College Algebra Definitions and Procedures
Cartesian Coordinate System
x-intercept is a point (a, 0)
y-intercept is a point (0, b)
x is the independent variable, the abscissa
y is the dependent variable, ordinate
Quadrant II
Quadrant I
Quadrant III
Quadrant IV
Distance Formula
The distance between two points (x1,y1) and (x2, y2) is
__________________
d = √ (x2 – x1)2 + (y2 – y1)2
d
(x2, y2)
| y2 – y1|
The Midpoint Formula
If the endpoints of a segment are (x1, y1) and (x2, y2),
then the coordinates of the midpoint are
x1 + x2 , y1 + y2
2
2
(x1,y1)
| x2 – x1 |
Function
domain
A correspondence between a first set, the domain,
9
and the second set, the range, such that each member
3
of the domain corresponds to exactly one member of the range.
4
The Vertical Line Test
If it is possible for a vertical line to cross a graph more
than once, then the graph is not the graph of a function.
range domain range
2
9
4
3
2
1
4
Not a function
Function
Linear Function
A function f is a linear function if it can be written as
f(x) = mx + b
m = slope
b = y-intercept
Slope
The slope m of a line containing points (x1, y1) and
(x2, y2) is given by the diagram
m = rise = the change in y = y2 – y1
run
the change in x
x2 – x1
Horizontal and Vertical Lines
Horizontal lines y = b or f(x) = b
The slope of a horizontal line is zero.
Vertical Lines x = a
The slope of a vertical line is not defined.
m = y2 – y1
x2 – x1
(x2, y2)
y2 – y1
(rise)
(x1, y1)
x2 – x1
(run)
x1
m=0
●
(x1, y1)
= __0__
x2 – x1
●
x2
m is not defined
● (x2, y2)
m = y2 – y1
x2 – x1
● (x1, y1)
(x2, y2)
= y2 – y1
0
The Point-Slope Equation
The point-slope equation of a line with slope m passing through (x1, y1) is y – y1 = m(x – x1)
Parallel Lines
Parallel lines have the same slope and different y-intercepts.
y = 3x + 4 and y = 3x – 7
Perpendicular Lines
Perpendicular lines have reciprocal and opposite slopes.
y = ½ x + 4 and y = -2x – 7
Relative Maxima and Minima
A point f(c) is a relative maximum if it is
the highest point within some open interval.
A point f(c) is a relative minimum if it is the
lowest point within some open interval.
relative
maximum
relative
mimimum
Composition of Functions
The composite function f ◦ g, the composition of f and g, is defined as
(f ◦ g)(x) = f( g(x) )
where x is in the domain of g and g(x) is the domain of f.
Algebraic Tests of Symmetry
x-axis: If replacing y with –y produces an equivalent equation,
then the graph is symmetric with respect to the x axis.
y-axis: If replacing x with –x produces an equivalent equation,
then the graph is symmetric with respect to the y-axis.
Origin: If replacing x with –x and y with –y produces an equivalent
equation, then the graph is symmetric with respect to the origin.
y = x2 + 2
(-y) = x2 + 2 is y = -x2 – 2
not symmetric to the x-axis
y = (-x)2 + 2 is y = x2 + 2
is symmetric to the y-axis
(-y) = (-x)2 + 2 is y = -x2 – 2
not symmetric to the origin
Even and Odd Functions
Even Function: If the graph of a function f is symmetric with respect
to the y-axis. For each x in the domain of f, f(x) = f(-x)
Odd Function: If the graph of a function f is symmetric with respect
to the origin. For each x in the domain of f, f(-x) = -f(x)
h(x) = 5x7 – 3x3 – 2x
h(-x) = 5(-x)7 – 3(-x)3 – 2(-x) =
-5x7 + 3x3 + 2x
not even
7
3
g(x) = 5x – 3x – 2x
-g(x) = -(5x6 – 3x3 – 2x) =
-5x7 + 3x3 + 2x
odd
Transformations of y = f(x)
Vertical Translation: y = f(x) + b for b > 0
y = f(x) + b up b units y = f(x) – b down b units
Horizontal Translation: y = f(x + d)
y = f(x – d) right d units y = f(x + d) left d units
y = f(x)
y = f(x) + b
Reflections
Across the x-axis: y = -f(x) for y = f(x)
Across the y-axis: y = f(-x) for y = f(x)
Vertical Stretching or Shrinking: y = af(x)
Stretch vertically for |a| > 1
Shrink vertically for 0 < |a| < 1
For a< 0, the graph is also reflected across the x-axis
Horizontal Stretching and Shrinking: y = f(cx)
Shrink horizontally for |c| > 1
y = f(-x)
y = af(x)
Stretch horizontally for 0 < |c| < 1
|a| > 1
For c < 0, the graph is also reflected across the y-axis
Copyright Professor Shotsberger 1/2006
y = f(x – d)
y = f(cx)
0< |c| <1
The complex numbers:
a + bi
7, -8.7, ½ , ¼ i, π, 5 + 2i
Imaginary numbers:
a + bi, b ≠ 0
½ - i, ¼ i, 5 + 2i
Imaginary numbers:
a + bi, a ≠ 0, b ≠ 0
½ - i, 5 + 2i
The Number i
Real numbers:
a + bi, b = 0
7, -8.7, ½, √3, - 18
Pure imaginary numbers:
a + bi, a = 0, b ≠ 0
¼ i, 8 i
Irrational numbers:
√2, π, 3√5
Rational numbers:
7, -8.7, ½, -2
____
i = √ - 1 and i2 = - 1
Conjugate of a Complex Number
The conjugate of a complex number a + bi is a – bi.
