Solving and Graphing Quadratic functions.ppt

QUADRATIC FUNCTIONS
Learning Targets:
I can…
1.understand and apply the concepts in any related
real – life situations and problems in the future.
2.represent and identify a quadratic function using
table of values, graph and equation.

A QUADRATIC FUNCTION
vertex
Axis of
symmetry
y = x2
y = -x2
A quadratic function has a form
y = ax2
+ bx + c
where a ≠ 0.
The graph of a quadratic function is U-
shaped and is called parabola.
The lowest or highest point on the graph of
a quadratic function is called the vertex.
The graphs of y = x2
and y = -x2
are
symmetric about the y-axis, called the axis
of symmetry.

THE GRAPH OF A QUADRATIC
FUNCTION
vertex
Axis of
symmetry
y = x2
y = -x2
 The parabola opens up if a>0 and
opens down if a<0.
 The parabola is wider than the
graph of y = x2
if |a| < 1 and
narrower than the graph of y = x2
if |
a| > 1.
The x-coordinate of the vertex is
-b/(2a).
The axis of symmetry is the
vertical line x = -b/(2a).

Quadratic Functions
The graph of a quadratic function
is a parabola.
A parabola can open up or down.
If the parabola opens up, the
lowest point is called the vertex.
If the parabola opens down, the
vertex is the highest point.
NOTE: if the parabola opened
left or right it would not be a
function!
y
x
Vertex
Vertex

Graphing Quadratic Functions
y = ax2
+ bx + c

y = ax2
+ bx + c
The parabola will open down
when the a value is negative.
The parabola will open up
when the a value is positive.
Standard Form
y
x
The standard form of a
quadratic function is
a > 0
a < 0

y
x
Line of
Symmetry
Line of Symmetry
Parabolas have a symmetric
property to them.
If we drew a line down the
middle of the parabola, we
could fold the parabola in half.
We call this line the line of
symmetry.
The line of symmetry ALWAYS
passes through the vertex.
Or, if we graphed one side of
the parabola, we could “fold”
(or REFLECT) it over, the line
of symmetry to graph the other
side.

Find the line of symmetry of
y = 3x2
– 18x + 7
Finding the Line of Symmetry
When a quadratic function is in
standard form
The equation of the line of
symmetry is
y = ax2
+ bx + c,
2
b
a
x 

For example…
Using the formula…
This is best read as …
the opposite of b divided by the
quantity of 2 times a.
 
18
2 3
x 18
6
 3

Thus, the line of symmetry is x = 3.

Finding the Vertex
We know the line of symmetry
always goes through the vertex.
Thus, the line of symmetry
gives us the x – coordinate of
the vertex.
To find the y – coordinate of the
vertex, we need to plug the x –
value into the original equation.
STEP 1: Find the line of symmetry
STEP 2: Plug the x – value into the
original equation to find the y value.
y = –2x2
+ 8x –3
8 8
2
2 2( 2) 4
b
a
x   
 
 
 
y = –2(2)2
+ 8(2) –3
y = –2(4)+ 8(2) –3
y = –8+ 16 –3
y = 5
Therefore, the vertex is (2 , 5)

A Quadratic Function in Standard Form
The standard form of a quadratic
function is given by
y = ax2
+ bx + c
There are 3 steps to graphing a
parabola in standard form.
STEP 1: Find the line of symmetry
STEP 2: Find the vertex
STEP 3: Find two other points and reflect
them across the line of symmetry. Then
connect the five points with a smooth
curve.
Plug in the line of
symmetry (x – value) to
obtain the y – value of the
vertex.
MAKE A TABLE
using x – values close to
the line of symmetry.
USE the equation
2
b
x
a
-
=

STEP 1: Find the line of
symmetry
Let's Graph ONE! Try …
y = 2x2
– 4x – 1
( )
4
1
2 2 2
b
x
a
-
= = =
y
x
Thus the line of symmetry is x = 1

y = 2x2
– 4x – 1
STEP 2: Find the vertex
y
x
( ) ( )
2
2 1 4 1 1 3
y = - - = -
Thus the vertex is (1 ,–3).
Since the x – value of the
vertex is given by the line of
symmetry, we need to plug
in x = 1 to find the y – value
of the vertex.

