Finals Flashcards
degrees → radian
degree • pi/180
radians → degree
radian • 180/pi
evaluate sin x, cos x, and tan x for a value x
- if necessary, convert x to degrees
- Check what quadrant it’s in
- Do whatever the quadrant says to do
- Use special/quadrantal angles tables to solve
special angles table
quadrantal angles table
y = sin x graph → 0 1 0 -1 0
y = cos x graph → 1 0 -1 0 1
tan x = sinx/cosx
state the number of revolutions for an angle
- convert to radians
- radians/2pi
convert to decimal degree form
Ex: 152o15’29”
152 + (15/60) + (20/3600) =
152.26o
decimal degree form
156.33o
DMS form
122o25’51”
convert to DMS form
Ex: 24.240
24o(.24*•60)’ (.4**•60)”
24o14’24”
*.24 is from 24.24
**.4 is from (.24•60 = 14.4)
1’ = ?
one minute = (1/60)(1o)
1” = ?
one second = (1/60)(1’) = (1/3600)(1o)
quadrant rules
terminal ray
the pipe cleaner
in which quadrant does the terminal side of each angle lie when it is in standard position?
- convert to degrees
- if negative, + 360; if over 360, - 360
- find which quadrant it’s in
find the exact value of sin/cos/tan x. no calculator.
- if radians, convert to degrees
- use quick charts
use a calc to approximate sin/cos/tan x to four decimal places
- if degree, change calc to Degree mode
- if radians, change calc to Radians mode
sketch w/out a calculator a sin/cos/tan curve
xmin = -2pi
xmax = 2pi
xscl = pi/2 (unless stated otherwise)
ymin = -5
ymax = 5
yscl = .5
sin, cos
- if y = c + cos(x) → period & amplitude same; c = pos moves max/min up, c = neg moves max/min down
- if y = a sin(x) → period same; move max to a
- if y = sin(bx) → max/min/amp same; normal period/b
- if y = sin(x + b) → if b = pos, move left; if b = neg, move right
**tan, cot, sec, csc - **no ampl/min/max
- y = c + tan(x) → if c = pos, move up; c = neg, move down (easier to just move x-int’s)
- y = a csc(x) → move min/max’s to a
- y = cot(bx) → period/b
period
- for sin, cos curves → the shortest distance along the x-axis over which the curve has one complete up-and-down cycle
- for tan, distance b/w consecutive x-intercepts
amplitude
max - min
vertical asymptotes
lines that the graph approaches but doesn’t cross
periodic
repeating
Ex: tan function
csc, sec, cot
csc = 1/sin
sec = 1/cos
cot = 1/tan
what happens to y = csc(x) whenever y = sin(x) touches the x-axis?
vertical asymptote
why are y = sin(x) and y = csc(x) tangent whenver x is a multiple of x
they r reciprocals, so csc’s max is at sin’s min, and csc’s min is at sin’s max
sine function: y = sin(x)
“wave”
amplitude - 1
period - 2pi
frequency = 1 cycle in 2pi radians (1/2pi)
max - 1
min - (-1)
one cycle occurs between 0 and 2pi with x-int @ pi
cosine function: y = cos(x)
also “wave”
amplitude = 1
period = 2pi
period = 2pi
frequency = 1 cycle in 2pi radians (1/2pi)
max = 1
min = -1
one cycle occurs b/w 0 & 2pi w/ x-int’s @ pi/2 & 3pi/2
tangent function: y = tan(x)
amplitude = none, go on forever in vertical directions
period = pi
one cycle occurs b/w -pi/2 and pi/2 (x-int = 0)
cotangent function: y = cot(x)
amplitude = none
period = pi
one cycle occurs b/w 0 and pi (x-int pi/2)
relationship b/w tan & cot graphs
the x-int’s of y = tan(x) are the asymptotes of y = cot(x)
and vice versa
cosecant function: y = csc(x)
amplitude = none
period = 2pi
one cycle is between 0 & 2pi, with the center being @ pi/2
relationship b/w sin & csc graphs
the maximum values of y = sin(x) are min values of the pos sections of y = csc(x)
the min values of y = sin(x) are the max values of the neg sections of y = csc(x)
the x-int’s of y = sin(x) are the asymptotes for y = csc(x)
secant function: y = sec(x)
amplitude = none
period = 2pi
one cycle occurs between -pi/2 and pi/2, with the center being at 0
relationship b/w cos & sec graphs
the max values of y = cos(x) are the min values of the pos sections of y = sec(x)
the min values of y = cos(x) are the max values of the neg sections of y = sec(x)
the x-int’s of y = cos(x) are the asymptotes for y = sec(x)
algorithm
a set of step-by-step directions for a process
simple sequence of steps that u follow in order
loop
a group of steps that r repeated for a certain # of time or until some condition is met
solving algorithms in two ways
- algebra - just solve
- graph - for inequalities
- graph both sides of equation separately
- estimate x-coord of intersection
- Ex: - 3x + 9 < 4
- all values of x for which y = -3x + 9 is BELOW y =4 (remember to use sign! and flip if necessary!!)
