Maths-C1

The Basic laws of Indices:

a^m x aⁿ = a^m+n

a^m ÷ aⁿ = a^m-n

(a^m)ⁿ = a^mn

a⁰ = 1

a^-m = ¹/_am

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

_\/_a = a^1/2

The Basic Laws of Surds:

_\/_a x _\/_a = a

a_\/_c x b_\/_d = ab_\/_cd

_\/_a + _\/_a = 2_\/_a

Expanding Brackets with Surd:

The same as normal brackets but using the laws of surds.

Rationalising the Denominator:

Multiply the top and bottom by the bottom but reversing the '+' or '-' 

Completing the Square Formula:

(x+b)²+c

Using this to find the maximum point of an equation on a graph.

x² ­+ 4x – 5

(x+2)² – 4 – 5

(x+2)² - 9

This is a positive quadratic expression. If this was an equation where y = x² + 4x - 5 the maximum point would be (-2,-9).

The Discriminant Formula:

b² – 4ac

Using the Discriminant for the Amount of Roots:

If b² – 4ac > 0 then there are 2 distinct roots on the graph. Touches the x-axis twice.

If b² – 4ac = 0 then there are no real roots on the graph. Doesn't touch the x-axis.

If b² – 4ac < 0 then there is repeated or equal roots. Touches the x-axis once.

If asked to find the set of values that satify BOTH equations, add the number line to the sketch of the graph. Where the line and shaded area intersect, that is where they are both satisfied.

Quadratic Graphs

When a>0 the line is positive.

When a<0 the line is negative.

To find the x-axis, set y=0

To find y-axis, set all x=0

Minimum/Maximum point will be found from completing the square.

Cubic Graphs

Positive cubic graphs (when a>0 ) will start in the 4th quadrant and leave in the 1st. Negative cubic graphs (when a<0 ) will start in the 2nd quadrant and leave in the 4th . 

To find x-axis, set y=0

To find y-axis, set all y=0 and mutiply the answers together.

Some have repeated roots e.g. y = (x-2)²(x-3) so the line will only touch the x-axis at 2 then go through 3.

Reciprocal Graphs

A reciprocal graph never touches the x or y-axis. The general formula is y=¹/_x.

As the x value gets bigger, the y value gets smaller and vice versa.

If x>0 then the reciprocal is in the 2nd and 3rd quadrant.

If x<0 then the reciprocal is in the 1st and 4th quadrant.

x Direction Transformations:

f(x+a) - moves to the left by 'a' (negative is right)

f(ax) - stretch in the x direction (divide x by a)

-f(x) - reflection in the x-axis

y Direction Transformations:

f(x)+a - moves up by 'a' (negative is down)

af(x) - stretch in the y direction (multiply y by a)

f(-x) - reflection in the y-axis

Finding the Equation of a Line

Use ‘y₂-y₁ / x₂-x₁’ to find m

Use 'y-y₁ = m(x-x₁)' to find the equation.

Perpendicular and Parellel Lines

2 lines that are parellel have the same gradient.

A line perpendicular to the first line will have a gradient -¹/_m (m being gradient from l₁)

Mid-point between 2 Points

Mid-Point = ^x1+^x2/_2, ^y1+^y2/₂

Distance Between 2 Points / Length of a Line

_\/(x₁-x₂)² + (y­­₁-y₂)²

Recurrence Relations

U_n+1 means the next term

 5  <- term we end with

      K  <-the formula for the nth term

k=1  <- term we start with

Arithmetric Sequences and Series

Finding a Term in a Sequence

a₂ - a₁ = a₃ - a₂ <- solving what the letter is in a sequence

u_n­ = a + (n-1)d <- soliving the nth term in a sequence

a = starting term / n = term we're finding / d = difference between terms

Example

Given the first 3 terms of an arithmetic sequence are: x-1 , 2x-3 and 4x-11, find the value of the 12^th term.

find out x: a₂-a₁=a₃-a₂ so 2x-3-(x-1)=4x-11-(2x-3) -> x-2=2x-8 -> x=6

a=5 (first term is x-1->6-1)

n=12 (finding the 12^th term)

d=4 (second term is 9[2x-3])

a₁₂ = 5+ (12-1)4 -> a₁₂ = 5 + 44 -> a₁₂ = 49

Finding the Sum of a Series

S_n = ⁿ/₂(2a + {n-1}d) 

Example

a = 5 / n = 25 / d = -3. Use this information to find S₂₅

S₂₅ = ²⁵/₂(2(5) + {25-1}(-3))

S₂₅ = ²⁵/₂(10-72)  -->  S₂₅ = 25(-31)  -->  S₂₅ = -775 

Differentiation

To differentiate, you need to times the power by the co-efficient then minus one from the power.

3x⁵ + x⁴ becomes 15x⁴ + 4x³

The Gradient of the Tangent at a Point on a Curve using Differentiation

When finding the gradient of the tangent to a curve, differentiate the equation (y=...) of the curve. <-(this is called gradient function ^dy/­_dx­)

Equation of a Tangent

Use ^dy/­_dx­ to get the gradient of the tangent then substitute the values into the equation of a straight line.

Example

Find the equation of the tangent of the curve f(x) = 2x⁴ - 1 at the point P(1,y)

We need to find the gradient and the y coordinate of the point P. Start with the y coordinate of point P:

f(x) = 2x⁴-1  ->  f(1) = 2(1)-1  ->  f(1) = 1

Point P is (1,1). Now differentiate the curve equation to get the gradient.

f(x) = 2x⁴-1 then f’(x) = 8x³. f’(1) = 8(1). Gradient = 8.

Now use this to find the equation.

y-y₁ = m(x-x₁)  ->  y-1 = 8(x-1)  ->  y = 8x-7.

Equation of a Normal

The normal is a straight line perpendicular to the tangent.

Example - Find the equation of the normal of the curve y=5­_\/x-2 at the point P(1,3)

We are given a point so need to find gradient. Differentiate to get a gradient: -> y=5­_\/x-2 -> y=5x^1/2-2 -> ^dy/_dx=⁵/₂x^-1/2. Substitute x value into ^dy/_dx to get a gradient of ⁵/₂ at point P. Perpendicular is the negative reciprocal so the gradient of the normal is -²/₅. Use the equation of a line to get the equation of the normal.

y-y₁ = m(x-x₁) where x₁=1 y₁=3 and m=-²/₅

y-3 = -²/₅(x-1)  ->  5y-15 = -2x+2  ->  2x + 5y – 17 = 0 

Integration

To integrate, you do the opposite thing in the opposite order. Add one to the power then divide the co-efficient by the power. After integrating you must put '+c' after as you dont know that value.

3x⁵ + x⁴ becomes ¹/­₂x⁶ + ¹/₅x⁵ + c

Working out 'C'

When given coordinates to a point, substitute the coordinates into the intergrated equation and solve.