lenox Public Schools

To all upcoming honors pre-calculus student:
It is important that you don't lose ground on your math skills over the summer.
To help you out with this, here is a packet that covers some prerequisite skills for honors pre-calc.
Please complete the eyen exercises at the end of each section
-
P.1 through P.7
Where a graphing calculator is required, you may use an online grapher if you do not own a calculator
We will address questions from the packet on the first two days of school'
The completed packet will count as your first homework grade of the quarter and you will be assessed
on this material in the second week of school.
Please email me @ [email protected] with questions.
Have a great summer!
Prerequisites
P.l
Real Numbers
Converting Between lntervals and lnequalitles
Convert interval notation to inequality notation or vice versa. Find tle endpoints and
state whether the interval is bounded, its type, and graph the interval.
(a) (-3,21
O) (2,
(c)-l<x<4
-)
SOLUTION
(a) The interval (-3, 2] corresponds to -3 < x < 2 and is bounded and half-open (see
Figure P.la). The endpoints are-3 and 2.
(b) The interval (2, -) corresponds to x > 2 and is unbounded and open (see Figure P.lb).
The only endpoint is 2.
(c) The inequality I < x < 4 corresponds to the half-open, bounded interval [-1, 4) (see
Figure P. I c). The endpoints are -1 and 4.
(a)(f I t t | | F I I t Flr
-5-4-3-2-1 0 I 2 3 4 5
o,
(c)(l I t I r | | | I ) l)r
-5-4-3-2-1 0 I 2 3 4 5
Figure P.I
Graphs of the intervals of real numbers in Alternate Example 4.
Exercises for Alternate Example 4
In Exercises 1-6, convert interval notation to inequality notation or vice versa. Find the endpoints and state whetlrer the interval is bounded
and its type. Then gr-aph the interval.
t.[-2,31
2.
(*,0)
3.
4<x<4
4-
2<x<5
5.
-3<x<5
6.
[3,
-)
Using the Distributive Property
(a) Write the expanded form of 2b(x
-
3).
(b) Write the factored form of 5r + l0ry.
SOLUTION
(a\
2b(x-3)=2b(x)-2bQ)
=2bx-6b
@) 5x+
l0xy=5x+2.5ry
=5r.1 +SxQl)
=5x(l +
2y)
I
2
Chapter
P
Prereguisites
Exercises for Alternate Example
In Erercises
7.
13.
7-ld
use the
b(i -3)
ligh+
7gz
5
disfibutive property to expaod or factor"
s. 2r{3 -c)
e.
Qi+rn
fi.344+2n\
ll.3n+6mi
,
la. t*b+,w
,, ..*
14. n'p +2n3t
,
D
ailpiiiyr;s expressioii invdiu"g:ir6wen
Write each expression in simplified form with positive exponents.
,[+) '
@*:
n-z -
(a\ (7nm\3nzm2)
soLuiloN
(a) Qnm3)d3n2*2) = (7 .3){n . n2ym3 .
=2ln3m5
m27
)
@
"uL-;=4'+
n-zn- z-
,",
(-2)
= nz-l =nz
[+)
=(m4)-2
3a
_ fr-8
3a
=!m8
_9
m8
Exercises for Alternate Example 7
i
ln Exercises 15--20, simplifr the expression. Assume that the variables in denominators are nonzero.
rs.t*i:xqfy\
n. ,o!!5
16. edb1x-5;b6)
rB.
cJd
I
x3Y2
,'yu
19.
[+)'
*(s) '
I
Uslng Scientlf lc Notatlon
simprify
o,fi;t#f9)
SOLUrlON
a-t-la^-!-r
l -!fL--ltrfrlrrtr.frr,
l}rtttlo! Ar
rrSttrg
a
(2,400X130)-(2.4)103
12,000
x
(l:3)102
Write in scientific notation.
6.zyaa
x
=€lXl.J).ld3
l'2
104
=2-6.l}l
:26
to2
Apply lawsof exponentr
\r
Section
P.2
Cartesian,CoordinateSystem
Uslng a Calculator
Figure P.2 shorrs two ways to perform the computation. In the fust, the numbers are
entered in decimal form. In the second, the numbers are entered in scientific notation.