box and whisper plot
gives data in 4 parts - each part = 25% of the data
scatter plot
break-even point
when income = expenses
- separate expenses from income
- write equations to model the situation for I and E
- find break-even point, when I = E
- use a graph
- graph the equations on same set of axes
- x-coord of itnersection = BEP
- use algebra
- use a graph
matrices
x + y = 10,000
7x + 15y = 86,000
[A] [xy] = [B]
A = coefficient matrix, B = constant matrix
[17115] [xy] = [10,00086,000]
use calc 2nd matrix → edit
[A]-1 [B]
linear equation with 3 variables
ax + by + cz = d
assigning
- connect
- look for something that connects with only one other thing
- narrow down
diagramming
- put the thing w/ the most things in the center
- when 2 vertices r connected by 2+ edges, draw @ least 1 edge as a curved line
- draw arrows to show “direction”
matrixing
each row = departure
each column = destination
1 = on, 0 = off
vertex of a network
a dot in the network
edge of a network
line connecting 2 dots in a network
maximizing and minimizing
x-value of vertex = -b/2a
finding the shortest route
- Draw a network diagram that models the map. Each vertex represents a city. Each edge represents an interstate highway. Distances do not need to be drawn to scale.
- Label the starting point with the ordered pair (-,0)
- For each edge that connects a labeled and an unlabeled vertex, find this sum:
- s = y-value of ordered pair for labeled vertex + length of edge
- Choose the edge from Step 3 that has the minimum sum s. Label the unlabeled vertex of that edge with this:
- (label of the other vertex of the edge, s)
- Repeat Steps 3 & 4 until the vertex for the destination is labeled. (Go all the way back to find shortest distance).
- When the vertex for Danville is labeled, use the ordered pairs to find the shortest route. (backtrack)
constraint
any condition that must be met by a variable or by a linear combination of variables
x > 0 → The # of AM ads can’t be negative
y > 0 → the # of PM ads can’t be negative
x + y < 20 → the total # of ads must be less than/equal to 20
200x + 50y < 2200 → the total cost of the ads must be less than or equal to $2200
feasible region
the graph of the solution of a system of inequalities that meets all given constraints
graphing feasible regions
use system of inequalities from “constraint” definition
- Since x > 0 and y > 0, the feasible region is in the first quadrant.
- Graph x + y < 20 in the first quadrant.
- Identify points inthe blue shaded region that also make 200x + 50y < 2200 true.
- The feasible region consists of all pts on or inside quadrilateral ABCO. u can find the coord’s of each vertex by solving a system of equations from the intersecting lines
- the origin (0,0) is the solution of the system: x = 0, y = 0
- solve this system: x + y = 20, x = 0 to get (0,20)
- Solve x + y = 20, 200x + 50y = 2200 to get (8,12)
- solve 200x + 50y = 2200, y = 0 to get (11,0)
linear programming
can be used
- when u can represent the constraints on the variables with a system of linear inequalities
- when the goal is to find the max/min value of a linear combo of the variables
corner-point principle
any max/min value of a linear combo of the variables will occur at one of the vertices of the feasible region
using the corner-point principle
AM ads heard by 90,000; PM ads heard by 30,000
@ most 20 ads, @ least as many AM ads as PM ads, at least 720,000 listeners
hows many of each ad should u run to minimize the total cost? how much will the ads cost?
- Represent the constraints with a system of linear inequalities
- Let x = the number of AM ads, y = # of PM ads
- x + y < 20
- x > y
- 90,000x + 30,000y > 720,000
- x > 0
- y > 0
- Graph and find vertices
- Write a linear combo that represents the total cost of the ads: 200x + 50y
- Use the corner-pt principle. Find the total cost for the combo of ads represented by each vertex.
- min turns out to be A(6,6)
even function
a function if its graph = symmetric w/ respect to y-axis
f(-x) = f(x)
odd function
a function if its graph is symmetric with respect to the origin
turn upside down to test - 180o
f(-x) = -f(x)
holes
when a value of x sets both the denom & the numer of a rational function equal to 0, there is a hole in the graph
a single pt in which the function has no value
to find:
f(x) = (x2[x-2])/(x-2)
look for repeating thingies like x -2
therefore, x = 2
finding asymptotes, holes, 0’s, x-int’s, & y’int’s
f(x) = x2+x-6 / x+3
- vertical asymptote
- set denom to 0
- x + 3 = 0 → x = -3
- holes
- factor out the numerator to (x+3)(x-2)
- x+3 is found on both the num & denom
- therefore, x = -3
- zeros (where y = 0)
- set f(x) to 0 and solve for x
- x-int’s → same as zeros
- y-int (where x=0)
- set all x’s to 0 and solve
control variable
a variable that determines, or controls, another variable
x
domain
dependent variable
a variable that is determined by, or depends on, another variable
y
range
function
a relationship for which each value of the control variable is paired with only one value of the dependent variable
domain
all possible values of the control variable
x-axis
range
all possible values of the dependent variable
y-axis
values of a function
the numbers in its range
vertical line test
if no vertical line crosses a graph in >2 points, the graph represents a function
one-to-one function
a function in which each member of the range = paired with exactly one member of the domain
horizontal line test
if no horizontal line crosses the graph, it’s
- a one-to-one function
- has an inverse
many-to-one function
a member of the range may be paired with more than 1 member of the domain
linear function
a function that has an equation of the form f(x) = mx + b, where m is the slope of its linear graph & b is the y-intercept
domain = all real #’s
range = all real #’s
f(x) = x + b → pos moves left, neg moves right
f(x) = mx → 0 < m < 1 less steep, m > 1 more steep
f(x) = -x → reflection
slope
change in f(x)
____________
change in x
long division
if no 3a, would still have to put 0a!!!
