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Objectives: Be able to…..
• Distinguish between rational and irrational
numbers.
• Understand what is meant by a surd.
• Simplify expressions involving surds.
• Perform arithmetic with surds.
Objectives: Be able to…..
Use language of coordinate geometry.
Find the distance between two given points.
Find the mid points of a line segment joining two given points.
Find, use and interpret the gradient of a line segment.
Know the relationship between the gradients for parallel and
perpendicular lines.
Rearrange equation to show line in for Ax + By + C = 0 and vice versa.
Find the equations of straight lines given
– (a) the gradient and a point
– (b) two points
Verify, given their coordinates, that points lie on a line.
Find the coordinates of a point of intersection of two lines.
Find the fourth vertex of a parallelogram given the other three.
Chapter 4: Quadratic functions and there graphs.
Learning objectives: To be able to…
Solve quadratic equations.
Complete the square of a quadratic expression.
Understand the effect of a translation on a quadratic
graph.
Find the equation of a parabola when translated by a
given vector.
Understand the term discriminant.
What do these have in common?
They all have similar curved flight through the air. The curve is called a
parabola and it is generated by a quadratic function.
A general function for this type of curve is given by:𝑓 𝑥 = 𝑎𝑥 2 + 𝑏𝑥 + 𝑐
Quadratic expressions
A quadratic expression is an expression in which the highest power of the variable is 2.
For example:
t2
x2 – 2
w2 + 3w + 1
4 – 5g2
2
The general form of a quadratic expression in x is:
ax2 + bx + c
(where a ≠ 0)
x is a variable.
a is the coefficient of x2.
b is the coefficient of x.
c is a constant term.
Factorising quadratic expressions
Factorising an expression is the inverse of expanding it.
Expanding or multiplying out
x2 + 3x + 2
(x + 1)(x + 2)
factorising
When we expand an expression we multiply out the brackets.
When we factorise an expression we write it with brackets.
Factorising Quadratics – The Process
Step 1:
Find factor pairs
of 3x2 that sum -7
3x2 - 7x + 2
3x2 -1x -6x + 2
Step 2: Rewrite
expression
with factor pairs
Step 3: Factorise
the two halves
x(3x -1) -2(3x-1)
(x-2) (3x-1)
Step 4: Put
Into brackets
6
-1&-6
Factorising Quadratics – The Process
Step 1:
Find factor pairs
of 4x-3 that sum +1
4x2 + x - 3
4x2 -3x +4x -3
Step 2: Rewrite
expression
with factor pairs
Step 3: Factorise
the two halves
x(4x -3) +1(4x-3)
(x+1) (4x-3)
Step 4: Put
Into brackets
-12
-3&4
Factorising Quadratics – The Process
-8x2 -14 x +15
-8x2 +6x -20x+15
Step 1:
Find factor pairs
of-8x15 that sum -14
Step 2: Rewrite
expression
with factor pairs
Step 3: Factorise
the two halves
-2x(4x -3) -5(4x-3)
(-2x-5) (4x-3)
Step 4: Put
Into brackets
-120
-6&20
Significant points on a
Quadratic Curve that crosses
the x axis. Sketch the graph
The ROOTS = the x intercepts.
(Where Y = 0)
Find from table and or graph.
The Y INTERCEPT
= C /constant. (Where X = 0)
Find from equation.
The VERTEX = the
maximum or minimum point
on the graph. Find by
substituting the X value, that
is the  way between the roots,
to find Y.
-2
-1
0
1
2
3
4
5
x2
4
1
0
1
4
9
16
25
x2 -3x
10
4
0
-2
-2
0
4
10
x2 -3x – 4
6
0
-4
-6
-6
-4
0
6
y
6
0
-4
-6
-6
-4
0
6
4
5
7
6
5
4
3
2
1
0
-3
-2
-1 -1 0
y values
y = x -3x – 4
2
x
1
2
-2
-3
-4
-5
-6
-7
x values
3
6
y = -x2 + 1
2
1
0
-4
-3
-2
-1
y values
The quadratic curve
-1 0
-2
-3
-4
-5
-6
-7
-8
-9
x values
1
2
3
4
Perfect squares
Some quadratic expressions can be written as perfect squares. For example:
x2 + 2x + 1 = (x + 1)2
x2 – 2x + 1 = (x – 1)2
x2 + 4x + 4 = (x + 2)2
x2 – 4x + 4 = (x – 2)2
x2 + 6x + 9 = (x + 3)2
x2 – 6x + 9 = (x – 3)2
Completing the square
Adding 16 to the expression x2 + 8x to make it into a perfect square is called completing the
square.