The calculator uses '2.6E1" to stand for 2.6 x 101.
Flgure P.2 Be sure you undenstarid how your calculator displays scientific notation
Exercises for Aiternate Example 9
ln Exercises 2l-26, simpliff using scientific notation.
,r
..
3.2x107
1.6 x l0)
180
720
3.600x
..
aL-
98,000 x 50
^.
(4.5)10-4x(3.0)10-l
23. Q.2x 104111.5 x
14,000
1031
24. (6.0 x t01111.8 x t0-2;
I
9x102
-
P.2 Cartesian Coordinate System
Finding Standard Form Equations of Circles
Find the standard form equation ofthe circle.
(a) Center
e,1),
(b) Center (0,0), radius 7
radius 6
soLuTtoN
(r- n12+1y-tcf :i
(x-(-3)'+O-(a)f :62
(a)
(x+
3)2 +
Standard form equation
Substitute values for h, k, and r. h = -3, k = -2, r = 6
O+2)2:36
(b)(r- nf +g-tcf
:l
Standard form equation
(r-o)'+g-of :t2
Substitute values for
i+f:+s
h
k, and
r
h = O, k = O, r = 7
Exercises for Alternate Example 6
In Exercises
l.
l-6, find
the standard form equation ofeach circle.
Center (0,4), radius 4
2.
Center (1,
l), radius l0
3.
Center(O, 0),radius I
4.
Center (-4, 0), radius 2.5
i
.t
I
5. center
I
(*,;),radius4
I
6. center(-3,-3),radius8
I
I
't
J
I
I
Verifylng Right Triangles
I
'.t
.l
tle converse of the fythagorean theorem and the distance formula to show
points (-6, 8), (8, 6), and (2, 0) determine a right triangle.
Use
tlat the
ChapterP Prerequisites
SOLUTION
The three points are plotted in Figure P.3. We need to show that the leneths of
the sides
of the triangle satisff the Pythagorem relationshiup o, + b, = c2. epplying the distance
formula, we find that
6={G -2)z+(6-o)z =.J72
Figure P.3
The riangle in
l_-
Alternate Example 8.
Jt*-8)2+(8- qz =J2cr,
The triangle is a right triangle because
,'+ b' = (V*)'
. (Ji)'
=128 +72=2oo=c2
Exercises for Alternate Example 8
In Exercises 7-12, use the converse ofthe Pythagorean theorem and the distance formula to show that
the given points determine a right
triangle.
7. (0,0), (0, l), (1,0)
8. (1, l), (13, l), (1,6)
11. (0, 0), (-1, 2), (6, 3)
12. (4, 0), (6, 2), (6, -2)
e. @,0), (-4, -3), (0, 0)
r0. (-3,
4), (2,4), (2,10)
I
Using the Mldpoint Formuta
It
is a fact from geometry that the midpoints of the diagonals of a rectangle bisect each
other. Prove this with a midpoint formula.
SOLUTION
We can position a rectangle in the rectangular coordinate plane as shown in Figure p.4.
oc and Ats, we
find that
Applying the midpoint formula for the coordinate plane to segments
'
Figure P.4 The coordinates ofC
must be (a, b) in order for
quadrilateral OBCA to be a rectangle.
b q+!'l= g. g)
( 2", 2 ) |,\2',2)
midpoint of segment oc
=(
midpoint of segmen t AB
=(
9!s, q+!
\ 2
g. g
2 )) = |,\2',2)]
The midpoints of segments oc and AB are the same, so the diagonals of the rectangle
OBCA meet at their midpoints and thus bisect each other.
Exercises for Alternate Example 9
In Exercises
I
3-l
8, prove that the diagonals of the figure determined by the points bisect each other.
