Maths edexcel Core 1 AS notes

(a^m) x (a^n) = (a^(m+n))

(a^m) / (a^n) = (a^(m-n))

(a^-m)= 1/(a^m)

(a^(1/m) = msqr(a)

(a^(n/m)) = msqur(a^n)

(a^m)^n) = a^(mn)

a^0=1

x^2-y^2 = (x+y)(x-y)                      !This is the difference of two squares!  

!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

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*

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.

equations of lines

• y= mx+c
• ax+by+c=0

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)

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

so

2Sn = (2a +(n-1)d) +(2a +(n-1)d)....+ (2a +(n-1)d)

so 2Sn= n(2a +(n-1)d)

and

Sn= (n/2)(2a +(n-1)d)

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)

!functions!

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)

!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.