```9.2: QUADRATIC FUNCTIONS:
Quadratic Function: A function that can
be written in the form y = ax2+bx+c where
a ≠ 0.
Standard Form of a Quadratic: A function
written in descending degree order, that
is ax2+bx+c.
Parabola: The graph of the function
f(x) = x2.
Axis of Symmetry: The line that divide
the parabola into two identical halves
Vertex: The highest or lowest point of
the parabola.
Minimum: The lowest point of the
parabola.
Maximum: The highest point of the
Line of Symmetry
parabola.
Vertex = Minimum
GOAL:
FINDING THE VERTEX OF
2
ax +bx+c:
To find the vertex of a quadratic equation
where a ≠ 1, we must know the following:
1)
−

2)
−

3) y(
= axis of symmetry .
= the x value of the vertex of the
parabola.
−
)

= the y value of the vertex of the
parabola, that is; min or max
Ex:
Provide the graph, vertex, domain and
range of:
y=
2
3x -9x+2
SOLUTION: To graph we can create a
table or we can find the vertex with other
two points  Faster.
y = 3x2-9x+2
1)
−

= axis of symmetry .  a = 3, b = -9
−(−)
−

()





1.5
Thus x = 1.5 = axis of symmetry
SOLUTION: (Continue)
2)
−

3) y(
y=
= .  = x value of the vertex  ( 1.5, )
−
)

= the y value of the vertex of the
parabola y = 3x2-9x+2
− 2
−
3( ) -9( )+2

 y = 3(1.5)2-9(1.5)+2
 y = 3(2.25)-9(1.5)+2
 y = 6.75-13.5+2
 y = -4.75
Vertex = ( 1.5, -4.75)
SOLUTION: (Continue)
Vertex = ( 1.5, -4.75)
Now:
choose one value of x on the left of 1.5:  x=0
x = 0  3(0)2-9(0)+2
y=2
 (0, 2)
choose a value of x on the right of 1.5:  x=3
x = 3  3(3)2-9(3)+2
y=2
 (3, 2)
We not plot our three point:
(0,2), vertex(1.5, -4.75) and (3,0)
SOLUTION:
Vertex:
( 1.5, -4.75)
(0, 2)
(3, 2)
Domain: (-∞, ∞)
Range: (-4.75, ∞)
REAL-WORLD:
During a basketball game halftime, the heat
uses a sling shot to launch T-shirts at the
crowd. The T-shirt is launched with an initial
velocity of 72 ft/sec. The T-shirt is caught 35
ft above the court. How long will it take the
T-shirt to reach its maximum height?
What is the maximum height? What is the
range of the function that
models the height of the
T-shirt over time?
SOLUTION:
To solve vertical motion problems we must
know the following:
An object projected into the air reaches
the following maximum height:
h= -16
Where
2
t
+ vt + c
t = time,
v = initial upward velocity
c = initial height.
SOLUTION:
During ….The T-shirt is launched with an initial velocity of
72 ft/sec. The T-shirt is caught 35 ft above the court. From
the picture we can see that initial height is 5ft.
−

Thus t = unknow, 
v = initial upward velocity = 72 ft/sec
c = initial height= 5 ft
h= -16
Now:

2
t
+ vt + c  h= -16
−
t=

−
t=
−
2
t
+ 72t + 5
 a = -16, b = 72  t =
 t = 2.25
−()
(−)
SOLUTION: (Continue)
2)
−

3) y(
y=
= 2.25 = x value of the vertex: ( 2.25,
−
)

)
= the y value of the vertex:
y = -16t2+72t+5
− 2
−
-16( ) +72( )+5

y = -16(2.25)2+72(2.25)+5
y = -16(5.06)+72(2.25)+5
 y = -81+162+5
 y = 86
Vertex = ( 2.25, 86)
SOLUTION: (Continue)
Vertex = ( 2.25, 86)
Now:
choose one value of x on the left of 1.5:  x=0
x=0
 -16(0)2+72(0)+5
 y = 5  (0, 5)
choose a value of x on the right of 1.5:  x=3
x = 4.5  -16(4.5)2+72(4.5)+5  y =  (4.5, 5)
We not plot our three point:
(0,5), vertex(2.25, 86) and (4.5, 5)
SOLUTION:
Vertex:
( 2.25, -86)
(0, 5)
(4.5, 5)
Domain: (-∞, ∞)
Range: (-∞, 86)
Maximum : 86 ft.
VIDEOS:
Graphs and Their Properties
VIDEOS: