Dustin's Final Reflection

This year has been a tough year, but a good year. Although my grades weren't great, I learned alot. The major things we covered were trig and calc. I loved trig when it was just triangles, then this year I saw a different side of trig. It was difficult, but I understood it in the end. It will help me in Calculus and it will help me in college, so it was worth the hard work.

We also learned a lot of calculus towards the end of the year. That was the one thing that I just perfectly understood at the end of the year. This year was great and I loved my class. I could've done much better and I will remember to do that next year after seeing how my grades turned out.

Devin's Reflection

How to Find the Inverse of a Function:

• Replace f(x) with y
• Reverse the roles of x and y
• Solve for y in terms of x
• Replace y with f-1(x)
• check: should equal to x

Example:

f(x) = √x + 4

(x)^2 = (√y + 4)^2

x^2 = y + 4

y = x^2 - 4

f-1(x) = (x^2 - 4)

f(f-1(x)) = f(x^2 - 4) = √(x^2 - 4) + 4 = x

f-1(f(x)) = f-1(√x + 4) = (√x + 4)^2 - 4 = x + 4 - 4 = x

Logarithm Properties:

• logb MN = logb M + logb N
• logb M/N = logb M - logb N
• logb M^K = K logb M
• logb b^k = k
• b^logb^k = k

Changing Bases: (Done when you can't solve a log)

• Rewrite it as an exponential
• Take the log of both sides
• Move the variable to the front
• then solve

Example:

log5 10 = x

5^x = 10

log 5^x = log 10

x log 5 = 1

x = 1/log 5

Devin's Reflection

Completing the Square:

You can use completing the square to solve a quadratic equation when factoring doesn’t work. This method can only work when 1 is the coefficient of x².

For example:

x² + 6x - 2 = 0

x² + 6x = 2

x² + 6x + 9 = -2 + 9

(x + 3)² = 7

x + 3 = √7

x = -3 ± √7

(-3 + √7,0) (-3 -√7,0)

Rational Root therom:

Example: f(x)= 2x^3 + 3x^2 - 8 + 3

Step 1: find all possible roots..
p: factors of 3: 1, -1, 3, -3
q: factors of 2: 1, -1, 2, -2

*p is the leading constant term & q is the leading coefficient
possible roots are (p/q): 1, -1, 1/2, -1/2, 3, -3, 3/2, -3/2

Step 2: plug roots in calc & the zeros will be: 1, 1/2, -3

Step 3: synthetic division: (x - 1) (2x^2 + 5x + 3)

Step 4: slove further (factor): (x - 1) (2x^2 + 5x + 3)= (x - 1) (2x - 1) (x + 3)

Answer: x = 1, 1/2, -3

Domain & Range of functions:

Polynomials-domain of all polynomials is (−∞, ∞).

Fractions-you set the bottom to zero, solve for x, and then set up intervals

Square Roots-domain: set the inside = to zero, then set a # line, try values on either side of each #, and get ride of the negatives-range:graph

Absolute Value-domain: (- ∞ , + ∞)-range: [0 , + ∞)

Devin's Reflection

SOLVING TRIG EQUATIONS

for any line m = tan alpha
m = slope , alpha = angle of inclination
For a conic: tan 2 alpha = B/A-C
if A=C then pi/4
A = coefficient of x^2, B = coefficient of xy, C = coefficient of y^2

y=Asin(Bx-h)+C
amplitude is hight
b is period p=2π/b
h is horizontal shift
c is vertical shift

Identities:

1. check identities
2. algebra
3. identities

Reciprocal Relationships

• cscΘ=1/sinΘ
• secΘ=1/cosΘ
• cotΘ=1/tanΘ

Relationships with Negatives

• sin -Θ= -sinΘ and cos -Θ= -cosΘ
• csc -Θ= -cscΘ and sec -Θ= -secΘ
• tan -Θ= -tanΘ and cot -Θ= -cotΘ

Pythagorean Relationsihps

• sin²Θ+cos²Θ=1
• 1+tan²Θ=sec²Θ
• 1+cot²Θ=csc²Θ

Cofunction Relationships

• sinΘ=cos(90°-Θ) and cosΘ=sin(90°-Θ)
• tanΘ=cot(90°-Θ) and cotΘ=tan(90°-Θ)
• secΘ=csc(90°-Θ) and cscΘ=sec(90°-Θ)

Devin's Reflection

Reference Angles (must be between 0° and 90°)
1)find which quadrant angle is in
2)determine the sign in that quadrant (+ve or -ve)
3)subtract 180° until the angle is between 0° and 90° (0 and π/2)

