MST124 BOOK A UNIT 2

y = mx + c
- equation of a straight line
- m is the gradient and c is the y-intercept

a line that slopes up from left to right has a positive gradient
a line that slopes down from left to right has a negative gradient 

The gradient of a straight line
- the gradient of the straight line through the points (x₁,y₁) and (x₂,y₂), where x₁ does not equal y₁, is given by
gradient = (y₂-y₁)/(x₂-x₁)
(change in rise over the change in run)

the gradient of a horizontal line is zero (rise is 0)
the gradient of a vertical line is undefined (run is 0, not possible to divide by 0)

- to find the x-intercept, change the equation of a straight line to equal 0 
(0 = mx + c)

the horizontal line with y-intercept c has the equation y = c
the vertical line with x-intercept d has the equation x = d

the equation of the straight line with gradient m that passes through the point (x₁,y₁) is
y-y₁ = m(x-x₁) 

Gradients of perpendicular lines
the gradients of any two perpendicular lines (not parallel to the axes) have product -1

Strategy
To solve simultaneous equations: substitution method
- rearrange one of the equations, if necessary, to obtain a formula for one unknown in terms of the other
- use this formula to substitute for this unknown in the other equation
- solve the equation in one unknown to find out its value
- substitute this value into an equation involving both unknowns to find the values of the other unknown
- confirm that the two values satisfy the original equations by subbing in the values to both

Strategy
To solve simultaneous equations: elimination method 
- multiply one or both of the equations by suitable numbers, if necessary to obtain two equations that can be added/subtracted to eliminate one of the unknowns
- add or subtract the equations to eliminate this unknown
- solve the equation in one unknown to find its value
- substitute this value into an equation involving both unknowns to find the value of the second unknown
- confirm that the two values satisfy the original equations by subbing the unknowns into both equations 

- if two lines have different gradients, then they intersect at a single point and there is one solution
- if the two lines have the same gradient but different y-intercepts, they are parallel and do not intersect (there are 0 solutions)
- if the two lines have the same gradient and the same y-intercept, the equations are identical and there are an infinite number of solutions 

graph of the equation y = ax² + bx + c has a parabola shape
- is a is positive, then the parabola is u-shaped
- is a is negative, then the parabola is n-shaped 
- in both cases, the parabola has a vertical axis of symmetry

- the lowest point on a u-shaped parabola or the highest point on an n-shaped parabola is called the vertex of the parabola

- a u or n shaped parabola can have one, two or 0 x-intercepts 
- there is always exactly one y-intercept because there is exactly one value of y for each value of x, including x=0 

Strategy
To factorise a quadratic of the form x² + bx + c
- start by writing: x² + bx + c = (x  )(x  )
- find the factor pairs of c (including both positive and negative factors)
- choose a factor pair with sum b
- write your factor pair p, q in position: x² + bx + c = (x+p)(x+q)

Strategy
To factorise a quadratic of the form ax² + bc + c
- take out any numerical common factors
- if the coefficient of x² is negative, take out the factor -1
- find the positive factor pairs of a
- for each factor pair d,e write down a framework (dx  )(ex  )
- find all the factor pairs of c, the constant term (both +ve and -ve)
- for each framework and each factor pair of c, write the factor pair in the gaps in the framework in both possible ways 
- for each of the resulting cases, calculate the term in x that you obtain when you multiply out the brackets
- identify the case where this term is bx, which is the required facotrisation 

- quadratic with no constant term can be factorised by taking x out as a common factor
- e.g., x² + 4x = x(x+4)

- a difference of two squares is any expression of the form A² - B² 
- A² - B² = (A+B)(A-B)

Simplifying a quadratic equation
- if necessary, rearrange the equation so that all the non-zero terms are on the same side 
- if the coefficient of x² is -ve, then multiply the equation through by -1 to make it +ve
- if the coefficients have a common factor, then divide the equation through by this factor
- if any of the coefficients are fractions, then multiply the equation through by a suitable number to clear them 

- if the product of two or more numbers is 0, then at least one of the numbers must be 0

Strategy
To complete the square in a quadratic of the form x² + bx (+ c)
- write down (x  )², filling the gap with the number that's half of b, including its + or - sign
- subtract the square of the number that you wrote in the gap
- if there is a c value, collect the constant terms 

Strategy
To complete the square in a quadratic of the form ax² + bx + c, where a does not equal 1
- rewrite the quadratic with the coefficient a taken out of the expression as a factor
- this generates a pair of brackets
- complete the square of the quadratic inside the brackets, which generates a second pair of brackets inside the first
- multiply out the outer brackets
- collect the constant terms  

Quadratic Formula
The solutions of the quadratic equation ax² + bx + c = 0 are
x = (-b ± √(b² - 4ac))/2a

The number of real solutions of a quadratic equation
- the quadratic equation ax² + bx + c = 0 has:
- two real solutions if b² - 4ac > 0 (the discriminant is positive)
- one real solution is b² - 4ac = 0 (the discriminant is 0)
- no real solutions if b² - 4ac < 0 (the discriminant is negative) 

Properties of the graph of y = ax² + bx + c, where a doesn't equal 0
- the graph is a parabola with a vertical axis of symmetry 
- if a is positive it is u-shaped; if a is negative it is n-shaped 
- it has two, one or no x-intercepts 
- it has one y-intercept 

Strategy
To sketch the graph of y = ax² + bx + c, where a doesn't equal 0
- find whether the parabola is u-shaped or n-shaped
- find its intercept, axis of symmetry and vertex
- plot the features found, and sketch the parabola
- label the parabola with its equation, intercepts and the coordinates of the vertex 

- to find any x-intercepts, put y = 0
- the axis of symmetry lies halfway between the x-intercepts
- the vertex lies on the axis of symmetry 
- if there is just one x-intercept, then the vertex of the parabola is at this x-intercept 

- when there are no x-intercepts, a method of finding the equation of the axis of symmetry is to find any two points on the parabola that have the same y-coordinate, and the axis of symmetry lies halfway between them 
(example 20 on page 180 of book A)