13. Square (-1, l), (2, l), (2, 4), (-1, 4)
15. Parallelogram (0,0), (-2,0), (-4, 6),(1,6)
17. Rectangle (-2, 8), (-2, 2), (2, 2), (2, 8)
14. Rectangle (-2, 2), (-2, 6),(8, 6), (8, 2)
16: Parallelogram (-6, 6), (1, 6), (2, 3), (-5, 3)
18. Square whose sides have length a
I
Section
P.3
P.3 Linear
Equations and lnequalities
Linear Equations and Inequalities
Solving a Linear Equation
Solve
-3(x + 2) + 2(x
-
t) = -4r + 4. Support the result with a calculator.
SOLUTION
-3{x
+
7) + 2{r
- l) = -45+4
-3x-6+b-2=-4x+4
-x*8=-4x+4
3x=
12
Combine like terms
Add
x=4
Figure P.5 The lgt line stores the
number 4 into the variable x.
Distributive property
I
and add
4x
Dvide by 3.
To support our algebraic work we can use a calculator to evaluate the original equation
forx:4. Figure P.5 shows that each side of the oiiginal equation is equal to -12 rf x = 4.
Exercises for Alternate ExamPle 2
In Exercises 1-6, solve the equation. S.eporftiraesait*l,iih:reai€€+ae
l. -(r- 5)-2(x-7):-2x+
4.
6(5x
+
1)
-2(x -2)
= 5r +
2. 5(2x)-2$x
r0
l0
S. -:t(2x -2)
3. x+2 + 5(x-
-3):-x-9
+ 5(x
- l) :
l)=7x+3
6. 2(3x + 6) + z(Sx
-10x
-
7)
:
25x
-
I
14
Solving a Linear Equation lnvolving Fractions
solve
2r! =, *+..
soLuTloN
The denominators are 3, 1, and 2. The LCD of the fractions is 6. (See Appendix A.3
if
necessary.)
5v+2
v
'
= 3l
2
J
.(,"#) =u(,.i)
e
.5Y+z =6.3+e .L
32
Multiplybythe LCD6.
Dstributive property
2(5y + 2\=18+3y
10y+4=18+3/
l}Y=14a3,
7Y=14
!=2
Simplify.
Subtract 4.
Subtract 3y.
Dvide by 7.
We leave it to you to check the solution either using pencil and paper or a calculator.
Exercises for Alternate ExamPle 3
ln Exercises 7-12, solve the equation.
i.5'+? =3 +t
32
i
I
!
,r.4x-5 = _2_L
42
8.
5x-l
t
22
-=+-b=l+L
lZ.
34
x
,.;-r=t-+
,0. -x+7 -- I
35
+r
I
,F
ChapterP Prerequisites
Solving a Llnear lnequailty
Solve4(r-2)+5>7x-6.
SOLUTION
a$-2)+ 527x-6
4x-8+527x-6
Dskibutive property
sjrnplity.
4x-32.7x..p
4x)-.7x
-
3
-!,-_
Add 3.
-3x>1
x<l
Subtract 7x
Multiply by
-
j.
Cn" in"Or"tity raerses.)
The solution set of the inequality is the set of all real numbers
less than or equal to
interval notation, the solution set is (_e", l].
l. In
Exercises for Alternate Example 4
In Exercises I 3_l 8, solve the inequality.
13.1(2x-l)-l>3.t-6
17.2(2x- l)+3x<8r-7
14.4(b-l)+5<x-13
18. -3(2x
-
1) + 2(x
-
l)
15.-x-15<3(x-6)-5
16.3(-2x+2)Z4x
I
< -3,r + 6
Solving a Doubte lnequatity
Solve the inequality and graph its solution set.
,=2I/.,
SOLUTION
t<b-l<s
3
3
<2x- I < 15
16
2<x<8
4 <2.x <
Muftiptyby3,
Add t.
oivideby2.
The solution set is the set ofall real numbers greater than or equal to
2 and less than g. In
interval notation, the solution set is [2, g). Its graph is shown in rigure
r.6.
_l 0123456789
Figure P.6
The graph of the solution set of the double inequality in Alrcrnarc
Example 6.
