Transcript The Quadratic Function

The Quadratic Function
A Quadratic function is a function of
the form
2
f ( x)  ax  bx  c
where a, b and c are real numbers
and
a0
The Quadratic Equation
A Quadratic equation is an equation equivalent
of the form
ax2  bx  c  0
where a, b and c are real numbers and a  0
Examples of Quadratic Equations
5x 2  4 x  2  0
2x 2  x  2
x2
1

0
2
2
Second Degree Equation
A Quadratic equation is also a Second Degree
equation because the highest exponent is
two.
The Standard Form of a quadratic equation is
ax2  bx  c  0
• where a, b and c are real numbers and a  0
Zeros of a Quadratic Function
The zeros of a quadratic function
f ( x)  ax2  bx  c
f ( x)  x 2  x  12
is the solution of the equation f(x) = 0
Since x = 4 and x= -3 are solutions of
2
f
(
x
)

x
 x  12 are x = 4 and x= -3.
the zeros of
We note that
f (4)  4 2  4  12  0
f (3)  (3) 2  (3)  12  0
Methods for Solving
Quadratic Equations
Factoring
Taking the square root
Using the quadratic formula
Completing the square
The Quadratic Formula
For any Quadratic equation ax2  bx  c  0
in standard form
The Quadratic formula is given by
 b  b 2  4ac
x
2a
We have derived this equation in class
The Discriminant
The expression
b  4ac
2
is called the discriminant of the
quadratic equation
The Discriminant
• The Discriminant of a Quadratic Equation
tells whether the quadractic equation has
any real solutions
• It also tells how many real zeros or xintercepts the equation has.
The Discriminant
Two distinct real solutions
The quadratic equation has 2 distinct
unequal real solutions if
b 2  4ac  0
The Discriminant
One distinct Solutions
The quadratic equation has 1distinct
or 2 multiply real solutions if
b 2  4ac  0
• No real solutions
The quadratic equation has two non real solutions if
b  4ac  0
2
Application
(Second degreee equation)
• Page 143 #87
Side of square metal sheet =x
Subtract 1 foot from each end of each side
Result - new side is x – 2
Depth of box is the inch taken from the sides
Area of bottom of box is (x – 2)(x – 2)
Volume of box is (side) (side) (depth)
Volume is (x – 2)(x – 2)(1)
Since box is to hold 4 cubic feet
Volume of box is 4 feet
2
V (x) = (x – 2)(x – 2)(1)= (x-2) = 4
Application
•
(Second degree equation)
Page 142 #90
s(t )  4.9t  20t
• Height of Object
• Initial height is 0, since the object is thrown
from the ground.
• When object is 15 meters above the ground,
the height is
2
2
s(t )  4.9t  20t  15
• Using the quadratic formula, we get t = .99
or t = 3 .09
Page 142 #90(cont.)
• b. When the object strikes the ground the
height is s(t )  4.9t 2  20t  0
• Thus t = 0 or t = 4.08. Thus the object will strike
the ground after 4.08 sec.
• c. When the object is 100 meter above the
ground, the height function is
s(t )  4.9t 2  20t  100
• There is no real solution so the object never
reaches 100 meters.
The Quadratic function
•
Opens up when a>o
opens down a < 0
Vertex
axis of symmetry