1)find the reference angle using chart or calculator
2)find what quadrant you need to be in based on the sign of the value
3)use notes to move to that quadrant
To Move:
I to IV = make negative and add 360°
I to III = add 180°
I to II = make negative and add 180°
II to IV = add 180°

Unit Circle

sin theta = y/r

cos theta = x/r

tan theta = y/x

csc theta = r/x

sec theta = x/y

cot theta = x/y

r=sqrtx^2+y^2

90 pi/2
180 pi
270 3pi/2
360 2pi

sin pos in one and two and neg in three and four

cos pos in one and four and neg in two and three

tan pos in one and three and neg in two and four

cot pos in one and neg in two three and four

sec pos in one and neg in two three and four

csc pos in one and two and neg in three and four

AREA OF A NON-RIGHT TRIANGLE
A=1/2(leg)(leg)sin(angle between)

RIGHT TRIANGLES

• hypotenuse opposite angle
• a=1/2bh
• SOHCAHTOA

sin theta=opposite/hypotenuse
cos theta=adjacent/hypotenuse
tan theta=opposite/adjacent

LAW OF SINES
sinA/a sinB/b sinC/c

• used when you know pairs in a non-right triangle
• you are setting up proportions

Right Triangles

• A=1/2bh
• SOHCAHTOA
• sinΘ=opposite leg/hypotenuse
• cosΘ=adjacent leg/hypotenuse
• tanΘ=opposite leg/adjacent leg

Law of Cosines
(opposite leg)²=(adjacent leg)² + (other leg)² - 2(adjacent leg)(adjacent leg)cos°
EG: x, 5, 6 angle = 35°
x²=5²+6²-2(5)(6)cos35°
x=√(5²+6²-2(5)(6)cos35°)
x≈3.443

Area of Inscribed Shapes
A=nr²sinΘcosΘ

Devin's Reflection

Triangle trigonometry:
sine = opp/hyp
cos = adj/hyp
tan = opp/adj
csc = hyp/opp
sec=hyp/adj
cot=adj/opp

Moving between quadrants:

I to IV = make it negative and add 360º
I to III = add 180º
I to II = make it negative and add 180º

II to IV = move 180º

Law of sines (sinA/a) = (sinB/b) = (sinC/c)
Law of cosines (opp leg)² = (other adj leg)² -2(adj leg) (adj Leg) cos (angle b/w)

sinΘ = y/r
cosΘ = x/r

tanΘ = y/x
cotΘ = x/y
cscΘ = r/y

secΘ = r/x

r=√(x² + y²)

Graphing Trig functions:

y=Asin(Bx-h)+C

A = amplitude or height

B determines period(p)

p = (2π/B)

h = ((phase shift-horizontal shift)/(opposite))

C= vertical shift

Trigonometric Identites:
csc = 1/sin x
sec = 1/cos x
cot x = 1/tan x
sin (-x) = -sin x
cos (-x) = cos x
csc (-x) = -csc x
sec (-x) = sec x
tan (-x) = -tan x
cot (-x) = -cot x
sin^2 x+cos^2 x = 1
1+tan^2 x = sec^2 x
1+cot^2 x = csc^2 x
sin x = cos(90*-x)
tan x = cot (90*-x)
sec x = csc (90*-x)
cos x = sin (90*-x)
cot x = tan (90*-x)
csc x = sec (90*-x)
tan x = sinx/cosx
cot x = cosx/sinx

Devin's Reflection

Ch. 6 Conics

The standard equation of a circle is (x-h)^2+(y-k)^2 .....the center is (h,k)

To find the intersection of a line and a circle:

1. solve the linear eqn for y.
2. substitute in the circle eqn.
3. solve for x.
4. plug the x value in to get the y value.

If your x value is imaginary, then there is no point of intersection.

EX: find the center and radius.(x-3)^2+(y+7)^2=19

c:(h,k)

center: (3,-7)

radius: square root of 19

--Parabolas have no major axis and no asymptotes.

Axis of symmetry x=-b/2a

Finding the vertex
(-b/2a, f(-b/2a))

or

complete the square to get vertex form
y=(x+a)^2+b a&b are #'s

(-a,b) vertex

focus: 1/4p= coeff of x^2 then add p.

directrix is p units behind vertex. subtract p.

EX: 1/8y^2

v(0,0)

Focus: p=2(2,0)

directrix: x=-2