Post on 15-Jan-2016
1Higher Maths 2 1 2 Quadratic Functions
Any function containing an term is called a Quadratic
Function.
The Graph of a Quadratic Function
2Higher Maths 2 1 2 Quadratic Functions
The graph of a Quadratic Function is
a type of symmetrical curve called a
parabola.
x
2
f ( x ) = ax 2 + bx + cGeneral Equation
of a Quadratic Function
a > 0
a < 0
Minimum turning
point Maximum turning
point
with a ≠ 0
turning point
Sketching Quadratic Functions
3Higher Maths 2 1 2 Quadratic Functions
Before it is possible to sketch the graph of a
quadratic function, the following information must be
identified:• the nature of the turning point
i.e.
• the coordinates of the y-
intercept• the zeroes or ‘roots’ of the function
i.e. the x-intercept(s), if any
minimum or maximum
• the location of the axis of
symmetry
and coordinates of the turning
point
a > 0 or a < 0
substitute x
= 0solve f ( x ) =
0 :ax 2 + bx + c = 0
evaluate f ( x ) at
axis of symmetry
The General Quadratic Equation
Note(Back)
Maths Intermediate 2 (Units 1, 2 and 3)
Friday, April 21, 2023
Any quadratic equation can be written in
the basic form
Examples
a The vertical size
of the parabola.
positive negative
Writing the equation in this way makes it easy to
sketch the parabola using , and .ba c
b The -coordinate of the vertex.x
( x – b )²y
x
( x + b )²y
x
positive negative
cThe -coordinate
of the vertex.
y
y
x( 4 , 3 )
y
x
( -3 , 7 )
y = ( x – 4 )² + 3
y = - ( x + 3 )² + 7Note that the equation of the axis
of symmetry is simply x = b .
y = ( x – )² +ba c
Maths Intermediate 2 (Units 1, 2 and 3)
Friday, April 21, 2023
If we know the equation of a parabola, it is also
possible to calculate the -intercept.y
Example
The General Quadratic Equation
Note(Back)
Calculating the -intercepty
Find the y-intercept of the
graph with equation
y
xy = 2( x – 3 )² – 1
as shown opposite.
At the y-intercept, x = 0 y = 2( 0 – 3 )² – 1
= 2 × ( -3 )² – 1
= 2 × 9 – 1
= 17
17
a The vertical size
of the parabola.
positive negative
b The -coordinate of the vertex.x
( x – b )²y
x
( x + b )²y
x
positive negative
cThe -coordinate
of the vertex.
y
Perfect Squares
6Higher Maths 2 1 2 Quadratic Functions
x 2 + 6 x +
2x
2 + 6 x
+ 2( )2
+ 9 – 9( ) x+ 3 – 7
A perfect square is an
expression that can be written
in the formExample
Complete the square
for
x 2 + 6 x +
2
= =
( ... )2
x 2 + 4 x +
4( )2x+ 2= x
2 + 5 x + 9= ( )2x+ ?
Step One:Separate number
term
Step Two:Try to form a perfect
square from remaining terms
Step Three:Remember to balance the extra number, then write
out result
Completing the Square
7Higher Maths 2 1 2 Quadratic Functions
y = x 2 – 8 x +
19
= x 2 – 8 x
+ 19 = ( )2
+ 16 – 16( )
x – 4 + 3
It is impossible for a
square number to be
negative.
The minimum possible
value of is
zero.
( x – 4 )2
The minimum possible value of
y is 3 .x = 4 .This happens
when
The minimum
turning point is at( 4 , 3 )
A turning point can often
be found by completing the
square.Example
y-coordinatex-coordinate
Find the minimum turning
point of
Quadratic Equations can be
solved in several different
ways:
• using a graph to identify
roots
• factorising
• completing the square
• using the quadratic formula
8Higher Maths 2 1 2 Quadratic Functions
Solving Quadratic Equations
Example
f ( x ) = 0Use the
graph to
solvex = -2 -2 5x =
5or
Example 2
Solve6 x 2 + x – 15 =
0
6 x 2 + x – 15 =
0(2 x – 3)(3 x + 5)
= 0
The trinomial can
be factorised...
or
2 x – 3 = 0
3 x + 5 = 0
2 x = 3
3 x = -5x =3
2 x = -53
If a Quadratic Equation cannot be factorised, it is
sometimes possible to solve by completing the
square.
