Download Quadratic functions

IB STUDIES
Graphing Quadratic
Functions
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
2
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
3
The simplest quadratic functions are of the form
f (x) = ax2 (a  0)
These are most easily graphed by comparing them with the
graph of y = x2.
Example: Compare the graphs of
1
y  x 2, f ( x)  x 2 and g ( x)  2 x 2
2
y  x2
y
1 2
f ( x)  x
2
5
g ( x)  2 x 2
x
-5
5
4
Example: Graph f (x) = (x – 3)2 + 2 and find the vertex and axis.
f (x) = (x – 3)2 + 2 is the same shape as the graph of
g (x) = (x – 3)2 shifted upwards two units.
g (x) = (x – 3)2 is the same shape as y = x2 shifted to the right
three units.
y
f (x) = (x – 3)2 + 2
g (x) = (x – 3)2
y = x2
4
(3, 2)
vertex
x
-4
4
5
Example: Graph the parabola f (x) = 2x2 + 4x – 1 and find the axis
of symmetry, x and y-intercepts and vertex.
f (x) = 2x2 + 4x – 1
Use your GDC to find the TP
(1, 3)
axis of symmetry
x  1
x-intercepts (-2.22,0) and (0.225,0)
y-intercept (0,-1)
6
Example: Graph and find the vertex, axis of symmetry
and x and y-intercepts
of f (x) = –x2 + 6x + 7.
a < 0  parabola opens downward.
vertex (3,16)
y-intercept (0,7)
x-intercepts (7, 0), (–1, 0)
axis of symmetry x=3
equivalent forms:
f ( x)  ( x  3) 2  16
f ( x)  ( x  7)( x  1)
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Vertex of a Parabola
The vertex of the graph of f (x) = ax2 + bx + c (a  0)
 b
is   ,
 2a
 b 
f   
 2a  
Example: Find the vertex of the graph of f (x) = x2 – 10x + 22.
f (x) = x2 – 10x + 22 original equation
a = 1, b = –10, c = 22
 b  10

5
At the vertex, x 
2a 2(1)
 b
2
f
  f (5)  5  10(5)  22  3
 2a 
So, the vertex is (5, -3).
8
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.
9
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.
9
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.
10
Example: Find an equation for the parabola with vertex (2, –1)
passing through the point (0, 1).
y
y = f(x)
(0, 1)
x
y  ax 2  bx  c
(2, –1)
Substitute point (0,1) to find c.
1 0 0 c  c 1
1 2
f ( x)  x  2 x  1
2
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