Exercises for Alternate Example d
In Exercises 19-24, solve the inequality and graph the solution
set.
D. -2<3xA2 <4
24.
-t
<b;t
<3
zo.
4<sx{2 <s
-
21.
1<5x+2 <)
--
6
22. -1<4x
17
<3
8.3<b{3ss
t
Section
P.4
P.4
Lines in the
Plane
7
Lines in the Plane
Using the Point-Stope Form
Use the point-slope form to find an equation for the line that passes through (-5, 2) and
has slope -3.
SOLUTION
Substitute 11 : -5 and
resulting equation.
/r = 2, and m = l
l-!t=m(x-x1)
into the point-slope form, and simpliff the
Point-slope form
y-2='1(x-(-5\
y-2=4x- (-3X-5)
y -2 =*3x - 15
Y =-3x - 13
xr=-5, yr=),
p=-t
Dstributive property
A common simplified form
Exercises for Alternate Example 2
In Exercises l-6, find a point-slope form equation for the line through the point with given slope.
L $,6);slope-2
2. (0, -2); slope 2
4. (6,0); slope
5.
5
3. (a,$;
( ,A); slope-1
slope 1.5
6. (5,4);
slope
I
-10
Use a Graphing
Draw the graph of 3x - 2y =
Utility
A.
SOLUTION
First we solve fory.
3x-2Y = $
4y = -3* + 6
y =| x - 3
,2
Figure P.7 shows the graph
equation 3x
-2y:6
Solve
Divide by
of
#
-2.
I =)* -3,
or equivalently, the graph of the linear
in the [-5.1, 5.1] by 14.7,4.71 viewing window.
[-s.1, s.1] by I-4.7,4.7)
Flgure P.7 The
for y.
:
- 2y 6. The points (0, -3) (y-intercept) and (2, 0) (r-intercept)
appear to lie on the graph and, as pairs, are solutions ofthe equation, providing visual support that
the gmph is correct.
graph of 3x
"-!
'')$l
.l'
I
i
i
I
I
I
I
I
!
I
Exercises for Alternate ExamPle 4
?5nJ ? -€2{ "
10.3x-v=4
9.3x-2y:4
In Exercises 7-12, graph the linear equation on qgra?hc".
7.
b-y=3
8.
-2x+2y=4
ll.
-x+2Y=6
12.4x-4Y=l) I
ChapterP Prerequisites
Finding an Equation of a paraltel Llne
Find an equation of the line through p(2,
3x +
2y:
5.
-l)
that is parallel to the line z with equation
SOLUTION
ofl
We find the slope
by writing its equation in slope-intercept form.
3x+2y=J
Equation for
2y=-3x+5
y = -], * |
'22
'2 ofl
The slooe
is
-f
I
Subtract3x.
Divide by 2.
.
The line whose equation we seek has slope
-3l2 and contains the point
Thls, the point-slope form equation for the line we seek is
y
- (-r)
y
(2,
-l).
-]u -rl
3^ J
r=-;x+
y+
Exercises for Alternate Example
=
(xr,y):
Distributiveproperty
=_ix +2
5
In Exercises I 3-l 8, find an equation for the line passing through the point and parallel
to the given line.
B. P(A,a);3x-2y=4
17. P(2,5);2x-y=g
14. P(0,1);-2x+2y=5
18. P(3,
4);4x
15.
P(-3,0);-r+3y=4
16. P(-5,
l);3r+ 4y=+
I
+ 8y = -a
Finding an Equatlon of a perpendlcular Llne
Find an equation of the line through p{2,1) that is perpendicular to the Iine
equation 3x - y :3. Support the result with a grapher.
r
with
SOLUTION
We find the slope of Lby writing its equation in slope-intercept form.
3x-y=3
Equation for L
!=3x- 3
The slope
Subtract3x.Muttiplyby -1.
ofZ is 3.
The line whose equation we seek has slope
-ll3 and passes through (xr, y) = (2, -3).