synthetic division
ignore green
inside numbers from coefficients
once have an x2 in answer, just factor normally to get x
degree
greatest power
horizontal asymptotes
- if degree of num > degree of denom → NO hor. asympt.
- n < d → hor. asympt. at y = 0 (x-axis)
- if n = d, there is hor. asympt. @ y = an/bm where an = leading coeffcient of num & bm = leading coefficient of denom.
piecewise function
a function defined by 2+ equations
each equation applies to diff part of the function’s domain
Ex:
1/2lb or less → $13
more than 1/2 lb but less than 1 lb → $20
1lb + → $25
w = weight, c(w) = charge
c(w) = {13 if 0 < w < 1/2
{20 if 1/2 < w < 1
{25 if 1 < w
absolute value functions
|x| = x if x > 0; -x if x < 0
f(x) = |x| + b → pos moves up, neg moves down
f(x) = |x - 3| → neg moves right, pos moves left
f(x) = m|x| → 0 < m < 1 less steep, m > 1 more steep
f(x) = -|x| → reflection
graph: y = |x + 2|
y = x + 2 if x + 2 > 0 → x > -2
-(x + 2) if x + 2 < 0 → x < -2
Graph y = x + 2 and y = -(x + 2)
quadratic functions
f(x) = ax2 + bx + c
where a, b, and c are constants and a doesn’t equal 0
parabola
sign of a determines whether parabola opens upward or downward
f(x) = x2 + b → pos. up, neg. down
f(x) = (x - c)2 → c neg move right, pos. move left
f(x) = ax2 → same as others
f(x) = -x2 → same as others
height and distance formulas
t = time, v0 = initial speed, A = angle at which obj is thrown, h0 = initial height
distance fallen d(t) = 16t2
height after being thrown h(t) = -16t + (v0 sin A)t + h0
distance problems
a person drops a penny from a height of 50ft. express height of penny above ground as a function of time
let h(t) = height after t secs
(distance fallen) + (height above ground) = (initial height)
16t2 + h(t) = 50ft
h(t) = -16t2 + 50
domain = 0 < t <
range = 0 < h(t) < 50
height problems
lacrosse: ball leaves player’s stick from initial height of 7ft @ speed of 9ft/s & @ angle of 30o w/ respect to horizon
1. express height of ball as function of time
h(t) = -16t2 + (V0 sin A)t + h0
- 16t2 + (90 sin 30)t + 7
- 16t2 + 45t + 7 <- (quadratic)
1. When is the ball 25ft above ground?
h(t) = 25 → 25 = -16t2 + 45t + 7
0 = -16t2 + 45t - 18
- graph & find intersecting pts [h(t) = 25]
- quadratic formula
t = .5 sec and 2.3 sec
direct variation
has the form f(x) = kxn
where n is a pos integer & k doesn’t equal 0
the exponent n is also **degree **of the function
direct variation problems
solve for k to write a function
ball bearing varies directly w/ cube of radius, r = .4cm and w = 2.1g
write direct var function that describes weight in terms of radius
r = radius W(r) = weight
W(r) = kr3
2.1 = k(.4)3
k = 32.8
W(r) = 32.8r3
polynomial function
sum of 1+ direct variation functions
Ex: P(x) = x4 + 16x3 + 5x2 - 13x + 6
zeros of a function
values of x that make f(x) = 0
radical problems
sqrt(x + 7) - 1 = x
sqrt(x + 7) = x + 1
(sqrt[x+y])2 = (x+1)2
x + 7 = (x+1)(x+1)
x + 7 = x2 + 2x + 1
0 = x2+ x - 6
x = -3, x = 2
**check for extraneous **
x = 2
radical functions
functions that have a variable under a radical symbol
f(x) = √x + b → pos up, neg down
f(x) = √(x-c) → neg right, pos left
f(x) = a√x → 0 < a < 1 steeper, a > 1 less steep
f(x) = -√x → reflection
Properties of Radicals
n√b = x when xn = b
n√(bn) = {b when n is odd} { |b| when n is even}
n√(ab) = n√a • n√b {when n is odd, for all values of a & b; when n is even, for pos. values of a & b}
rational function
a function defined by a rational expression
rational expression
a quotient of 2 polynomials
graphing: find asymptotes