maths revision, c1 as maths

• a^m x aⁿ = ^am+n
• a^m/aⁿ = a^m-n
• a^-m = 1/a^m
• a⁰ = 1
• a^m/n = ⁿ√(a)^m
• (a^m)ⁿ = a^m*n

• √(a) * √(a) = a
• √(a) * √(b) = √(ab)
• √(a/b) = √(a) /√(b)
• -√(a)²  = a
• -√(a)³ = -√(a) * -√(a) * -√(a) = -a√(a)

• Discriminant tells you how many solutions an equation has
• Discriminant → the b² – 4ac part of a quadratic equation
• Roots → another name of solutions of an equation
• b² – 4ac > 0 → 2 solutions/roots
• b² – 4ac = 0 → 1 solution/roots/repeated roots
• b² – 4ac < 0 → no solutions/roots

EG (2x – 4) (x+3) > 0

• Step 1: solve quadratic-

x = (4/2) = 2  or  x = -3

• Step 2: sketch graph 
• Step 3: check regions for positive and negative

Therefore:

x < -3  and  x > 2

because within this region y in below 0

• (difference in y)/(difference in x) → (dy/dx)
• y = mx + c
• 2 lines are parallel if they have the same gradient
• 2 lines are perpendicular if their gradients multiply to give -1

·         y = f(x)

f(x) + c

Moves up or down by c

f(x + c)

Moves left or right by –c

af(x)

Stretched in y direction by scale factor of a

f(ax)

Stretches in x direction by scale factor of (1/a)

-f(x)

Reflects in x axis

f(-x)

Reflects in y axis

• Used to find gradients in a curve (eg for rate of change)
•  (dy/dx) is the equation for differentiation
• If y = axⁿ; then (dy/dx) = nax^n-1
• Multiply by the power; take 1 away from the power

• The reverse process of differentiation
•  Add 1 to the power
• Divide by the new power
• y = axⁿ; ∫ axⁿ dx = (axⁿ⁺¹)/(n+1)

Position

0

1

2

3

Term

-2

2

6

10

1st difference

+4

+4

+4

To work out the formula for this sequence first find out what the 0^th term is.

U_n = 4n – 2

The nth term for an arithmetic progression is : a + (n − 1)d; where a is the first term, n is the number of terms and d is the common difference.

 S_n = a + (a + d) + (a + 2d) + . . . + (L − 2d) + (L − d) + L

S_n is the sum of the arithmetic series, a is the first term, d is the common difference and L is the last term.

S_n = a + (a + d) + (a + 2d) + . . . + (L − 2d) + (L − d) + L

S_n = L + (L − d) + (L − 2d) + . . . + (a + 2d) + (a + d) + a

Copy out the formula but reverse it. Then add the two equations together:

2Sn = (a + L) + (a + L) + (a +L) + . . . + (a + L) + (a +L) + (a + L)

2S_n = n (a+L)

S_n = (n/2) (a+L)

L = a + (n − 1)d

S_n = (n/2) (2a+(n − 1)d)

_n

∑^Un

^r

• n is the end point of the sequence
• r is the starting point of the sequence
• Un is the n^th term
• ∑ is the symbol for sum of.