Thus, the point-slope forrn equation for the line we seek is
y-(1)=-LrA-rl
* 3 = -i, *?
I'33
[-s.1, s,l] by 1a.7,4.71
Flgure P.8 The graphs ofy:3randy: (-ll3)x-78 in this square
3
viewing window appear to intersect at
a right angle.
v---1-'33
t..
Distributive property
7
Figure P.8 shows the graphs of the two equations in a square viewing window and
suggests that the graphs are perpendicular.
Section
P.5
Sotving Equations Graphically, Numerically, and
Algebraically 9
Exercises for Alternate Ex.rmple 6
InB(qcls€sl9.24,findanequationforttrelinepassingthroughthepointandperpendiculartotregiverrline.@
22. P(3,0);x+4y:6
21. P(-S,a);5x+2y=6
20. P(2,1);-x+2y:-l
- 2y :'3
24. P(5,7\;5x-l0y=':l
23. 49,4);-3x+3Y=
19. P(-3, -4); -3x
1
So
lv in g E qu ations
-
I
erapkicafiy, Nuurerie alty,
ebraicall
P.5
Solving by Findtng x'lntercePts
1
Solve the equation
I
2* -7*
+ 3 : 0 graphically.
SOLUTION
I
Solve Graphically
: 6t - 7 x + 3 (Figure P,9). We use TRACE to see
that (0.5,0) and (3, 0) arex-intercepts ofthis graph. Thus, the solutions ofthis equation
are x:0.5 and x : 3. Answers obtained graphically are really approximations, although
in general they are very good approximations.
Find the r-intercepts of the graph of y
I
.J
'l
i
I
l
[-s.1, s.1] bY l'4.7,4.71
Figure P.9 ltappearsthat(0.5,0)and(3,0)arex-interceptsofthegraph
of
y--?i -7x+3.
Solve AlgebraicallY
ln this case, we can use factoring to find exact values.
-7x +3 =O
(2x- 1)(x-3)=$
2x2
Factor-
We can conclude that
2x-
So,; =
.,1
'i
.t
or
x-3:0
,=Lz
or
x:3
1,/2 and
J
'i
Exercises for Alternate ExamPle
.tI
i
I
'4
.l
i
I
I
I
I
I
i
I
't
l:0
l
x:
3 are the exact solutions of
tie original
equation-
I
In Exercises l-5, sotve the equationgraphiedfi€oa,f,m by using factoring to solve the equation.
t. *-*=o
6. 5l+ l9x+ 12=o
2.3 -lb=o
3.
* -x'6=0
4.
2* -x-3:A
s.
z*-tx+l=o
I
ChapterP Prerequisites
Solve the equation
Soiving Using the Quadrailc Formula
Sx:2.
* -
SOLUTION
First we subtract 2 from each side of the equation to put it in the form
Zf -
Sx
-Z:
0. We can see that
a:2, b:
-5,
and
c:
.=6 * t[ 6z- 4o"
x=__-ra
-(-s)
r
a*
+
6y.
+ g = g.
-e..
Quadratic formula
4(2)(-2)
-
a-2, b=5, c=-2
Simplify.
t-5.!,
5.
ll
,=tj
by l-s.7,
=r.r, or r=L.E=r.r,
3.71
Figure P.lO Thegraphof
y-2* -5x-2.
The graph of y = v,t - 5x -2 in Figure
approximately -0.35 and 2.85.
p.l0 supports that the r-intercepts
are
Exercises for AJternate Example 4
In Exercises 7-12, solve the equation by using the quadratic formula.
7.2*-5x:O
8.3x2-r=2
g.3f-7x+t=O
lt. *-6x:5
10.4x2*6x:3
12.
l}f -x:2
I
Solving Graphically
Solve the equation x3 + 2x
- l:
0 graphically.
SOLUTION
Figure P. I I a suggests that x : 0.45339765 is the solution we seek. Figure p.l lb
provides numerical support that x: 0.4s339765 is a close approximationto the
solution because, when x : 0 -45i397 65, f + 2x 1 :
- -3.g67bg x l0-eor
-0.00000000396779, which is nearly zero.
