College Algebra Definitions and Procedures

College Algebra Definitions and Procedures

Cartesian Coordinate System

x-intercept is a point (a, 0)Quadrant IIQuadrant I

y-intercept is a point (0, b)

x is the independent variable, the abscissa

y is the dependent variable, ordinateQuadrant IIIQuadrant 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

Function domain range domain range

A correspondence between a first set, the domain,92 9

and the second set, the range, such that each member 34 3 2

of the domain corresponds to exactly one member of the range.41 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 = slopeb = 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 Functionsh(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 respectg(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| <1

For c < 0, the graph is also reflected across the y-axis

The complex numbers:

a + bi

7, -8.7, ½ , ¼ i, π, 5 + 2i

Imaginary numbers:Real numbers:

a + bi, b ≠ 0 a + bi, b = 0

½ - i, ¼ i, 5 + 2i 7, -8.7, ½, √3, - 18

Imaginary numbers: Pure imaginary numbers:Irrational numbers:Rational numbers: a + bi, a ≠ 0, b ≠ 0 a + bi, a = 0, b ≠ 0 √2, π, 3√5 7, -8.7, ½, -2

½ - i, 5 + 2i¼ i, 8 i

____

The Number i 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.x2 – 6x – 10 = 0

2. Find the new constant. (b/2)2 = (-6/2)2 = (-3)2 = 9x2 – 6x = 10

3. Add the new constant to each side.x2 – 6x + 9 = 10 + 9

4. Factor the left side and combine the right side.(x – 3)2 = 19

The Quadratic Formula ______

The solutions of ax2 + bx + c = 0, a ≠ 0 are: x =-b + √ b2 – 4ac

Discriminant

For ax2 + bx + c = 0

b2 – 4ac = 0One real-number solution

b2 – 4ac >0Two 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 = 0Find the intercepts: (-1,0) (4,0)

(x – 4)(x + 1) = 0

x – 4 = 0 x + 1 = 0

x = 4 x = -1Find the discriminant:

Find the solution: g(-1) = 0(-3)2 – 4(1)(-4) = 9 + 16 = 25

g(4) = 0Since 25> 0 it has 2 real-number solutions.

Quadratic Functionsf(x) = x2 + 10x + 23

f(x) = ax2 + bx + c or f(x) = a(x – h)2 + kf(x) = x2 + 10x + 25 – 25 + 23

Vertex (h, k)f(x) = (x + 5)2 - 2

Axis of symmetry x = hVertex: (-5, -2); Axis of symmetry: x = -5

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

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 > 0If 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 behavior2x3 has odd degree of 3 and 2 > 0

2. Find the zeros of the function by solving f(x) = 0Factor: 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

Synthetic Division

Divide f(x) = x3 + 2x2 – 5x – 6 by x – 3_3 |12-5-6

______31530

151024

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

The same as the number of variations of sign in P(x)P(-x) = 2(-x)5 – 5(-x)2 – 3(-x) + 6 =
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

The same as the number of variations of sign in P(-x)
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 denominatorHorizontal 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.

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)

Replace f(x) with y.y = 2x – 3
Interchange x and y.x = 2y – 3
Solve for y.x + 3 = 2y (x + 3)/2 = y
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 = logax as that number y such that x = ay, 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 xln 3 = log e 3

ln 1 = 0 and ln e = 1, for the logarithmic base e.

The Change-of-Base Formula

For any logarithmic bases a and b, and any positive number M,

log b M = log a Mlog 5 8 = log 108 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 Nlog 5 (2 • 4) = log 5 2 + log 5 4

The Power Rule: log a M p = p log a Mlog 5 23 = 3 log 5 2

The Quotient Rule: log aM = log a M – log a Nlog 516 = log 5 16 – log 5 2

N 2

The Logarithm of a Base to a Power

log a ax = x ax = axlog 5 58 = 8

Base-Exponent Property

ax = ay ↔ x = y2 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 x10log 7t = 10log10 7t = 7t

Property of Logarithmic Equality

log a M = log a N ↔ M = Nlog3 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, IndependentInconsistent, IndependentConsistent, Dependent

One solutionNo solution, parallel linesSame lines, infinite solutions

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

p<0 the parabola opens downward

directrix: x = h – p

(x – k)2 = 4p(x - h)

Vertex:(h, k)

Focus:(h + p, k)

Directrix:x = h – p vertex: (h, k) focus: (h + p, k)

p>0 the parabola opens right axis of symmetry ●

p<0 the parabola opens left

● (x, y)

Circle r

(x – h)2 + (y – k)2 = r2 ● (h,k) center

Center: (h, k)

Radius: r

Ellipse

(x – h)2 + (y – k)2 = 1, a > 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)

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 byThe 14thterm in the above sequence:

an = a1 + (n – 1)d,for any integer n ≥ 1a1 = 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 sequenceThe sum of the first 14 terms in the above:

is given by:Sn = 14 (4 + 69) = 7(73) = 511

Sn = n(a1 + an) 2

2Can 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

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 rn-1a1 = 3, n = 14, r = 2

For any integer n ≥ 1an = 3 (214-1) = 3(213) = 24576

Sum of the First n Terms

The sum of the first n terms of a geometric sequenceThe 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 – rCan 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 – rS∞ = ___-2___ = _-2 = - 4

Pascal’s Triangle 1 – (- ½ ) 3/2 3

(a + b)01

(a + b)111

(a + b)2121

(a + b)31331

(a + b)414641

(a + b)515101051

The Binomial Theorem Using Pascal’s TriangleExpand (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-1are from the = 16t4 + 96t3 + 216t2 + 216t + 81

(n + 1)st rowof Pascal’s Triangle