We can write
x2 + 8x = x2 + 8x + 16 – 16
If we add 16 we then have to subtract 16 so that both sides are still equal.
By writing x2 + 8x + 16 we have completed the square and so we can write this as
x2 + 8x = (x + 4)2 – 16
In general:
2
b b

x  bx   x     
2  2

2
2
Completing the square
Complete the square for x2 – 10x.
Compare this expression to (x – 5)2 = x2 – 10x + 25
x2 – 10x = x2 – 10x + 25 – 25
= (x – 5)2 – 25
Complete the square for x2 + 3x.
Compare this expression to
( x + 32 )2 = x 2 + 3 x + 94
x 2 + 3 x = x 2 + 3 x + 94  94
= ( x + 32 )2  94
Completing the square
How can we complete the square for
x2 – 8x + 7?
Look at the coefficient of x.
This is –8 so compare the expression to (x – 4)2 = x2 – 8x + 16.
x2 – 8x + 7 = x2 – 8x + 16 – 16 + 7
= (x – 4)2 – 9
In general:
2
2
b b

x  bx  c   x       c
2 2

2
Completing the square
When the coefficient of x2 is not 1, quadratic equations in the form ax2 + bx + c can be
rewritten in the form a(x + p)2 + q by completing the square.
Complete the square for 2x2 + 8x + 3.
Take out the coefficient of x2 as a factor from the terms in x:
2x2 + 8x + 3 = 2(x2 + 4x) + 3
By completing the square, x2 + 4x = (x + 2)2 – 4 so
2x2 + 8x + 3 = 2((x + 2)2 – 4) + 3
= 2(x + 2)2 – 8 + 3
= 2(x + 2)2 – 5
Solving quadratics by completing the square
Quadratic equations that cannot be solved by factorisation can be solved by completing the
square.
For example, the quadratic equation
x2 – 4x – 3 = 0
can be solved by completing the square as follows:
(x – 2)2 – 7 = 0
(x – 2)2 = 7
x–2=
x=2+
x = 4.65
7
or
 7
x=2–
7
x = –0.646 (to 3 s.f.)
Sketching graphs by completing the square
In general, when the quadratic function y = ax2 + bx + c is written in completed square form
as
a(x + p)2 + q
𝑏
𝑏
where: p= 2 𝑎𝑛𝑑 𝑞 = −(2)2 + c
The coordinates of the vertex will be (–p, q).
The axis of symmetry will have the equation x = –p.
Also:
If a > 0 (–p, q) will be the minimum point.
If a < 0 (–p, q) will be the maximum point.
Plotting the y-intercept, (0, c) will allow the curve to be sketched using symmetry.
Sketch the graph of y = x2 + 4x – 1 by writing it in completed square form.
x2 + 4x – 1 = (x + 2)2 – 5
The least value that (x + 2)2 can have is 0 because the square of a number cannot be negative.
(x + 2)2 ≥ 0
Therefore
(x + 2)2 – 5 ≥ – 5
The minimum value of the function y = x2 + 4x – 1 is therefore y = –5.
When y = –5, we have,
(x + 2)2 – 5 = –5
(x + 2)2 = 0
x = –2
The coordinates of the vertex are therefore (–2, –5).
The equation of the axis of symmetry is x = –2.
x = –2
y
Also, when x = 0 we have
y = x2 + 4x – 1
y = –1
0
(0, -1)
So the curve cuts the y-axis at the point (0, -1).
Using symmetry we can now sketch the graph.
Solving (x + 2)2 – 5 = 0 gives the roots as x = ±√5 – 2.
x
(–2, –5)
Translations of Parabola
(Movement of Curve)
Completed square =
(x ± 𝑎) ± 𝑏
Movement of graph
Vector
+↔ −
𝑎
±𝑏
+↔ −
±
Translations of Parabola
(Movement of Curve)
Completed square =
x±𝑎 2±𝑏
Movement of graph
Vector
+↔ −
𝑎
±𝑏
+↔ −
±
Quadratic Graphs
• y = x2
x
x2
y
-2
4
4
-1
1
1
0
0
0
1
1
1
2
4
4
• y = x2+1
y
y = x2 +1
5
4
x
-2
-1
0
1
2
3
x2
4
1
0
1
4
2
x2+1
5
2
1
2
5
1
y
5
2
1
2
5
• y=x2-1
x
-2
-1
0
1
2
x2
4
1
0
1
4
2-1
3the curves?
0
-1
oticexabout
y
3
0
-1
0
3
0
3
-2 -1 0
-1
y = x2
y = x2 - 1
1
2
x
What do you notice about where the
lines cross the y-axis?