[-6.1, 4.1] by 1a.7,4-7-]
(4)
(b)
The graph ofy : ,3 1 2t l. (a) shows that (0.4533976,0) is an appmximatioo
to t}e x-intercept ofthe graph. (b) supports this
Figure
P.ll
Exercises for Alternate Example
lnExercisesl3-l8,solvetheequationgraphically.
13.x3-x+2=0
18. x3 + 6x+2=O
14.21+x--o
-
conclusion.
5
a4*;'=\ot'
tt* *-il tr:Sr- a-r sr-\.* 2t-<*-t
rs.l+x-4:0
16.rr+6r-1=j
fi.-s+2x+3=0
I
l2
Chapter
P
Prerequisites
P.6 Complex Numbers
Adding and Subtracting Comptex Numbers
Add or subtact.
(a).(6-5r)+(-3-7r)
(b)(2+
lrr-(-8-2,)
SOLUTION
fd)"(6"-5i) + (-3
(b)
-7)
o (6 + (_3)) + (_s + (_7)),
=
3 + (_12)i
= 3 _ tzi
(2+llr-(-8 -zi)=(2-(-8)-(ll - (i))i = l0+t3i
Exercises for Alternate Example
I
In Exercises l-8, write the sum or difference in the standard form a + bi.
r. (-r -3r)+(-n +4
s.
(4 +
30r)+
ctt
-29i)
2. (-10 + 2r) -i (-l - r)
6. (-8 + 18,) - (21 + 12,)
3.
(7 + 7i) + (2
+ 2i)
_.ti)
+ (12 + 32i)
7. (t
4.
(18-3r)-(l -9,)
- (t7 -79i\
8. (7 + 13'
I
Raising a Complex Number to a power
,,
,=
S *$r,findzz
andi.
SOLUTION
,,=(*.+,)(*.+,)
=f;*|i*f,i*f,*
=|*|i*f,i*]er>
-i
zt=22.2
=,(+.+,)
=*,.*u
={,*$u,
=-fr*&,
22
Exercises for Alternate Example 3
In Exercises 9-14, write the complex number in standard form.
t.1t+zif
ro.
1-l++43
1r. (6-8D3
n. ga+tf
tt.1-z-t43
14. (-3_5,),
I
Section
P.6
Complex Numbers
t3
Di"viding Gomplex Numbers
Write the complex number in standard form.
(o)#
(")*
SOLUTION
3_ 3
(a)
'' 2+i 2+i',2-i
2-t
_
(b) 3+3i _ 3+3i
2-i 2-i .2+t
2+i
3(2-i)
(2+i)(2-i)
_ (3+3rx2+r)
(2-i)(z+i)
_ 6-2i
n2
z-t
-6+3i+6i+3i2
'2
22
_ 6-2i
4-(-l)
-i2
_ 6+9r+3(-l)
4-(-1)
6-2i
4+l
_6-2i
_
=51-2i:-3
4+l
_3+9i
5
5
62i
55
3
.9i
55
Exercises for Alternate Example 4
In Exercises l5-20, write the complex number in standard form.
15.
1
2i
rc.
4+i
2i
fi.3-2i
l+zi
18.
3-i
3+i
t9. 4+3i
3+4i
20 6-8'
3 -4i
solving a euadratic Equation
@
Solve2x2+2r+
l:0.
SOLUTION
Solve Algebralcally
Using the quadratic formula with a
x=
1+
22
:
2, b
:
2, and c :
l,
we obtain
-4Q)0)
2Q)
4
*-2r.J
-4
= 1L2i
4
=-1+1;
2- 2'
So,
tle solutions
are
-l*li*a -*-i,
,a complex conjugate pair.
I
ChapterP Prerequisites
Confirm Numerically
Substituting
-1.*,
into the original equation, we obtain
,(-+.+,)' .r(-;.+,)*,=(-,)+(-r
+i)+ I =o
By a similar computation we can confirm the second solution.