a + bi and a – bi are complex conjugates.
Completing the Square
1. Move the constant to the other side.
2. Find the new constant. (b/2)2 = (-6/2)2 = (-3)2 = 9
3. Add the new constant to each side.
4. Factor the left side and combine the right side.
x2 – 6x – 10 = 0
x2 – 6x
= 10
x2 – 6x + 9 = 10 + 9
(x – 3)2 = 19
The Quadratic Formula
________
The solutions of ax2 + bx + c = 0, a ≠ 0 are: x = -b + √ b2 – 4ac
2a
Discriminant
For ax2 + bx + c = 0
b2 – 4ac = 0 One real-number solution
b2 – 4ac >0 Two different real-number solutions
b2 – 4ac < 0 Two different imaginary-number solutions, complex conjugates.
Zeros, Solutions, and Intercepts
g(x) = x2 – 3x – 4
Find the zeros: x2 – 3x – 4 = 0
(x – 4)(x + 1) = 0
x–4=0 x+1=0
x=4
x = -1
Find the solution: g(-1) = 0
g(4) = 0
Find the intercepts:
(-1,0)
(4,0)
Find the discriminant:
(-3)2 – 4(1)(-4) = 9 + 16 = 25
Since 25> 0 it has 2 real-number solutions.
Quadratic Functions
f(x) = x2 + 10x + 23
2
2
f(x) = ax + bx + c or f(x) = a(x – h) + k
f(x) = x2 + 10x + 25 – 25 + 23
Vertex (h, k)
f(x) = (x + 5)2 - 2
Axis of symmetry x = h
Vertex: (-5, -2); Axis of symmetry: x = -5
Copyright Professor Shotsberger 1/2006
Graphs of Functions
Linear Function
f(x) = mx + b
Constant Function
f(x) = c
Quadratic Function
f(x) = ax2 + bx + c
Square-root Function
f(x) = √x
Cube Function
f(x) = ax3
Rational Function
f(x) = 1
x
The Leading Term Test
If anxn is the leading term of a polynomial, then the behavior of the graph as x → ∞ or as x → -∞ will be:
If n is even, and an > 0
If n is even, and an < 0
If n is odd, and an > 0
Graphing a Polynomial Function
1. Use the leading-term test to determine end behavior
2. Find the zeros of the function by solving f(x) = 0
3. Use the x-intercepts to divide the x-axis into intervals.
Choose a test point in each interval to determine + or -.
4. Find f(0), the y-intercept.
5. Find additional function values to determine the shape.
6. Check: does it have at most n x-intercepts and at most
n – 1 turning points.
The Intermediate Value Theorem
For any polynomial function P(x) with real coefficients,
suppose that for a ≠ b, P(a) and P(b) are of opposite signs.
Then the function has a real zero between a and b.
If n is odd, and an < 0
f(x) = 2x3 + x2 – 8x – 4
2x has odd degree of 3 and 2 > 0
Factor: x2(2x + 1) – 4(2x + 1) = 0
(2x + 1)(x – 2)(x + 2) = 0
The zeros are: - ½ , 2, and -2
3
-∞ -2 - ½
2
+∞
Test points:
(-3, -25) (-1, 3) (1, -9) (3, 35)
-below
+above -below +above
f(x) = 2x3 + x2 – 8x – 4
Is there a real zero between -1 and 1?
f(-1) = 2(-1)3 + (-1)2 – 8(-1) – 4 = 3
f(1) = 2(1)3 + (1)2 – 9(1) – 4 = -10 YES
The Remainder Theorem
If a number c is substituted for x in the polynomial f(x),
Then the result f(c) is the remainder that would be obtained
by dividing f(x) by x – c.
