```THE GRAPH OF A QUADRATIC
FUNCTION
A quadratic function is a function of the
form
f(x) = ax 2 + bx + c
where a, b & c are real numbers and a  0
The domain of a quadratic function is all
real numbers.
FUNCTIONS
As we’ve already seen, f(x) = x2 graphs
into a PARABOLA.
This is the simplest quadratic function
we can think of. We will use this one as
a model by which to compare all other
VERTEX OF A PARABOLA
All parabolas have a VERTEX, the lowest
or highest point on the graph
(depending upon whether it opens up or
down.)
AXIS OF SYMMETRY
All parabolas have an AXIS OF
SYMMETRY, an imaginary line which
goes through the vertex and about
which the parabola is symmetric.
HOW PARABOLAS DIFFER
Some parabolas open up and some
open down.
Parabolas will all have a different vertex
and a different axis of symmetry.
Some parabolas will be wide and some
will be narrow.
Example
Y-intercept
Roots
or x
intercepts
Vertex
y = (x - 3)2 - 4
x
y
6
5
5
0
4
3
-3
2
1
0
-3
0
-4
5
FUNCTIONS
The general form of a quadratic function
is:
f(x) = ax2 + bx + c
The position, width, and orientation of a
particular parabola will depend upon
the values of a, b, and c.
FUNCTIONS
Compare f(x) = x2 to the following:
f(x) = 2x2 f(x) = .5x2
f(x) = -.5x2
If a > 0, then the parabola opens up
If a < 0, then the parabola opens down
FUNCTIONS
Now compare f(x) = x2 to the following:
f(x) = x 2 + 3
f(x) = x 2 - 2
Vertical shift up
Vertical shift down
FUNCTIONS
Now compare f(x) = x2 to the following:
f(x) = (x + 2)2
f(x) = (x – 3)2
Horizontal shift to
the left
Horizontal shift to
the right
FUNCTIONS
When the general form of a quadratic
function f(x) = ax2 + bx + c is changed to the
vertex form:
f(x) = a(x - h) 2 + k
We can tell by horizontal and vertical
shifting of the parabola where the vertex
will be.
The parabola will be shifted h units
horizontally and k units vertically.
FUNCTIONS
Thus, a quadratic function written in the form
f(x) = a(x - h) 2 + k
will have a vertex at the point (h,k).
The value of “a” will determine whether the
parabola opens up or down (positive or
negative) and whether the parabola is narrow
or wide.
FUNCTIONS
f(x) = a(x - h) 2 + k
Vertex (highest or lowest point): (h,k)
If a > 0, then the parabola opens up
If a < 0, then the parabola opens down
FUNCTIONS
Axis of Symmetry
The vertical line about which the graph
of a quadratic function is symmetric.
x=h
where h is the x-coordinate of the
vertex.
FUNCTIONS
So, if we want to examine the
characteristics of the graph of a
quadratic function, our job is to
transform the general form:
f(x) = ax2 + bx + c
into the vertex form:
f(x) = a(x – h)2 + k
FUNCTIONS
This will require to process of
completing the square which is a little
different than completing the square to
Factor x2 + 6x + 9
Perfect Square
Trinomial
(x + 3)(x + 3) or (x + 3)2
The factors are in the form
(x + a)2 or (x - a)2.
Note the relationship between the middle term and the last term.
The last term is one-half the middle term squared.
2
1
  6
2
=
3
=9
2 
Find the value of the last term that will make the following
perfect square trinomials.
49
x2 + 14x + _______
49
4
x2 + 7x + _______
9
x2 - 3x + _______
4
(x + 7)2
2
x  7 
 2 
2
x  3 
 2 
Changing from Standard Form to Vertex Form
Write y = x2 + 10x + 23 in the form y = a(x - h)2 + k. Sketch the graph.
y = ( x2 + 10x ) + 23
y = (x2 + 10x + ____
25 - ____)
25 + 23
y=
(x2
+ 10x + 25) - 25 + 23
y = (x + 5)2 - 2
1. Bracket the first two terms.
2. Add a value within the
brackets to make a perfect
square trinomial. Whatever
to keep the value of the
function the same.
3. Group the perfect square
trinomial.
(-5, -2)
4. Factor the trinomial and
simplify.
