# Core 1 - All you need to know

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• Created by: lava-mite
• Created on: 17-10-12 19:53

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ALL YOU NEED TO KNOW
CORE 1
1. Coordinate geometry
Always draw a diagram of the given coordinates; you will be able to
see if the gradient of a line is positive or negative so check your
calculations.
Midpoint is found by finding average of 2 x-values and 2 y-values.
To find the gradient = difference of y
difference of x
To find differences work from left to right point.
Parallel lines have same gradient.
Perpendicular gradient = -1/ gradient
Equation of line y = mx + c
Use simultaneous equations to find a point where 2 lines meet.
Lengths found using Pythagoras a2+b2 , may need to simplify surds.
Could find area of triangle if perpendicular lines, half base x height.
2. Surds
To simplify surds find the largest square number 150 25 6 5 6
Multiply by the bottom with opposite sign.
34
5 3
3 4 5 3
= the bottom is difference of squares
5 3 5 3
5 3 3 20 4 3 8 3 23
=
52 ( 3 ) 2 22
3. Completing the square
This will be in the form (x-a)2+b or a(x-b)2+c or b-(x-a)2
x 2 6x 5 (x 3) 2 32 5 (x 3) 2 14 Least value y - 14 when x -3
2x 16x 3 2[x 8x] 3 2[(x 4) 4 ] 3
2 2 2 2
2(x - 4) 2 32 3 2(x 4) 2 29 Least value y - 29 when x 4
8 10x - x 2 [x 2 10x] 8 [(x 5) 2 5 2 ] 8
- (x - 5) 2 25 8 - (x - 5) 2 33 Greatest value y 33 when x 5
Sketch the graph of the quadratic; make sure you mark the vertex (the 2
values from above) and where it cuts y axis (value from original
function).
If given 1/f(x) use y value from above.

## Other pages in this set

### Page 2

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Intersection of curve and line
This can involve 3 different topics, or they may be separate questions.
Simultaneous equations ­ use substitution by putting line into curve,
rearrange the equation to get = 0.
If a quadratic equation is made then solve it and find 2 values of x and y,
write these as coordinates.
If the quadratic equation involves a letter to find, c, use number of roots.…read more

### Page 3

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First principles
There are 5 lines of working needed, you will drop marks if you don't set
working out correctly. If y = 2x2 -5x+3
dy lim [2(x x) 2 5(x x) 3] [2x 2 5x 3]
x 0
dx x
lim 2x 4xx 2x 5x 5x 3 2x 5x 3
2 2 2
x 0
x
lim 4xx 2x 5x
2
x 0
x
x 0 4x 2x - 5
lim
4x - 5
7.…read more

### Page 4

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Stationary points
Find dy/dx and put it = 0.
This will give a quadratic equation, factorise and solve finding 2
values of x.
Substitute both x in to original equation to find y, write as
coordinates.
d2y
Find , substitute in both values of x.
dx 2
d2y d2y
If >0 is a minimum, and 2 <0 is a maximum
dx 2 dx
Remember what the shape of a positive cubic graph should
look like. The maximum is to the left.…read more

### Page 5

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Binomials
The formula are at the top of first page of formula booklet.…read more

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