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c1 notes
Fichas del conjunto (25)
- 1difference of squaresx^2 - y^2 = (x+y)(x-y)
- 2how is a number is written exactly?in the surd form
- 3square root of a prime number is ?a surd
- 4quadratic formulacomplete
- 5completing the squarecomplete
- 6sketching quadratic graphs- decide whether the parabola is a u shape (a>0) or umbrella shape (a<0)
- work out where curve crosses x and y axis by putting x=0 to find y and y=0 to find x
- consider the discriminant to find the roots - 7discriminant: b^2 - 4acdiscriminant > 0 : 2 roots
discriminant = 0 : 1 root
discriminant < 0 : no roots - 8inequalitymultiply or divide an inequality by a negative number, change sign
- 9to solve quadratic inequality- solve the corresponding quadratic equation
-sketch the graph of quadratic function
- use graph to find set of values
-if values overlap write it together
if not overlap write it separately - 10graph transformation: what effects what?inside () : effects x - moves left/right
outside () : effects y - moves up /down - 11graph transformation multiplication-f(ax), multiplication inside () affects x only - scale factor of 1/a so do x * 1/a
-af(x), multiplication outside () affects y only - vertical stretch of scale factor o f a so do y * a - 12graph transformation addition/ subtraction-x misbehaves:
f(x+a) horizontal translation of -a so do x-a
-y behaves:
f(x)+a vertical translation of +a so do y+a - 13coordinate geometry finding gradienty2 - y1 / x2 - x1
- 14coordinate geometry equation of line: when you have a coordinate and gradienty - y1 = m (x - x1)
- 15coordinate geometry equation of line: 2 sets of coordinatesy-y1/y2-y1 = x-x1/x2-x1
- 16nth term of arithmetic seriesa + (n-1) d
- 17sum of arithmetic seriesSn = n/2 [2a + (n-1)d]
Sn = n/2 (a + L) - 18gradient of a curve at a specific point is defined as being the same as the gradient of ........?the tangent to the curve at that point
- 19f(x) differentiated is f ' (x) y differentiated is dY/dxintegrating f ' (x) gives you f(x)
- 20rate of a function at a particular pointfind f ' (x) then substitute the value of x
- 21equation of tangent1) differentiate
2) substitute the value of x to get gradient
3) find y coordinate by sub x into original equation
4) use y-y1 = m(x-x1) - 22equation of normal1) perpendicular so differentiate to find gradient
2) sub x value to find gradient value
3) find reciprocal gradient that adds to get -1
4) use y-y1 = m(x-x1) - 23integrationadd 1 to the power then divide value by the power then simplify
*NB add + C - 24finding the + C valuesubstitute x and y value from the coordinate at which the curve passes through to find C
- 25—
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- A level (C1) Mathematics Fichas sobre c1 notes , creado por Angel Jose el 03/05/2017.
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- Angel Jose
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