- Created by: bronwen :)
- Created on: 08-02-12 17:43
chapter 1 - Alegbra and functions
(a^m) x (a^n) = (a^(m+n))
(a^m) / (a^n) = (a^(m-n))
(a^(1/m) = msqr(a)
(a^(n/m)) = msqur(a^n)
(a^m)^n) = a^(mn)
x^2-y^2 = (x+y)(x-y) !This is the difference of two squares!
chapter 2 - Quadratic functions
!completing the square!
x^2 +bx= (x+(b/2))^2 - (b/2)^2
!the quadratic formula!
((-b) (+or-) squr (b^2-4ac))/ (2a)
equal roots when b^2 = 4ac
real roots when B^2> 4ac
no real roots when b^2<4ac
chapter 3 -equations and inequalitites
You can solve linear silmultaneous equations by elimination or substitution
you can use the substitution method to solve simultaneous equations , where one equation is linear and the other is quadratic. You usually start by finding an expression for x or y from the linear equation.
when you multiply or divide an inequality by a negetiv number, you need to change the inequality sign to its opposite
to solve a quadratic inequality you :
- solve the corressponding quadratic equation
- sketch the graph of the quadratic function
- use your sketch to fiind the required set of values.
- look back in pride -- you just did matsh ! *round of appaluse*
chapter 4 - sketching graphs
cubic curves =
- (ax^3) + (bx^2) + (cx) +d
- hit x axis three times.
example = sketch the curve with the equation y=(x-2)(x-1)(x+1)
- 1. when y = 0 x = 2,1,-1
- 1. so crosses x axis at (2,0) (1,0) (-1,0)
- 2. when x= 0 y= -2x-1x1= 2
- 2. so curve crosses y axis at (0,2)
- 3.when x is large a positive y is large and positive
- 3. so when x is positive majority of time y is positive propotionatly
- 4. draw that graph
with y = x^3 you have to do a table to work out values.
y= k/x is a reciprocal function so has asymptotes.
CHAPTER 5 - coordinate geometry in the (x,y) plane
equations of lines
- y= mx+c
y-y1=m(x-x1) ---> finding equation of line in the format y=mx+c
you can find the equation of the line that passes through the point with coordinates (x1,y1) and (x2,y2) by using
- (y-y1)/(y2-y1) = (x-x1)/(x2-x1)
if a line has a gradient of a line perpendicular to it hs a gradient of -(1/m)
chapter 6 - sequences and series
the nth term of an arithmetic series is (a+(n-1)d)
! prooving the sum of the ariothmetic sequence!
sn= a+ (a+d) + (a+2d) +... +(a+(n-2)d) + ( a+(n-1)d)
sn= ( a+(n-1)d)+(a+(n-2)d) +...+ (a+2d) +(a+d) +a
2Sn = (2a +(n-1)d) +(2a +(n-1)d)....+ (2a +(n-1)d)
so 2Sn= n(2a +(n-1)d)
Sn= (n/2)(2a +(n-1)d)
chaoter 7 = differention
the gradiant of a curve at a specific point is defined as being the same as the gradient of the tangent of the curve at that point.
if f(x) = n f'(x) = nx^(n-1)
the equation of the tangent to a curve at a point (a,f(a)) is y-f(a) =F'(a)(x-a)
chapter 4 - n2
f(x+a) is a horizontal translation of -a
f(x)+a is a vertical translation of +a
*every thing is backwards in x land*
f(ax) i a horizontal stretch pf scale factor 1/a so you mutiply the x coords by 1/a )y is unchanged)
af(x) is a vertical stretch of a scale factor a, so you multiply the y coords by a (x left alone)
chapter 6 - n2
!The formula for the sum of an arithmetic series !
Sn= (n/2)(2a +(n-1)d)
or Sn = (n/2)(a+L)
where a is the first term, d is the common difference, n is the number of terms and L is the last term in the series.