That is, if f(x) = (x – c) • Q(x) + R, then f(c) = R
f(x) = x3 + 2x2 - 5x – 6
_______x2 + 5x + 10
x - 3 ) x3 + 2x2 - 5x - 6
x3 - 3x2
5x2 - 5x
5x2 - 15x
10x - 6
10x - 30
3
2
F(3) = (3) + 2(3) – 5(3) – 6 = 24
24 = R
Copyright Professor Shotsberger 1/2006
Synthetic Division
Divide f(x) = x3 + 2x2 – 5x – 6 by x – 3
_3 |
1
2
-5
-6
______3
15
30
1
5
10
24
The Fundamental Theorem of Algebra
Every polynomial function of degree n, with n ≥ 1, has at least one zero in the system of complex numbers.
The Rational Zeros Theorem
Let P(x) = anxn + an- 1xn – 1 + . . . + a1x + a0
All the coefficients are integers.
The numbers p and q are relatively prime
(have no common factors besides 1).
If p/q is a zero of P(x), then p is a factor of a0 and q is a factor of an.
Descartes’ Rule of Signs
Let P(x) be a polynomial function with real coefficients and
a nonzero constant term.
The number of positive real zeros of P(x) is either:
1. The same as the number of variations of sign in P(x)
2. Less then the number of variations of sign in P(x) by
a positive even integer.
The number of negative real zeros of P(x) is either:
3. The same as the number of variations of sign in P(-x)
4. Less then the number of variations of sign in P(-x)
by a positive even integer.
A zero of multiplicity m must be counted m times.
P(x) = 3x4 + 2x3 – 4x + 6
p=6 q=3
p/q = ±1 or ±2 or ±3 or ±6
±1, ±3
Possibilities: -1, 1, -2, 2, -3, 3, -6, 6,
-2/3, 2/3, -1/3, 1/3
P(x) = 2x5 – 5x2 – 3x + 6
2 to – 5 and -3 to 6 are 2 variations.
So, there are 0 or 2 positive real zeros
P(-x) = 2(-x)5 – 5(-x)2 – 3(-x) + 6 =
-2x5 - 5x2 + 3x + 6
-5 to 3 is one variation in sign
So there is exactly one negative real zero
Asymptotes of Rational Function
Vertical Asymptotes:
f(x) = __2x + 3__
Occur at any x-values that make the denominator zero.
3x2 + 7x – 6
Horizontal Asymptotes:
3x2 + 7x – 6 = (3x – 2)(x + 3)
The horizontal asymptote is y = 0 (the x-axis)
Vertical Asymptotes: x = -3 and x = 2/3
degree of numerator < degree of denominator
The horizontal asymptote is not the x-axis when
numerator degree is 1 < denominator degree is 2
degree of numerator = degree of denominator
Horizontal Asymptote: y = 0
Oblique Asymptote:
Degree of numerator is 1 greater than degree of the denominator.
There can be only one horizontal asymptote or one oblique asymptote and never both.
Graphing a Rational Function
1. Find the real zeros of the denominator.
Sketch the vertical asymptotes.
2. Find and sketch the horizontal or oblique asymptotes.
3. Find the zeros of the numerator – the x-intercepts.
4. Find f(0) the y-intercept.
5. Find other values to determine the general shape.
Copyright Professor Shotsberger 1/2006
Inverse Relation
Interchanging the first and second coordinates of each ordered pair in a relation produces the inverse relation.
One-to-One Functions
A function f is one-to-one if different inputs have different outputs
if a ≠ b, then f(a) ≠ f(b)
Or a function f is one-to-one if when the outputs are the same, the inputs are the same, that is:
if f(a) = f(b), then a = b
Properties of One-to-One Functions and Inverses
 If a function is one-to-one, then its inverse is a function.
 The domain of a one-to-one function f is the range of the inverse f -1
 The range of a one-to-one function f is the domain of the inverse f -1
 A function that is increasing over its domain or is decreasing over its domain is a one-to-one function.