Quadratic Graphs
• y = x2
x
x2
y
-2
4
4
-1
1
1
0
0
0
1
1
1
2
4
4
• y = x2
8
7y
6
5
4
x
-2
-1
0
1
2
3
x2
4
1
0
1
4
2
x2
2

0

2
1
y
2

0

2
• y=2x2
y = 2x2
-2 -1 0
-1
y = x2
y = x2
1
2
x
x
-2
-1
0
1
2
x2
4
1
0
1
4
2x2
8
2
0
2
8
What do you notice curves?
y
8
2
0
2
8
What do you notice about where the
lines cross the y-axis?
Sketching a curve - process
• Intercept
• Roots (if they exist)
• as 𝒙 → ∞, what happens to 𝒚
Objective
• Practise and consolidate graph sketching.
• Be able to translate cubic graphs.
• Sketch curves of functions showing points of
intersection.
Key points in sketching quadratic graphs
ax + bx + c
• Shape a Positive Pothole / Minus Mountain
• Y Intercept = c
• X Intercept (Roots) from Discriminant / Solving
– b2 – 4ac > 0 two roots (cuts x axis) factorise/formula/square
– b2 – 4ac = 0 one repeated root (vertex touches x axis)
– b2 – 4ac < 0 no real roots (vertex above/below x axis)
• Vertex (turning point → positive/min or negative/max)
– Symmetry / half way between roots for x then substitute for y.
– Completing the square
The quadratic formula
• ax2 + bx + c
•
•
•
•
•
•
•
𝑏
𝑐
+ x+
𝑎
𝑎
𝑏
2
x + x
𝑎
𝑏 2
𝑏 2
(x + ) – ( )
2𝑎
2𝑎
𝑏 2 𝑏2
(x + ) - 2
2𝑎
4𝑎
𝑏 2
(x + )
2𝑎
𝑏 2
(x + )
2𝑎
𝑏
x +
2𝑎
x2
• x
=0
=0
𝑐
=𝑎
𝑐
= 𝑎
𝑐
=𝑎
2
𝑏
𝑐
=
2
4𝑎
𝑎
2 −4𝑎𝑐
𝑏
=
4𝑎2
2 −4𝑎𝑐)
±
(𝑏
=
4𝑎2
2 −4𝑎𝑐)
−𝑏
±
(𝑏
=
2𝑎
Learning Objectives: Be able to…
• Understand the vocabulary of polynomials
• Add subtract and multiply polynomials.
• Find, compare and equate coefficients of
polynomials.
Vocabulary, definitions and conventions.
An expression comprising constants, variables and powers / indices.
The powers can only be positive integers : x2 − 4x + 7 is a polynomial, but x2 − 4/x + 7x is not.
The degree or order of the polynomial is given by the highest power of the variable, so;
CONSTANT (Polynomial of degree 0)
LINEAR (Polynomial in m of order 1)
QUADRATIC (Polynomial in x of order 2)
CUBIC (Polynomial in g of order 3)
QUARTIC (Polynomial in f of order 4)
=5
= 2m + 3
= 3x2 + 4x – 4
= 5g3 – 8g – 3
= f4 + 3f + 7
The normal convention is that the polynomial is written in descending powers of the variable.
Polynomials can be added or subtracted by collecting like terms.
They can be multiplied using a variety of methods.
Coefficient means the constant preceding a variable.
Adding and subtracting polynomials.
Simply follow normal rules of column addition /
subtraction and simplify by collecting like
terms.
Make sure that polynomials are lined up in
correct order in columns.
Be careful when handling negative numbers.
+
X4
4
+ 3x + 7
2
3x + 4x – 4
2
X +3x + 7x + 3
-
X4
4
+ 3x + 7
2
3x + 4x – 4
2
X -3x - x + 11
Multiplying polynomials – using a grid.
Multiply x² + 3x – 2 by 2x² – x + 4
x²
2x²
-x
4
3x
-2
Multiplying polynomials – using a grid.
Multiply x² by
2x²
x²
2x²
-x
4
3x
-2
Multiplying polynomials – using a grid.
Multiply x² by
2x²
x²
2x²
-x
4
2x4
3x
-2
Multiplying polynomials – using a grid.
Multiply 3x by
2x²
x²
2x²
-x
4
2x4
3x
-2
Multiplying polynomials – using a grid.
Multiply 3x by
2x²
2x²
-x
4
x²
3x
2x4
6x³
-2
Multiplying polynomials – using a grid.
Fill in the rest of the table in the same
way
2x²
-x
4
x²
3x
2x4
6x³
-2
Multiplying polynomials – using a grid.