Exercises for Alternate Example
5
ln Exercises 2l-28, solve the equation.
*+5x+7=0
ZS.*-x+13:0
zt.
22.3*-x+4:0
26.2x2-lgx+13:0
zs.
q*-y+5-0
24.4x2+3=0
27.x2+3x+3=6
zs.
I
sl+20:o
Solving Inequalities Algebraically
P.7
anrl Granhirrllrr
uru
vtul/rrr\urrr
Solving Another Absotute Value lnequality
Solve l2-r + 1l > 4.
SOLUTION
The solution of this absolute value inequality consists of
inequalities.
2x+l
<4
2x <-5
x
5
<-:
2
or
or
2x+1>4
2x>3'
or
,rtr2
Subtract
tle solutions of both
1.
Divide by 2.
The solution consists of all numbers that are eirher one of the two intervals
[-5.1,
s.l]
Figure P.l4 The graphs of
y:l2;E+llandy:4.
*n,.n may be written u, ( *,
[*,
j
)
- rhe notation ..u,, is read
(
) " ;, )
as "union." Figure P.14 shows that points on the graph of y :
l2x + 1l are above the
points on the graph of y: 4 for values ofx to the left of
-f and to the risht or ] .
o. (
;,
by l-4.7,4.7)
these
-
),
-;
Exercises for Alternate Example 2
In Exercises l-6, solve the absolute value inequality.
i.
ix*-+,
I
lri>
i
2-
w+21>2
3.
lt-zlsl
4.
l-r-31>3
5.
l2r-5ls
r
6. l3x+
ll>2
I
Section
P.7 Solving lnequalities Algebraically and Graphically l5
I
i
\
Solving a Quadratic lnequality
I
Solvex2-x-6>0.
I
i
I
I
SOLUTION
I
First we solve the corresponding equation.
x2-x-6=0
(r-3)(x +4=0
x-3=0
x=3
t-s.l,
5.
ll
by l-6.7, 2.7)
Factor.
or
or
x
t2= 0
x=-2
ab=0t
Sofue
for
a=Oorb=O
x
The solutions ofthe corresponding quadratic equation are 1 and 3, and they are
solutions of the original inequality because 0 > 0 is true. Figure p.l5 shows that the
points on the graph of y: ,z - x - 6 are on or above the x-axis for values ofx at or to
the left of-2 and at or to the right of3.
Flgure P.l5 The graph of
/= * -x-6appears to cross the
x-axisatx:--2andx:3-
Exercises for Alternate Example
3
in Exercises 7-12, solve the inequaiiiy.
7.;-9>o
8. 12-5x+4<o
g.
f+3x-4>o
lo. x2-6x+9>6
11.x2+7x+l)>g
I
12.2i +7x+3 <0
Solving a Cubic lnequality
Solvex3
-3*+
I >0graphically.
SOLUTION
We can use the graph of7 : x3 - 3*'+ 1 in Figure P. 16 to show that the solutions of the
corresponding equation xl - 3x2 + I : 0 are approximately -0.53, 0.65, and 2.88. The
points on the graph of y : ,t * 3* + 1 are above the x-axis for values ofx between {.53
and 0.65, and for values ofx to the right of2.88.
The solution of the inequality is [-0.53, 0.65] u [2.88, -y. We use square brackets
i - 3x2 + t : 0 are also solutions ofthe inequality.
because the zeros of
[-6.1, 4.1] by {-a.7 ,4.t1
:
Figure P.16 The graph ofy 13 3x2 + 1 appears to be on or above the x-axis between
the negative ,x-intercept a:rd the srnaller of the tvrc pcsitive x-intercepts and to -.he right of ..he
-
larger x-intercept.
Exercises for ALternate Example 7
In Exercises 13-18, solve the cubic inequality graphically.
13. x3+x-3>0
18. x3- 5;+4>o
t1
I
I
I
I
I
u.
I
-lx>o
15. x3 +
5.r2
+.r > 0
16.
-x3+l+r>0
17. -x3 + 4; +2x>
O
r

Download Report