Download unit-2-quadratic-equations-and-functions

f (x )  ax  bx  c
2
Quadratic equations
and functions
PROF. M. ALONSO
 b  b  
,f 



 2a  2a  
 Solve
quadratic equations
 Graph a quadratic function
 Identify the vertex of a parabola
 Write the equation of the axis of symmetry
 Identify
the domain and range
 Identify the intercept(s) for a given
parabola
 Solve verbal problems.
A
quadratic equation has the following form:
ax2 + bx + c = 0
 where
x represents the variable and a,b, c
are real numbers, a  0.
 In
order to be a quadratic equation, the
variable must have the exponent 2
 the variable can NOT be in a denominator.
the letter a represents the number that is
next to the variable x2.
 the letter b represents a number that is
beside the variable x.
 c represents a number that is alone

ax2 + bx + c = 0
Equation
Form: ax2 +bx + c =0
5x2 + 3x + 2 = 0
-3x2 + 4x = 7
-3x2 + 4x - 7 = 0
x2 + 6x = 0
3x2 = 10
-9x2 = 0.7x
3x2 -
10 = 0
-9x2 - 0.7x = 0
 To
solve a quadratic equation is to find the
value of the variable that makes the
equation true.
 There are several methods to find the value
of the variable
 Using
the square root
 Factoring
 Quadratic Formula
 Completing the square
 This
method can only be used if b = 0. Thus,
if there is no middle term (b = 0) then get
the x2 alone and square root both sides.
 Example Solve
x2 - 20 = 0
2
x  20  0
 Solution:
x  20
2
x   20
2
x  2 5
 Example:
Solve x2 + 49 = 0
 Solution:
x 2  49  0
x 2  49
x   49
2
x  7 i
Remember
complex
numbers
 Example:
 Solution:
Solve 2x2 = 30
2 x  30
2
2
2x
30

2
2
x   15
2
x   15
 Example:
 Solución:
Solve
(x + 4)2 - 16 = 0
( x  4)  16  0
2
( x  4)  16
2
( x  4)   16
2
x  4  4
x44
x0
o
o
x  4  4
x  8
 Example
 Solution:
Solve (x – 2)2 = 20
( x  2)  20
2
( x  2)   20
2
x2  4 5
x  2  2 5
x  2 5  2
 See
if it will factor
 Example Solve
x2 - 49 = 0
 Solution;
Set each
factor equal
to zero
 Example
 Solution
Solve
x2 + 6x + 8 = 0
 Example
 Solution:
Solve
2x2 - 5x -3 = 0
 Example
 Solution:
Solve
2x2 - 20x = 0
 The
quadratic formula can be used to solve
any quadratic equation whether it factors or
not. If it factors, it is generally easier to factor,
but this formula would give you the solutions
as well.
 The quadratic formula is a formula based on
the values of a, b and c of the quadratic
equation
 b  b  4ac
x
2a
2
 Substituting
the values of a, b, and c in the
formula yields the value of the variable.
 In order to use this formula, the equation
must be equal to zero
 Example:
 Solution:
2x2 - 5x – 3 = 0
a = 2, b = -5, c = -3
Solve
 b  b  4ac
x
2a
2
 (5)  (5) 2  4(2)(3) 5  25  24 5  49 5  7
x



2(2)
4
4
4
5  7 12
5  7 2
1
x

3
x


4
4
4
4
2
Two values
 Example
Solve 3x2 = -4
 Solution: First, 3x2 + 4 = 0 .
 a = 3,
b = 0, c = 4

0  0 2  4(3)(4)   48  (  1)( 16 )( 3 )  4i 3  2i 3
x




2(3)
6
6
6
3
The two values are
2i 3
x
3
o
 2i 3
x
3
 Example
Solve -5x2 = 2x
 Solution: First -5x2 –2x = 0.

a = -5 b = -2 c = 0.
 (2)  (2) 2  4(5)(0) 4  4  0 4  4 4  2
x



2(5)
 10
 10
 10
The two values are
42
6
3
x


10 10
5
42
2
1
x


10 10
10
 Example
x2 + 4x – 5 = 0
 Solution:
 First,
get the constant term on the other
side by adding 5 to both sides
 x2
2
2
+ 4x = 5
b 4
 Add to both sides       4
2
+ 4x + 4 = 5 + 4
 Factor the left side
2
 x2
(x + 2)2 = 9
 (x
+ 2)2 = 9
 Use the square root method
( x  2) 2   9
 Simplify
x+2 =3
 The
x values are :
x+2=3
x=1
x + 2 = -3
x= -5
 Example
Solve x2 + 6x +25 = 0
 Solution