Changing from Standard Form to Vertex Form
Write y = 2x2 - 12x -11 in the form y = a(x - h)2 + k. Sketch the graph.
y = ( 2x2 - 12x ) - 11
y = 2(x2 - 6x) - 11
y = 2(x2 - 6x + ____
9
- ____)
9 - 11
y = 2(x2 - 6x + 9) - 18 - 112  9  18
y = 2(x - 3)2 - 29
Multiply, when you remove
this term from the brackets.
1. Bracket the first two terms.
2. Factor out the coefficient
of the x2- term.
3. Add a value within the
brackets to make a perfect
square trinomial. Whatever
to keep the value of the
function the same.
4. Group the perfect square
trinomial. When grouping
the trinomial, remember to
distribute the coefficient.
(3, -29)
5. Factor the trinomial and
simplify.
Completing the Square
y = -3x2 + 5x - 1
y = (-3x2 + 5x ) - 1
5
y=
- x) - 1
3
25
25
5
2
36
36
y = -3(x - x + ______
- ______
)
3
25
5
75
2
y = -3(x - x + 36 )  36 - 1
3
5 2 25
y = -3(x - ) +
-1
6
12
-3(x2
5 2
25 12

y = -3(x - ) +
6
12 12
5 2
13
y = -3(x - ) +
6
12
-1
5 13
Vertex is
,
6 12
Completing the Square - The General Case
Using the general form, y = ax2 + bx + c,
complete the square:
y = ax2 + bx + c
y = (ax2 + bx ) + c
b
y  a(x  x)  c
a
b
b2
b2
2
y  a(x  x  2  2 )  c
a
4a
4a
b 2 4ac  b 2
y  a( x  ) 
2a
4a
2
2
2
b
b
b
y  a(x 2  x  2 )  a( 2 )  c
a
4a
4a
b 2
b2
y  a( x  )  c 
2a
4a
The vertex is
b 4ac  b2
( ,
).
2a
4a
This IS the vertex BUT it is
easier just to remember that
the x-value is  b and then
2a
plug that in to the equation to
get the y-value for the vertex.
Using the Vertex Formula
Find the vertex and the maximum or minimum value of
f(x) = -4x2 - 12x + 5
b
b 
using the axis of symmetry, the vertex is 
, f ( )
2a 
 2a
Find the x-value of the vertex: Find the y-value of the vertex:
b
x
2a
(12)
x
2(4)
3
x
2
The vertex is
3 ,14.
 2

b
y f( )
2a
 3
 3
y  f    4    12
2
2
2
 3
 5
2
Therefore there is a maximum of
3
y = 14, when x = .
2
Direction of the Parabola
If the coefficient
If the coefficient of
of x2 is positive
x2 is negative the
the parabola will
parabola will
open up.
open down.


CHARACTERISTICS OF THE
FUNCTION
f(x) = ax2 + bx + c
-b
 - b 
VERTEX 
, f
 
 2a  
 2a
-b
AXISOF SYMMETRY: x 
2a
Parabola opens up and has a
minimum value if a > 0.
Parabola opens down and has a
maximum value if a < 0.
EXAMPLE
Determine without graphing whether the
given quadratic function has a maximum
or minimum value and then find the
value. Verify by graphing.
f(x) = 4x2 - 8x + 3
g(x) = -2x2 + 8x + 3
THE X AND Y INTERCEPTS OF A
1. Find the x-intercepts by setting the quadratic function
equal to zero and solve by whatever method is easiest.
2. If the discriminant b2 – 4ac > 0, the graph of f(x) =
ax2+ bx + c has two distinct x-intercepts and will cross
the x-axis twice.
3. If the discriminant b2 – 4ac = 0, the graph of f(x) = ax2 + bx
+ c has one x-intercept and touches the x-axis at its
vertex.
4. If the discriminant b2 – 4ac < 0, the graph of f(x) = ax2 + bx
+ c has no x-intercept and will not cross or touch the xaxis.
5. Find the y-intercept by substituting x=0 into function.
FUNCTIONS
Graph the functions below by hand by
determining whether its graph opens up
or down and by finding its vertex, axis of
symmetry, y-intercept, and x-intercepts,
if any. Verify your results using a
graphing calculator.
f(x) = 2x2 - 3
g(x) = x2 - 6x - 1
h(x) = 3x2 + 6x
k(x) = -2x2 + 6x + 2
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