College Algebra Definitions and Procedures
College Algebra Definitions and Procedures
Cartesian Coordinate System
x-intercept is a point (a, 0) Quadrant II Quadrant I
y-intercept is a point (0, b)
x is the independent variable, the abscissa
y is the dependent variable, ordinate Quadrant III Quadrant IV
Distance Formula
The distance between two points (x1,y1) and (x2, y2) is d (x2, y2)
__________________
d = √ (x2 – x1)2 + (y2 – y1)2 | y2 – y1|
The Midpoint Formula (x1,y1)
If the endpoints of a segment are (x1, y1) and (x2, y2), | x2 – x1 |
then the coordinates of the midpoint are
x1 + x2 , y1 + y2
2. 2
Function domain range domain range
A correspondence between a first set, the domain, 9 2 9
and the second set, the range, such that each member 3 4 3 2
of the domain corresponds to exactly one member of the range. 4 1 4
The Vertical Line Test Not a function Function
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.
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 (x2, y2)
The slope m of a line containing points (x1, y1) and y2 – y1
(x2, y2) is given by the diagram (rise)
m = rise = the change in y = y2 – y1 (x1, y1)
run the change in x x2 – x1 x2 – x1
(run)
Horizontal and Vertical Lines
Horizontal lines y = b or f(x) = b x1 x2
The slope of a horizontal line is zero.
Vertical Lines x = a
The slope of a vertical line is not defined. m = 0 m is not defined ● (x2, y2)
m = y2 – y1 m = y2 – y1
x2 – x1 ● ● x2 – x1 ● (x1, y1)
(x1, y1) (x2, y2)
= __0__ = y2 – y1
x2 – x1 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 relative
A point f(c) is a relative maximum if it is maximum
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
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 y = x2 + 2
x-axis: If replacing y with –y produces an equivalent equation, (-y) = x2 + 2 is y = -x2 – 2
then the graph is symmetric with respect to the x axis. not symmetric to the x-axis
y-axis: If replacing x with –x produces an equivalent equation, y = (-x)2 + 2 is y = x2 + 2
then the graph is symmetric with respect to the y-axis. is symmetric to the y-axis
Origin: If replacing x with –x and y with –y produces an equivalent (-y) = (-x)2 + 2 is y = -x2 – 2
equation, then the graph is symmetric with respect to the origin. not symmetric to the origin
Even and Odd Functions h(x) = 5x7 – 3x3 – 2x
Even Function: If the graph of a function f is symmetric with respect h(-x) = 5(-x)7 – 3(-x)3 – 2(-x) =
to the y-axis. For each x in the domain of f, f(x) = f(-x) -5x7 + 3x3 + 2x not even
Odd Function: If the graph of a function f is symmetric with respect g(x) = 5x7 – 3x3 – 2x
to the origin. For each x in the domain of f, f(-x) = -f(x) -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 y = f(x – d)
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) y = f(cx)
Stretch horizontally for 0 < |c| < 1 |a| > 1 0< |c| 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 Find the intercepts: (-1,0) (4,0)
(x – 4)(x + 1) = 0
x – 4 = 0 x + 1 = 0
x = 4 x = -1 Find the discriminant:
Find the solution: g(-1) = 0 (-3)2 – 4(1)(-4) = 9 + 16 = 25
g(4) = 0 Since 25> 0 it has 2 real-number solutions.
Quadratic Functions f(x) = x2 + 10x + 23
f(x) = ax2 + bx + c or f(x) = a(x – h)2 + 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 Constant Function Quadratic Function
f(x) = mx + b f(x) = c f(x) = ax2 + bx + c
Square-root Function Cube Function Rational Function
f(x) = √x f(x) = ax3 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 If n is odd, and an < 0
Graphing a Polynomial Function f(x) = 2x3 + x2 – 8x – 4
1. Use the leading-term test to determine end behavior 2x3 has odd degree of 3 and 2 > 0
2. Find the zeros of the function by solving f(x) = 0 Factor: x2(2x + 1) – 4(2x + 1) = 0
3. Use the x-intercepts to divide the x-axis into intervals. (2x + 1)(x – 2)(x + 2) = 0
Choose a test point in each interval to determine + or -. The zeros are: - ½ , 2, and -2