Get the constant term on the other side :
x2 + 6x = -25
 Add to both sides
2
2
b
6
    9
 2
 2

 x2
+ 6x + 9 = -25 + 9
 Factor the left side : (x + 3)2 = -16
 Use
the square root method
( x  3)    16
2
x + 3 =  4i
 The values of x are
 Simplify
x
+ 3 = 4i
x = -3 + 4i
x + 3 = -4i
x = - 3 – 4i
A quadratic function has the
following form:
f(x) = ax2 + bx + c
The domain of this
function are the real
numbers.
It also could be written as
y = ax2 + bx + c
A function must
have two
variables, x and y
Values of a,b,c
1.
2.
3.
4.
5.
f(x) = x2 + 6x + 8
f(x) = -3x2
f(x) = 7x2 + 3x + 2
f(x) = -4x2 + 12
Y = 5x2 - x
a
b
c
1
6
8
-3
0
0
7
3
2
-4
0
12
5
-1
0
 Click
here
 http://www.mathwarehouse.com/geometry/
parabola/
Vertex
This parabola opens up,
the lowest point is called
the vertex (minimum).
This
parabola
opens down,
the vertex is
the highest
point
(maximum)
.

The standard form of a quadratic function is:
y = ax2 + bx + c
The parabola will open up
when a >0 (positive).
The parabola will open down
when a < 0 negative.
 The
y intercept of the parabola is (0,c)
y = ax2 + bx + c
 To
find the x intercept of the parabola, solve
the quadratic equation
ax2 + bx + c = 0
Use any method!
 To
find the coordinates of the vertex of the
parabola use the formula
 b  b  
,f 



 2a  2a  
First, find the x value
b
x
2a
Substitute the x – value into the original
equation F(x) =ax2 + bx + c to find the
y –coordinate of the vertex.
 The
equation of the axis of symmetry is
b
x
2a
Let f(x) = ax2 + bx + c
 If a > 0, the parábola opens up, if a < 0 it opens
down
 The y - intercept is (0, c)
 To find the x - intercept, solve ax2 + bx + c = 0.
 The
vertex is  b ,f

 2a
 b  
 2a  
 
b
 The axis of symmetry is x 
2a
a=1
b=6
c=8
Graph the function f(x) = x2 + 6x + 8
Since a = 1 > 0 the graph opens up.

The y - intercept is (0 , 8)

To find the x - intercepts solve:
x2 + 6x + 8 = 0
(x + 4)(x + 2) = 0
x + 4 = 0
x + 2 = 0
x = -4
x = -2
The x - intercepts are:
(-4, 0) y (-2, 0)

 b
 ,f
 2a
First find x
 b  
 2a  
 
b
x
2a
6

 3
2(1)
Remember
a=1
b=6
c=8
Then y = f(-3) = (-3)2 + 6(-3) + 8
= 9 – 18 + 8
= -1
The vertex is (-3, -1)
The axis of symmetry is the line x = -3.
Axis of
symmetry
X=-3
Y -Intercept
(0,8)
9
8
8
7
6
X - Intercepts
5
4
3
2
1
-8
Vertex
(-3,-1)
-6
-4
-2
0
-1 0
-2
2
minimun
•Since a = -1 < 0 the graph opens down.
•The y- intercept is (0 , 9)
• The x - intercepts are:
9 - x2 = 0
(3 - x)(3 + x ) = 0
3–x=0
3+x=0
x=3
x = -3
x - interceptos (-3, 0) y (3, 0)
 b
 ,f
 2a
 b  
 2a  
 
b
x 
2a
0

2( 1)
0
Y = f(0) = 9 - (0)2 = 9
The vertex is (0, 9)
The axis of symmetry is x = 0.
f (x )  9  x
Vertex
(0,9)
Maximun
X -Intercept
(-3,0)
2
X - Intercept
(3,0)
 Quadratic
functions are used to model real
life situations where maximun or minimun
values are involved
 Remember that the x value of the vertex
indicates where a minimun or maximun
occurs
 Then, f(x) gives the minimun or maximun
value
If
an object is thrown vertically
upward, what is its maximum
height?
What should the dimensions of a
rectangular field be in order to
maximize its area?
What is the minimum cost of a
company?
To answer these questions a
quadratic function is needed
 Suppose
f(x) = ax2 + bx + c
 If a > 0 ( positive), then the function has a
minimun when x = b .
2a
 The
minimum value is
 b 
f

 2a 
.
 If a < 0 (negative) , then the function has a
maximum when x = b .
2a
The maximum value is f  b 
 2a 
Fireworks are thrown into the air. The function
A (t) = -16t2 + 200t + 8 models the height A ( in
feet) that they reach as a function of time t (in
seconds).
 If we want the fireworks to explode when they
reach their maximum height, when should they
explode?


At what height should they explode?
Remember the
function
A(t) = -16t
2
 Since
a = -16 < 0 then we get the
maximun value when
t=
 This
-b -(200) 200
=
=
= 6.25 segundos
2a 2(-16) 32
implies that at 6.25 seconds they
should explode because at this time
they reach their maximum height.
+200t + 8
 The
maximun height is
633 feet
Section
Remember to try the
exercises