9Higher Maths 2 1 2 Quadratic Functions
Solving Quadratic Equations by Completing the Square
Example
Solvex 2 – 4 x – 1 =
0The trinomial cannot be
factorised so complete the
square...x
2 – 4 x – 1 = 0
– 1 = 0( )
( )2
+ 4 – 4x 2 – 4 x
– 5 = 0x – 2
( )2 = 5x – 2
x – 2 = 5±
x = 2 5±
Now solve for
x ...
x ≈ 4.24 -0.24orNot
accurate!
most accurate answer
Quadratic Inequations
10
Higher Maths 2 1 2 Quadratic Functions
A Quadratic Inequation can be solved by using a
sketch to identify where the function is positive
or negative.Exampl
eFind the values of x for
which12 – 5 x – 2 x 2
> 0Factorise 12 – 5 x – 2 x
2
= 0 (4 + x)(3 – 2 x) = 0
The graph has
roots
x = -4 and
and a y-intercept at
12
-432
12Now sketch the graph:
The graph is
positive for-4 < x <and negative
forx < -4
32
x >and
32
32
The Quadratic Formula
11
Higher Maths 2 1 2 Quadratic Functions
x = -b b 2 – (4
ac )
±
2 a
with a ≠ 0
f ( x ) = ax 2 + bx + c
If a quadratic function has roots, it is
possible to find them using a formula.
This is very useful if the roots cannot be found
algebraically, i.e. by factorising or completing the
square.The roots
of
are given
by
root
If roots cannot be found
using the quadratic
formula, they are
impossible to find.
no roots
Real and Imaginary Numbers
12
Higher Maths 2 1 2 Quadratic Functions
36= ±6
-36 =
It is impossible to
find the square
root of a negative
number.
In Mathematics, the square
root
of a negative number still
exists
and is called an imaginary
number.
It is possible for a quadratic
equation to have roots which
are not real.
1 real
root
no real
roots
2 real
roots
• If
• If
The Discriminant
13
Higher Maths 2 1 2 Quadratic Functions
x = -b b 2 – (4
ac )
±
2 a= b
2 – (4 ac )
The part of the quadratic formula inside the square root is
known
as the Discriminant and can be used to find the nature of
the roots.
The Discriminant
• If b 2 – (4 ac ) > 0 there are two real
roots.b 2 – (4 ac ) = 0 there is only one real
root.b
2 – (4 ac ) < 0 the roots cannot be calculated
and are imaginary or non-real.
(‘real and
unequal’)(‘real and
equal’)
Using the Discriminant to Find Unknown Coefficients
14
Higher Maths 2 1 2 Quadratic Functions
Example
The quadratic equation2 x 2 + 4 x + p
= 0Find all possible values of
p.
has real
roots.
b2 – (4 ac ) 0a = 2 b = 4 c = p
For real roots,
16 – 8 p 0
16 8 p
8 p 16p 2
The equation has
real roots for
.
p 2
(for the roots
are
imaginary or non-
real)
p >
2
Finding Unknown Coefficients using Quadratic Inequations
15
Higher Maths 2 1 2 Quadratic Functions
Exampl
eFind q given that has non-real
roots.
x 2 + (q – 3) x + q =
0
a = 1 b = (q – 3) c = q
b2 – (4 ac ) < 0
For non-real
roots,
(q – 3)
2 – 4 q < 0
q2 – 6q + 9 – 4 q < 0
q2 – 10q + 9 < 0
(q – 9) (q – 1) < 0
Sketch graph of the inequation:
1 9
q2 – 10q + 9 < 0for 1 < q < 9
The roots of the original equationare non-real when1 < < 9q
Straight Lines and Quadratic Functions
16
Higher Maths 2 1 2 Quadratic Functions
When finding points of intersection between a line and a
parabola:
• equate the functions
• rearrange into a quadratic equation equal to zero
• solve for
• substitute back to find
x
Exampl
eFind the points of
intersection ofy = x 2 + 3 x
+ 2
x 2 + 2 x =
0
y = x + 2
and
x 2 + 3 x + 2 = x
+ 2
x ( x + 2 ) = 0
x = 0 x = -2or
y = 2 y = 0or
y
Points of intersection are (-2,0)
and (0,2).
Tangents to Quadratic Functions
17
Higher Maths 2 1 2 Quadratic Functions
The discriminant can be used to find the number of
points of intersection between a parabola and a
straight line. • equate the functions and rearrange into an equation equal to zero
• evaluate the discriminant of the new quadratic equation
b 2 – (4 ac ) > 0 Two points of
intersection
b 2 – (4 ac ) = 0 One point of
intersection
b 2 – (4 ac ) < 0 No points of
intersection
the line is a tangent