Horizontal Line Test
If it is possible for a horizontal line to intersect the graph
of a function more than once, then the function is not one-to-one
and its inverse is not a function.
not one-to-one
f(x)
Obtaining a Formula for an Inverse
If a function f is one-on-one, a formula for its inverse can generally be found by:
1. Replace f(x) with y.
y = 2x – 3
2. Interchange x and y.
x = 2y – 3
3. Solve for y.
x + 3 = 2y (x + 3)/2 = y
4. Replace y with f -1(x)
f -1(x) = (x + 3)/2
The graphs of f -1 and f are reflections across the line y = x.
y=x
f -1(x)
Inverse Functions and Compositions
If a function f is one-to-one, then f -1 is the unique function
2[(x+3)/2] – 3 = x + 3 – 3 = x f(f -1(x))
such that each of the following holds:
and
(f -1◦ f)(x) = f -1(f(x)) = x for each x in the domain of f
(2x – 3) + 3 = 2x = x
f -1(f(x))
-1
-1
-1
(f ◦ f )(x) = f(f (x)) = x for each x in the domain of f
2
2
-x
f(x) = a
f(x) = ax
Exponential Function
The function f(x) = ax , where x is a real number, a > 0 and a ≠ 1,
is called an exponential function, base a.
f(x) = ax – b
The Number e
e = 2.7182818284. . .
Logarithmic Function, Base a
We define y = log a x as that number y such that x = a y, where x > 0 and a is a positive constant other than 1.
log a 1 = 0 ↔ a0 = 1 ,
log a a = 1 ↔ a 1 = a , for any logarithmic base a.
Natural Logarithms
ln x means log e x
ln 1 = 0 and ln e = 1, for the logarithmic base e.
ln 3 = log e 3
Copyright Professor Shotsberger 1/2006
The Change-of-Base Formula
For any logarithmic bases a and b, and any positive number M,
log b M = log a M
log a b
log 5 8 = log 10 8 or loge 8 or ln 8 ≈1.292
log 10 5
loge 5
ln 5
Properties of Logarithms
The Product Rule:
log a MN = log a M + log a N
The Power Rule:
log a M p = p log a M
The Quotient Rule: log a M = log a M – log a N
N
log 5 (2 • 4) = log 5 2 + log 5 4
log 5 23 = 3 log 5 2
log 5 16 = log 5 16 – log 5 2
2
The Logarithm of a Base to a Power
log a ax = x
ax = ax
log 5 58 = 8
Base-Exponent Property
ax = ay ↔ x = y
2 3x – 7 = 32 ↔ 2 3x-7 = 25
3x – 7 = 5 and x = 4
A Base to a Logarithmic Power
alog a x = x
log a x = log a x
10log 7t = 10log10 7t = 7t
Property of Logarithmic Equality
log a M = log a N ↔ M = N
log3 3x = log3 20 ↔ 3x = 20
Exponential Growth Rate
P(t) = P0 ekt , k > 0
P0 = population at time 0, k = the exponential growth rate, t = time
The growth rate k and the doubling time T are related by
KT = ln 2, or k = ln 2
or T = ln 2
T
k
Exponential Decay: P(t) = P0 e-kt
Solving a System of Equations
Graphing:
x – y = 5 and 2x + y = 1
y=x–5
y = -2x + 1
Substitution: 2x + (x – 5) = 1
3x = 6, x = 2
y = (2) – 5 , y = -3 (2, -3)
Elimination: x – y = 5
2x + y = 1
3x
= 6, x = 2
2(2) + y = 1, y = -3 (2, -3)
Consistent, Independent
One solution
Inconsistent, Independent
No solution, parallel lines
Copyright Professor Shotsberger 1/2006
(2, -3)
Consistent, Dependent
Same lines, infinite solutions
CONIC SECTIONS
axis of symmetry
Parabola
(x – h)2 = 4p(y – k)
Vertex:
(h, k)
Focus:
(h, k + p)
Directrix:
y=k–p
p>0 the parabola opens upward
p<0 the parabola opens downward
focus: ● (h, k+p)
vertex: (h,k)
directrix: y = k – p
directrix: x = h – p
(x –
= 4p(x - h)
Vertex:
(h, k)
Focus:
(h + p, k)
