Section 1

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g(x) to -g(x)

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Last updated

7 years ago

Date created

Mar 1, 2020

Cards (94)

Section 1

(50 cards)

g(x) to -g(x)

Front

flip the graph over x-axis

Back

i^0

Front

1

Back

Y=f(X+k)

Front

Shifted left k units

Back

cosine 45 (square root above)

Front

2/2

Back

cosine 90

Front

0/2

Back

tangent of any angle is to corresponding to

Front

ratio of Y to X

Back

i^4 or 8 or 12

Front

1

Back

cosine 0 (square root above)

Front

4/2

Back

cosine 30 (square root above)

Front

3/2

Back

sin 45 (square root above)

Front

2/2

Back

odd+even=

Front

odd

Back

(a+b)(a-b)

Front

a^2-b^2

Back

(a-b)^2

Front

a^2-2ab+b^2

Back

i

Front

Square root -1

Back

tan 0

Front

0

Back

solution for (X-7)

Front

7

Back

inversely proportional

Front

f=K/X

Back

tan 30

Front

1/square root 3

Back

sin 0 (square root above)

Front

0/2

Back

sin 60 (square root above)

Front

3/2

Back

i^1

Front

i

Back

i^2

Front

-1

Back

(a+b)^2

Front

a^2+2ab+b^2

Back

two graph have no solution

Front

they are parallel

Back

Pythagorean theory

Front

1= sin^2 X + cos^2 X

Back

cosine 60

Front

1/2

Back

my improvement

Front

to read problems carefully

Back

tan 60

Front

square root 3

Back

sine of any coordinate is corresponding to

Front

Y axis

Back

3√X

Front

X^1/3

Back

sin 90 (square root above)

Front

4/2

Back

(^3√X)^4

Front

X^4/3

Back

even+even=

Front

even

Back

conversion factor

Front

180/pi

Back

tan 40

Front

1

Back

odd+odd=

Front

even

Back

cosine

Front

adjacent/hypothesis

Back

changing ratio

Front

changed - original / changed - original

Back

√X

Front

X^1/2

Back

i^3

Front

-i

Back

Cos ( ) = sin X

Front

pi/2 -X

Back

Sin

Front

opposite/hypothesis

Back

circle graph definition

Front

(x-h)^2 + (h-k)^2 = r^2

Back

cofficient

Front

number next to X

Back

sin 30 (square root above)

Front

1/2

Back

cosine of any angle is corresponding to

Front

X axis

Back

sin 30

Front

1/2

Back

y=a(x-h)^2+k

Front

(h,k)

Back

tan

Front

opposite/adjacent

Back

Sin ( ) = cosine X

Front

pi/2 - x

Back

Section 2

(44 cards)

"Approximately 19 percent"

Front

Cannont be beyond 19 percent

Back

"Which of the relationship represents relationship between h and c?

Front

Not necessary x, so be careful

Back

sin(π-x)

Front

sin(x)

Back

cos(2π-x)

Front

cos(x)

Back

X^2-2ax+a^2

Front

(X-a)(X-a)

Back

When looking at graph first thing you need to do

Front

See digit on X-axis

Back

Y=f(X-k)

Front

Shifted right k units

Back

cos(2π+x)

Front

cos(x)

Back

Initial

Front

Original

Back

sin(2π-x)

Front

-sin(x)

Back

tan(2π+x)

Front

tan(x)

Back

5, 4,

Front

3

Back

cos(π+x)

Front

-cos(x)

Back

important step to solve the problem

Front

highlight every terms in the sentence

Back

f(X)= ax^2 +bx+c Vertical axis of symmetry at

Front

X=-b/2a

Back

To find y intercept f(X)= ax^2 +bx+c

Front

Plug 0 to x

Back

cos π

Front

-1

Back

sin π

Front

0

Back

At least one solution?

Front

They are on graph

Back

To find y-interception?

Front

Plug in 0 to x

Back

tan(π+x)

Front

tan(x)

Back

when seeing triangle, judge whether it has

Front

a right angle

Back

Slope theorem

Front

Y^2-Y^1/x^2-x^1

Back

Y=f(X)+k

Front

Shifted up k

Back

X^2 - a^2

Front

(X+a)(X-a)

Back

sin 2π

Front

0

Back

cis 2π

Front

1

Back

X+2ax+a^2

Front

(X+a)(X+a)

Back

sin (π+x)

Front

-sin(x)

Back

sin π/2

Front

1

Back

12, 13,

Front

5

Back

When dividing i ?

Front

Use conjugatem

Back

To find X-interception?

Front

Foil and make it 0

Back

tan(2π-x)

Front

-tan(x)

Back

17, 15

Front

8

Back

(X-1)/(X+1) + (X+1)/(X+1)

Front

2x/X+1

Back

How to solve quadratic formula quickly

Front

-b+_square root b^2 -4ac / 2a

Back

sin(2π+x)

Front

sin(x)

Back

cos(π-x)

Front

-cos(x)

Back

(SinX+CosX)^2

Front

Sin^2X+2(sinx)(CosX)+cos^2x

Back

Sin^2X+Cos^2x

Front

(SinX+CosX)^2

Back

cos π/2

Front

0

Back

tan(π-x)

Front

-tan(x)

Back

cos(a+b)

Front

cos a cos b - sin a sin b

Back