4. Find f(0), the y-intercept.
5. Find additional function values to determine the shape. -∞ -2 - ½ 2 +∞
6. Check: does it have at most n x-intercepts and at most Test points:
n – 1 turning points. (-3, -25) (-1, 3) (1, -9) (3, 35)
-below +above -below +above
The Intermediate Value Theorem
For any polynomial function P(x) with real coefficients, f(x) = 2x3 + x2 – 8x – 4
suppose that for a ≠ b, P(a) and P(b) are of opposite signs. Is there a real zero between -1 and 1?
Then the function has a real zero between a and b. 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), f(x) = x3 + 2x2 - 5x – 6
Then the result f(c) is the remainder that would be obtained _______x2 + 5x + 10
by dividing f(x) by x – c. x - 3 ) x3 + 2x2 - 5x - 6
That is, if f(x) = (x – c) • Q(x) + R, then f(c) = R x3 - 3x2
5x2 - 5x
5x2 - 15x
10x - 6
10x - 30
F(3) = (3)3 + 2(3)2 – 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 P(x) = 3x4 + 2x3 – 4x + 6 p=6 q=3
All the coefficients are integers. p/q = ±1 or ±2 or ±3 or ±6
The numbers p and q are relatively prime ±1, ±3
(have no common factors besides 1). Possibilities: -1, 1, -2, 2, -3, 3, -6, 6,
If p/q is a zero of P(x), then p is a factor of a0 and q is a factor of an. -2/3, 2/3, -1/3, 1/3
Descartes’ Rule of Signs
Let P(x) be a polynomial function with real coefficients and P(x) = 2x5 – 5x2 – 3x + 6
a nonzero constant term. 2 to – 5 and -3 to 6 are 2 variations.
The number of positive real zeros of P(x) is either: So, there are 0 or 2 positive real zeros
1. The same as the number of variations of sign in P(x) P(-x) = 2(-x)5 – 5(-x)2 – 3(-x) + 6 =
2. Less then the number of variations of sign in P(x) by -2x5 - 5x2 + 3x + 6
a positive even integer. -5 to 3 is one variation in sign
The number of negative real zeros of P(x) is either: So there is exactly one negative real zero
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.
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 not one-to-one
of a function more than once, then the function is not one-to-one
and its inverse is not a function.
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: f -1(x)
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
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))
(f ◦ f -1)(x) = f(f -1(x)) = x for each x in the domain of f -1 2 2
f(x) = a-x 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 3 = log e 3
ln 1 = 0 and ln e = 1, for the logarithmic base e.
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 5 8 = log 10 8 or loge 8 or ln 8 ≈1.292
log a b log 10 5 loge 5 ln 5
Properties of Logarithms
The Product Rule: log a MN = log a M + log a N log 5 (2 • 4) = log 5 2 + log 5 4
The Power Rule: log a M p = p log a M log 5 23 = 3 log 5 2
The Quotient Rule: log a M = log a M – log a N log 5 16 = log 5 16 – log 5 2
N 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) (2, -3)
Elimination: x – y = 5
2x + y = 1
3x = 6, x = 2
2(2) + y = 1, y = -3 (2, -3)
Consistent, Independent Inconsistent, Independent Consistent, Dependent
One solution No solution, parallel lines Same lines, infinite solutions