Directrix:
x=h–p
p>0 the parabola opens right
p<0 the parabola opens left
k)2
vertex: (h, k)
axis of symmetry
●
● (x, y)
r
● (h,k) center
Circle
(x – h)2 + (y – k)2 = r 2
Center: (h, k)
Radius: r
Ellipse
(x – h)2 + (y – k)2 = 1, a > b > 0
a2
b2
Vertices:
(h - a, k), (h + a, k)
Foci:
(h - c, k), (h + c, k)
Length of minor axis: 2b, c2 = a2 – b2
Major Axis Horizontal
(h-c,k)
(x – h)2 + (y – k)2 = 1, a > b > 0
b2
a2
Vertices:
(h, k – a), (h, k + a)
Foci:
(h, k – c), (h, k + c)
Length of minor axis: 2b, c2 = a2 – b2
Major axis Vertical
Hyperbola
(x – h)2 – (y – k)2 = 1
a2
b2
Vertices:
(h – a, k), (h + a, k)
Asymptotes: y – k = (b/a)(x – h)
y – k = (-b/a)(x – h)
Foci:
(h – c, k), (h + c, k)
c2 = a2 + b2
(h+c,k)
●
(h, k)
●
●
(h, k+a)
(h-a,k)
●
(y – k)2 - (x – h)2 = 1
a2
b2
Vertices:
(h, k – a,), (h, k + a)
Asymptotes: y – k = (a/b)(x – h)
y – k = (-a/b)(x – h)
Foci:
(h, k – c), (h, k + c)
focus: (h + p, k)
(h+a,k)
(h, k+c)
(h, k) ●
●
(h, k-c)
(h, k-a)
(h-a,k)
(h-c,k) ●
(h+a,k)
●
(h,k)
● (h+c,k)
(h,k+c)
●
(h,k+a)
(h,k-a)
● (h,k)
● (h,k-c)
Copyright Professor Shotsberger 1/2006
Arithmetic Sequence
A sequence is arithmetic if there exists a number d ,
called the common difference, such that an+1 = an+ d
for any integer n ≥ 1
nth Term of an Arithmetic Sequence
The nth term of an arithmetic sequence is given by
an = a1 + (n – 1)d,
for any integer n ≥ 1
Sum of the First n Terms
The sum of the first n terms of arithmetic sequence
is given by:
Sn = n(a1 + an)
2
Geometric Sequence
A sequence is geometric if there is a number r,
called the common ratio, such that:
an+1 = r or an+1 = anr, for any integer n ≥ 1
an
nth Term of a Geometric Sequence
The nth term of a geometric sequence is given by,
an = a1 r n-1
For any integer n ≥ 1
Sum of the First n Terms
The sum of the first n terms of a geometric sequence
is give by:
Sn = a1 (1 – r n),
for any r ≠ 1
1–r
Limit or Sum of an Infinite Geometric Series
When |r| < 1, the limit or sum of an infinite geometric
Series is given by:
S∞ = __a1__
1–r
Pascal’s Triangle
(a + b)0
1
1
(a + b)
1
(a + b)2
1
2
3
(a + b)
1
3
(a + b)4
1
4
6
5
(a + b)
1
5
10
4, 9, 14, 19, 24 . . .
d = 5 (19 + 5 = 24)
The 14th term in the above sequence:
a1 = 4, n = 14, d = 5
an = 4 + (14 – 1)5 = 4 + 65 = 69
The sum of the first 14 terms in the above:
Sn = 14 (4 + 69) = 7(73) = 511
2
Can be written as: 14
Σ (5n – 1)
k=1
3, 6, 12, 24, 48, . . .
r = 24/12 = 2
The 14th term of the above sequence:
a1 = 3, n = 14, r = 2
an = 3 (214-1) = 3(213) = 24576
The sum of the first 14 terms in the above:
Sn = 3(1 – 214) = 3(-16383) = 49,149
1–2
-1
Can be written as: 14
Σ 3(2 n-1)
k=1
-2, 1, - ½ , ¼ , -1/8, . . .
r = 1 ÷ -2 = - ½ since |r| < 1 |- ½ | < 1
this series has a limit or sum
S∞ = ___-2___ = _-2 = - 4
1 – (- ½ )
3/2
3
1
1
3
1
4
10
The Binomial Theorem Using Pascal’s Triangle
For any binomial a + b and any natural number n,
(a + b)n = c0anb0 + c1an-1b1 + c2an-2b2 + . . .+ cn-1a1bn-1 + cna0bn
Where the numbers c0, c1, c2, . . .,cn-1 are from the
(n + 1)st row of Pascal’s Triangle
1
5
1
Expand (2t + 3)4 (use: 1 4 6 4 1)
= 1(2t)4(3)0 + 4(2t)3(3)1 + 6(2t)2(3)2 +
4(2t)1(3)3 + 1(2t)0(3)4 =
= 16t4 + 96t3 + 216t2 + 216t + 81
Copyright Professor Shotsberger 1/2006