5
–1
y = 2x2
– 4x – 1
( ) ( )
2
2 3 4 3 1 5
y = - - =
STEP 3: Find two other points
and reflect them across the line
of symmetry. Then connect the
five points with a smooth
curve.
y
x
( ) ( )
2
2 2 4 2 1 1
y = - - = -
3
2
y
x

Reference
http://www.chaoticgolf.com/pptlessons/g
raphquadraticfcns2.ppt

EXAMPLE
Graph y = 2x2
-8x +6
Solution: The coefficients for this function are
a = 2, b = -8, c = 6.
Since a>0, the parabola opens up.
The x-coordinate is: x = -b/2a, x = -(-8)/(2(2))
x = 2
The y-coordinate is: y = 2(2)2
-8(2)+6
y = -2
Hence, the vertex is (2,-2).

EXAMPLE(contd.)
Draw the vertex (2,-2) on graph.
Draw the axis of symmetry x=-b/(2a).
Draw two points on one side of the
axis of symmetry such as (1,0) and
(0,6). How were these points chosen?
Use symmetry to plot two more points
such as (3,0), (4,6).
Draw parabola through the plotted
points.
(2,-2)
(1,0)
(0,6)
(3,0)
(4,6)
Axis of symmetry
x
y

VERTEX FORM OF QUADRATIC
EQUATION
y = a(x - h)2
+ k
The vertex is (h,k).
The axis of symmetry is x = h.

GRAPHING A QUADRATIC
FUNCTION IN VERTEX FORM
(-3,4)
(-7,-4)
(-1,2)
(-5,2)
(1,-4)
Axis of symmetry
x
y
Example y = -1/2(x + 3)2
+ 4
where a = -1/2, h = -3, k = 4. Since a<0,
the parabola opens down.
To graph a function, first plot the vertex
(h,k) = (-3,4).
Draw the axis of symmetry x = -3
Plot two points on one side of it, such as
(-1,2) and (1,-4).
Use the symmetry to complete the
graph.

INTERCEPT FORM OF
QUADRATIC EQUATION
y = a(x - p)(x - q)
The x intercepts are p and q.
The axis of symmetry is halfway
between (p,0) and (q,0).

GRAPHING A QUADRATIC
FUNCTION IN INTERCEPT FORM
(-2,0)
(1,9)
(4,0)
Axis of symmetry
x
y
Example y = -(x + 2)(x - 4).
where a = -1, p = -2, q = 4. Since a<0
the parabola opens down.
To graph a function, the x-intercepts
occur at (-2,0) and (4,0).
Draw the axis of symmetry that lies
halfway between these points at x = 1.
So, the x - coordinate of the vertex is
x = 1 and the y - coordinate of the
vertex is: y = -(1 + 2)(1 - 4)= 9.

WRITING THE QUADRATIC
EQUATION IN STANDARD FORM
(1). y = -(x + 4)(x - 9)
= -(x2
- 9x + 4x - 36)
= -(x2
- 5x -36)
= -x2
+ 5x + 36
(2). y = 3(x -1)2
+ 8
= 3(x -1)(x - 1) + 8
= 3(x2
- x - x + 1) + 8
= 3(x2
- 2x + 1) + 8
= 3x2
- 6x + 3 + 8
= 3x2
- 6x + 11

MODELING WITH
QUADRATIC FUNCTIONS

QUADRATIC FUNCTION IN VERTEX
FORM
(2,-3)
(4,1)
x
y
1
1
Write the quadratic function for the parabola
shown.
Solution:
The vertex shown is (h,k) = (2,-3)
Using the vertex form of the quadratic function.
y = a(x-h)2
+ k
y = a(x-2)2
- 3
Use the other given point (4,1) to find a.
1 = a(4-2)2
- 3
1 = 4a - 3
4 = 4a
1 = a
Hence the quadratic function for the parabola is
y = (x-2)2
-3

QUADRATIC FUNCTION IN
INTERCEPT FORM
(-1,2)
(-2)
(3)
x
y
1
1
Write the quadratic function for the parabola
shown.
Solution:
The x intercepts shown are p = -2, q = 3
Using the intercept form of the quadratic
function.
y = a(x-p)(x-q)
y = a(x+2)(x-3)
Use the other given point (-1,2) to find a.
2 = a(-1+2)(-1-3)
2 = -4a
-1/2 = a
Hence the quadratic function for the parabola
is y = -1/2(x+2)(x-3)

Solving and Graphing Quadratic functions.ppt

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Solving and Graphing Quadratic functions.ppt