1. No category

Kerr
MPM2D
Applying Quadratic Models
5.3 Graphing Quadratics in Vertex Form
Every quadratic relation begins as y = x2. This is your basic parabola.
Notice the vertex is on the origin of the coordinate plane. Today we
will examine how all four transformations can change the shape of the
parabola and the location of the vertex. The transformations are
applied in the following order:
Transformation
Examine the
value of …
If …
a
a < -1 or a > 1
a
-1 < a < 1
a
a = -1
h
h<0
h
h>0
k
k<0
k
k>0
1. vertical stretch or
vertical compression
2. reflection in the
x-axis
3. horizontal
translation
4. vertical
translation
What
changes?
Parabola
vertical
becomes
stretch
more
narrow
Parabola
vertical
becomes
compression
more wide
reflection
Direction
in the xof opening
axis
translate
right
Location of
translate
vertex
left
moves
translate
down
translate up
Then …
The vertex form of a quadratic relation is:
1. vertical stretch (more
narrow) or vertical
compression (more wide)
2. reflection in xaxis if negative
3. moves vertex left
(+) or right (-)
y = a(x - h)2 + k
4. moves vertex up (+)
or down (-)
Kerr
MPM2D
Applying Quadratic Models
This is called "vertex" form is because we can very easily identify the
vertex of the quadratic relation. The vertex is (h, k), which means x =
h is the axis of symmetry and y = k is the optimum value. If the
parabola opens up (a > 0), the optimum value is a minimum. If the
parabola opens down (a < 0), the optimum value is a maximum.
Examples
1.
In order, describe the transformations you would apply to the
graph of y = x2 to create the graph of each relation. Then sketch
the graph.
a)
y = 5x2 – 4
______________________________
______________________________
This parabola opens __________.
The axis of symmetry is x = _____.
The vertex is __________.
b)
y = - (x + 8)2
5
4
______________________________
______________________________
______________________________
This parabola opens __________.
The axis of symmetry is x = _____.
The vertex is __________.
Kerr
MPM2D
Applying Quadratic Models
c)
y = (x – 1)2 - 2
______________________________
______________________________
This parabola opens __________.
The axis of symmetry is x = _____.
The vertex is __________.
d)
y=-
15
(x + 14)2 + 18
17
______________________________
______________________________
______________________________
______________________________
This parabola opens __________.
The axis of symmetry is x = _____.
The vertex is __________.
Kerr
MPM2D
Applying Quadratic Models
2.
Complete the table below.
Quadratic
Relation
y=
Vertical
Stretch or
Vertical
Compression
Factor
Reflection
in the
x-axis
Translations
(horizontal &
vertical)
Optimum
Value
(max or
min)
Axis of
Symmetry
3
(x - 3)2 - 1
4
y = -7(x + 6)2
None
No
HT 2 left
VT 4 down
y = -4
(min)
x = -2
None
Yes
HT 9 left
y=0
(max)
x = -9
y = -2x2 + 6
y = -3(x + 2)2 – 7
3.
Write the equation of the parabola that matches each
description.
a)
The graph of y = x2 is compressed by a factor of
about the x-axis and translated 3 units right.
1
, reflected
3
b)
_________________________
The graph of y = x2 is stretch by a factor of 7 and translated 6
units up.
c)
_________________________
The graph of y = x2 is reflected about the x-axis, translated 2
units right and 3 units up.
_________________________
Kerr
MPM2D
Applying Quadratic Models
The vertex form of quadratic relation is very helpful when solving
word problems because we can immediately identify the optimum value.
x-coordinate of the vertex &
axis of symmetry
if a is positive, parabola opens up
optimum value is a minimum
y = a(x - h)2 + k
if a is negative, parabola opens down
optimum value is a maximum
y-coordinate of the vertex &
optimum value
4.
A football is kicked into the air. Its height, in meters, after t
seconds is given by h = -4.9(t - 2.4)2 + 29.
a)
b)
c)
d)
What is the maximum height of the football.
At what time did the football reach its maximum height.
Sketch a graph that represents the height of the football.
When did the football reach a height of 10 m? Round your
answer to the nearest hundredth.
the axis of symmetry is x = _____
Solution:
h = -4.9(t - 2.4)2 + 29
the parabola opens ______ so the
optimum value is a _____________
a)
b)
c)
vertex (_____, _____)
the maximum height is y = _____
The maximum height of the football is __________.
The football reached its maximum height after __________.
Kerr
MPM2D
Applying Quadratic Models
d)
Translation: When h = 10, what is t?
h = -4.9(t - 2.4)2 + 29
Substitute h = 10 into the
original equation to isolate
and solve for t. Don’t forget
BEDMAS!
Therefore the football reached a height of 10 m after
________________________________________________.
How can you go from one form of a quadratic relation to another?
factored form
factored form
Expand and simplify.
Set the factors equal to zero.
Solve for the zeros.
Solve for axis of symmetry. (add and ÷ 2)
Substitute the axis of symmetry back into
the original equation to solve for the y-value
of the vertex.
Insert the given value of a, h(x) and k(y) into
vertex form.
standard form
vertex form
Kerr
MPM2D
Applying Quadratic Models
5.
Express the quadratic relation y =3(x + 5)(x - 7) in standard form
and vertex form.
Standard Form
y =3(x + 5)(x - 7)
Vertex Form
y =3(x + 5)(x - 7)
Hints for the homework:
 Question #15 is just like example #4.
 Question #17 is just like example #5.
Homework: Page 269 #1-4, 6, 7def, 8, 9, 11,
12cd, 13, 15, 17