Copyright Professor Shotsberger 1/2006
CONIC SECTIONS axis of symmetry
Parabola
(x – h)2 = 4p(y – k) focus: ● (h, k+p)
Vertex: (h, k)
Focus: (h, k + p)
Directrix: y = k – p vertex: (h,k) directrix: y = k – p
p>0 the parabola opens upward
p0 the parabola opens right axis of symmetry ●
p b > 0 (h-c,k) (h+c,k)
a2 b2 (h, k)
Vertices: (h - a, k), (h + a, k) ● ● ●
Foci: (h - c, k), (h + c, k)
Length of minor axis: 2b, c2 = a2 – b2 (h, k+a)
Major Axis Horizontal (h-a,k) (h+a,k)
● (h, k+c)
(x – h)2 + (y – k)2 = 1, a > b > 0
b2 a2
Vertices: (h, k – a), (h, k + a) (h, k) ●
Foci: (h, k – c), (h, k + c)
Length of minor axis: 2b, c2 = a2 – b2
Major axis Vertical ● (h, k-c)
Hyperbola (h, k-a)
(x – h)2 – (y – k)2 = 1 (h-a,k) (h+a,k)
a2 b2
Vertices: (h – a, k), (h + a, k) (h-c,k) ● ● ● (h+c,k)
Asymptotes: y – k = (b/a)(x – h) (h,k)
y – k = (-b/a)(x – h)
Foci: (h – c, k), (h + c, k)
c2 = a2 + b2 (h,k+c)
●
(y – k)2 - (x – h)2 = 1
a2 b2 (h,k+a)
Vertices: (h, k – a,), (h, k + a) (h,k-a) ● (h,k)
Asymptotes: y – k = (a/b)(x – h)
y – k = (-a/b)(x – h)
Foci: (h, k – c), (h, k + c) ● (h,k-c) Copyright Professor Shotsberger 1/2006
Arithmetic Sequence
A sequence is arithmetic if there exists a number d , 4, 9, 14, 19, 24 . . . d = 5 (19 + 5 = 24)
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 The 14th term in the above sequence:
an = a1 + (n – 1)d, for any integer n ≥ 1 a1 = 4, n = 14, d = 5
an = 4 + (14 – 1)5 = 4 + 65 = 69
Sum of the First n Terms
The sum of the first n terms of arithmetic sequence The sum of the first 14 terms in the above:
is given by: Sn = 14 (4 + 69) = 7(73) = 511
Sn = n(a1 + an) 2
2 Can be written as: 14
Σ (5n – 1)
Geometric Sequence k = 1
A sequence is geometric if there is a number r,
called the common ratio, such that: 3, 6, 12, 24, 48, . . . r = 24/12 = 2
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, The 14th term of the above sequence:
an = a1 r n-1 a1 = 3, n = 14, r = 2
For any integer n ≥ 1 an = 3 (214-1) = 3(213) = 24576
Sum of the First n Terms
The sum of the first n terms of a geometric sequence The sum of the first 14 terms in the above:
is give by: Sn = 3(1 – 214) = 3(-16383) = 49,149
Sn = a1 (1 – r n), for any r ≠ 1 1 – 2 -1
1 – r Can be written as: 14 Σ 3(2 n-1)
Limit or Sum of an Infinite Geometric Series k=1
When |r| < 1, the limit or sum of an infinite geometric -2, 1, - ½ , ¼ , -1/8, . . .
Series is given by: r = 1 ÷ -2 = - ½ since |r| < 1 |- ½ | < 1
S∞ = __a1__ this series has a limit or sum
1 – r S∞ = ___-2___ = _-2 = - 4
Pascal’s Triangle 1 – (- ½ ) 3/2 3
(a + b)0 1
(a + b)1 1 1
(a + b)2 1 2 1
(a + b)3 1 3 3 1
(a + b)4 1 4 6 4 1
(a + b)5 1 5 10 10 5 1
The Binomial Theorem Using Pascal’s Triangle Expand (2t + 3)4 (use: 1 4 6 4 1)
For any binomial a + b and any natural number n, = 1(2t)4(3)0 + 4(2t)3(3)1 + 6(2t)2(3)2 +
(a + b)n = c0anb0 + c1an-1b1 + c2an-2b2 + . . .+ cn-1a1bn-1 + cna0bn 4(2t)1(3)3 + 1(2t)0(3)4 =
Where the numbers c0, c1, c2, . . .,cn-1 are from the = 16t4 + 96t3 + 216t2 + 216t + 81
(n + 1)st row of Pascal’s Triangle
Copyright Professor Shotsberger 1/2006
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