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Let a, b, and c be real numbers a  0. The function
f (x) = ax2 + bx + c
is called a quadratic function.
The graph of a quadratic function is a parabola.
Every parabola is symmetrical about a line called the axis
(of symmetry).
y
The intersection point of the
parabola and the axis is
called the vertex of the
parabola.
f (x) = ax2 + bx + c
vertex
x
axis
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The leading coefficient of ax2 + bx + c is a.
a>0
When the leading coefficient
opens
is positive, the parabola
upward
opens upward and the
vertex is a minimum.
vertex
minimum
y
f(x) = ax2 + bx + c
x
y
x
vertex
When the leading
maximum
coefficient is negative,
the parabola opens downward
a<0
opens
and the vertex is a maximum.
downward
f(x) = ax2 + bx + c
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Case1: a>0
y
•Minimum value: k
•Range: [ k ,  )
•Increasing: [ h, )
a>0
opens
upward
x
vertex
minimum
•Decreasing: (  , h]
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Case2: a<0
y
•Maximum value: k
•Range: ( , k ]
•Increasing: (  , h]
x
vertex
maximum
a<0
opens
downward
•Decreasing: [ h, )
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Def:
1) The standard form for the equation of a quadratic
function is:
f (x) = a(x – h)2 + k (a  0)
where
h=-b/2a
k=f(h)=f(-b/2a)
2) Vertex : (h, k)
3) Axis of symmetry : x=h
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The standard form for the equation of a quadratic function is:
f (x) = a(x – h)2 + k (a  0)
The graph is a parabola opening upward if a  0 and opening
downward if a  0. The axis is x = h, and the vertex is (h, k).
Example: By completing the square method write the parabola
y
f (x) = 2x2 + 4x – 1in standard form and find the axis and vertex.
f (x) = 2x2 + 4x – 1
f (x) = 2x2 + 4x – 1
original equation
f (x) = 2( x2 + 2x) – 1
factor out 2
f (x) = 2( x2 + 2x + 1) – 1 – 2 complete the square
f (x) = 2( x + 1)2 – 3
x
standard form
a > 0  parabola opens upward .
h = –1, k = –3  axis x = –1, vertex (–1, –3).
x = –1 (–1, –3)
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Ex1: For the following functions
a) f(x)=2x2+4x+3
b) f(x)=-x2+2x+3
1. Write the function in the standard form
2. Find the vertex
3. Find the axis of symmetry
4. Find , if any, the maximum value of the function
5. Find , if any, the minimum value of the function
6. Find the range of the function
7. Find the interval(s) of increasing and decreasing
8. Sketch the graph of the function and show on the graph
the intercept(s), the vertex, and the axis of symmetry
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Example: Find an equation for the parabola with vertex (2, –1)
passing through the point (0, 1).
y
y = f(x)
(0, 1)
x
(2, –1)
f (x) = a(x – h)2 + k standard form
f (x) = a(x – 2)2 + (–1) vertex (2, –1) = (h, k)
Since (0, 1) is a point on the parabola: f (0) = a(0 – 2)2 – 1
1
1 = 4a –1 and a 
2
1
1 2
2
f ( x)  ( x  2)  1  f ( x)  x  2 x  1
2
2
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Example: A basketball is thrown from the free throw line from a
height of six feet. What is the maximum height of the ball if the
path of the ball is:
1 2
y   x  2 x  6.
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The path is a parabola opening downward.
The maximum height occurs at the vertex.
1 2
1
y
x  2x  6  a  , b  2
9
9
b
At the vertex, x 
 9.
2a
 b 
f
  f 9  15
 2a 
So, the vertex is (9, 15).
The maximum height of the ball is 15 feet.
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Example: A fence is to be built to form a
rectangular corral along the side of a barn
65 feet long. If 120 feet of fencing are
available, what are the dimensions of the
corral of maximum area?
barn
x
corral
x
120 – 2x
Let x represent the width of the corral and 120 – 2x the length.
Area = A(x) = (120 – 2x) x = –2x2 + 120 x
The graph is a parabola and opens downward.
The maximum occurs at the vertex where x   b ,
2a
 b  120
a = –2 and b = 120  x 

 30.
2a
4
120 – 2x = 120 – 2(30) = 60
The maximum area occurs when the width is 30 feet and the
length is 60 feet.
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Q85/227 Find two numbers whose sum is 8 and whose
product is a maximum.
Ex: If x is a real number, then find the maximum area of
a rectangle of length 3+2x and width 1-2x.
Ex: If x=3 is the axis of symmetry of the parabola
y=-2x2+cx+2, then find c
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