Fill in the rest of the table in the same
way
2x²
-x
4
x²
3x
-2
2x4
6x³
-4x²
Multiplying polynomials – using a grid.
Fill in the rest of the table in the same
way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
4
Multiplying polynomials – using a grid.
Fill in the rest of the table in the same
way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
4
Multiplying polynomials – using a grid.
Fill in the rest of the table in the same
way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
Multiplying polynomials – using a grid.
Fill in the rest of the table in the same
way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
Multiplying polynomials – using a grid.
Fill in the rest of the table in the same
way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
Multiplying polynomials – using a grid.
Fill in the rest of the table in the same
way
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
Multiplying polynomials – using a grid.
First the term in x4
Now add up all the terms in the
table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
(x² + 3x – 2)(2x² - x + 4) = 2x4
Multiplying polynomials – using a grid.
Now add up all the terms in the
table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
(x² + 3x – 2)(2x² - x + 4) = 2x4
+ 5x³
then the terms in x³
Multiplying polynomials – using a grid.
Now add up all the terms in the
table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
(x² + 3x – 2)(2x² - x + 4) = 2x4
+ 5x³
then the terms in x²
- 3x²
Multiplying polynomials – using a grid.
Now add up all the terms in the
table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
(x² + 3x – 2)(2x² - x + 4) = 2x4
+ 5x³
then the terms in x
- 3x²
+ 14x
Multiplying polynomials – using a grid.
Now add up all the terms in the
table
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
(x² + 3x – 2)(2x² - x + 4) = 2x4
+ 5x³
and finally the
constant term
- 3x²
+ 14x
-8
Multiplying polynomials – using a grid.
x²
3x
-2
2x²
2x4
6x³
-4x²
-x
-x³
-3x²
2x
4
4x²
12x
-8
(x² + 3x – 2)(2x² - x + 4) = 2x4 + 5x³ - 3x² + 14x - 8
Using a grid to find a specific coefficient in a term.
Eg: Find the coefficient of x2 in (x² + 2x – 2)(6x² -3x + 4)
x²
2x
6x²
-12x²
-3x
4
-2
-6x²
4x²
- 14x²
Equating coefficients: used to find either a specific
coefficient in a term or a unknown polynomial.
(x + 6)P(x) = x3 + 12x2 + 34x – 12
(x + 6)(ax2 + bx + c)
= x3 + 12x2 + 34x – 12
ax3 + (6a +b)x2 + (6b + c)x – 6c = x3 + 12x2 + 34x – 12
→ (equate coefficients of x3)
a=1
→ (equate coefficients of x2)
6a + b = 12 because a = 1 → b = 6
→ (equate coefficients of x)
6b + c = 34 because b = 6 → c = -2
→ (equate to constant)
6c
= -12
→ c = -2
The Factor Theorem
This is the Factor Theorem:
If (x – a) is a factor of a polynomial f(x) then f(a) = 0.
The converse is also true:
If f(a) = 0 then (x – a) is a factor of a polynomial f(x).
The Factor Theorem
Use the factor theorem to show that
(x + 2) is a factor of f(x) = 3x2 + 5x – 2.
Hence or otherwise, factorize f(x).
(x + 2) is a factor of 3x2 + 5x – 2 if f(–2) = 0
f(–2) = 3(–2)2 + 5(–2) – 2
= 12 – 10 – 2
=0
We can write
By inspection
And so
3x2 + 5x – 2 = (x + 2)(ax + b)
a = 3 and b = –1
3x2 + 5x – 2 = (x + 2)(3x – 1)
as required.
Factorizing polynomials
The factor theorem can be used to factorize polynomials by systematically looking for values
of x that will make the polynomial equal to 0. For example:
Factorize the cubic polynomial x3 – 3x2 – 6x + 8.
Let f(x) = x3 – 3x2 – 6x + 8.
f(x) has a constant term of 8. So possible factors of f(x) are:
(x ± 1),
(x ± 2),
(x ± 4)
or
(x ± 8)
f(1) = 1 – 3 – 6 + 8 = 0
 (x – 1) is a factor of f(x).
f(–1) = –1 – 3 + 6 + 8 ≠ 0
 (x + 1) is not a factor of f(x).
f(2) = 8 – 12 – 12 + 8 ≠ 0
 (x – 2) is not a factor of f(x).
f(–2) = – 8 – 12 + 12 + 8 = 0
 (x + 2) is a factor of f(x).
f(4) = 64 – 48 – 24 + 8 = 0
 (x – 4) is a factor of f(x).
We have found three factors and so we can stop.
The given polynomial can therefore be fully factorized as:
x3 – 3x2 – 6x + 8 = (x – 1)(x + 2)(x – 4)
The Factor Theorem
Factorize f(x) = x3 + 1
f(x) has a constant term of 1 so the only possible factors of f(x) are (x – 1) or (x + 1).
 (x – 1) is not a factor of f(x).
 (x + 1) is a factor of f(x).
f(1) = 1 + 1 ≠ 0
f(–1) = (–1)3 + 1 = 0
We don’t know any other factors but we do know that the expression x + 1 must be
multiplied by a quadratic expression to give x3 + 1. We can therefore write
x3 + 1 = (x + 1)(ax2 + bx + c)
We can see immediately that a = 1 and c = 1 so
x3 + 1 = (x + 1)(x2 + bx + 1)
= x3 + bx2 + x + x2 + bx + 1
= x3 + (b + 1)x2 + (b + 1)x + 1
Equating coefficients of x2 gives
b+1=0
b = –1
So x3 + 1 can be fully factorized as
x3 + 1 = (x + 1)(x2 – x + 1)
Dividing polynomials
Suppose we want to divide one polynomial f(x) by another polynomial of lower order g(x).
There are two possibilities. Either:
g(x) will leave a remainder when divided into f(x).
g(x) will divide exactly into f(x). In this case, g(x) is a factor
of f(x) and the remainder is 0.
We can use either of two methods to divide one polynomial by another. These are by:
using long division, or
writing an identity and equating coefficients.
Quotient
Divisor Dividend
+ Remainder
)
Dividend ÷ Divisor = Quotient + Remainder
Dividend = Divisor x Quotient + Remainder
For polynomials
Cubic
= Linear x Quadratic + Remainder
Remainder = 0 then linear is factor of cubic.
(factor theorem)
Remainder ≠0 then linear is not factor and has
remainder. (remainder theorem)
Dividing
polynomials
by
long
division
Using long division
The method of long division used for numbers can be applied to the division of polynomial
functions.
Let’s start by looking at the method for numbers.
For example, we can divide 5482 by 15 as follows:
3 6 5
15
5 4 8 2
– 4 5
9 8
– 9 0
8 2
– 7 5
7
This tells us that 15 divides into 5482 365 times,
leaving a remainder of 7.
We can write
5482 ÷ 15 = 365 remainder 7
or
5482 = 15 × 365 + 7
The
The
The
The
=
+
×
dividend
divisor
quotient
remainder
Dividing polynomials by long division
We can use the same method to divide polynomials.
For example:
What is f(x) = x3 – x2 – 7x + 3 divided by x – 3?
x2
x–3
+ 2x
This tells us that x3 – x2 – 7x + 3 divided by x – 3 is
x2 + 2x – 1.
–1
x3 – x2 – 7x + 3
x3 –
The remainder is 0 and so x – 3
is a factor of f(x).
3x2
2x2 – 7x
2x2 –
We can write
6x
x3  x 2  7 x + 3
= x2 + 2x  1
x 3
–x +3
–x+3
or
0
x3 – x2 – 7x + 3 =(x – 3)(x2 + 2x – 1)
Dividing polynomials by long division
Here is another example:
What is f(x) = 2x3 – 3x2 + 1 divided by x – 2?
2x2
x–2
+ x
+2
This tells us that 2x3 – 3x2 + 1 divided by x – 2 is 2x2
+ x + 2 remainder 5.
2x3 – 3x2 + 0x + 1
There is a remainder and so x – 2
is not a factor of f(x).
2x3 – 4x2
x2 + 0x
We can write
x2 – 2x
2x + 1
2x – 4
5
2 x3  3 x 2  1
5
2
 2x  x  2 
x2
x2
or
2x3 – 3x2 + 1 =(x – 2)(2x2 + x + 2) + 5
Objective: Sketching a cubic curve - steps
• Slope:
• Intercept:
• Roots:
+ve
-ve
where does it cross y axis (x = 0)
where it crosses x axis (y = 0)
• Factor Theorem
• Does quadratic factorise
• Quadratic Formula: Discriminant / Repeat roots
• Quadrant:
Sub for x close to y check slope.
y = (2x + 3)(x + 4)(5 – 2x)
• Slope:
• Intercept:
-ve
y = 60
3
,
2
5
2
• Roots:
x=-
• Quadrant:
sub x = 1, 75 > 60
-4 or
y = (x + 1)(x2 –x + 1)
•
•
•
•
Slope:
Intercept:
Roots:
Quadrant:
+ve
y=1
x = -1
na
y = (2x + 3)(x - 1)2
• Slope:
• Intercept:
+ve
y=3
3
,
2
• Roots:
x=-
• Quadrant:
sub x = -1, 4 > 3
1 repeated root