Download Solution

Objectives
Students will learn how to;
•Graph quadratic equations.
•Complete the square to graph quadratic equations.
•Use the Vertex Formula to graph quadratic equations.
•Solve a Quadratic Equation by factoring and square root
property.
•Complete the square to solve quadratics.
•Use the quadratic formula to solve quadratics.
•Solve for a Specified Variable
•Understand the properties of the Discriminant
Quadratic Function
A function  is a quadratic function if
f ( x )  ax 2  bx  c,
where a, b, and c are real numbers,
with a ≠ 0.
Simplest Quadratic
range
[0, )
y
x
–2
–1
0
1
2
(x)
4
1
0
1
4
4
3
2
–4 –3 –2
x
2
domain
(−, )
–2
–3
–4
3
4
f  x   x2
Simplest Quadratic
Parabolas are symmetric with respect to a line. The
line of symmetry is called the axis of symmetry of
the parabola. The point where the axis intersects the
parabola is the vertex of the parabola.
Opens up
Axis
Vertex
Vertex
Axis
Opens down
Applying Graphing Techniques to
a Quadratic Function
The graph of g(x) = ax2 is a parabola with vertex at the origin
that opens up if a is positive and down if a is negative. The
width of the graph of g(x) is determined by the magnitude of
a. The graph of g(x) is narrower than that of (x) = x2 if a>
1 and is broader (wider) than that of (x) = x2 if a< 1. By
completing the square, any quadratic function can be written
in vertex form
F ( x )  a( x  h )  k.
2
the graph of F(x) is the same as the graph of g(x) = ax2
translated hunits horizontally (to the right if h is positive
and to the left if h is negative) and translated k units
vertically (up if k is positive and down if k is negative).
Example 1
GRAPHING QUADRATIC
FUNCTIONS
Graph the function. Give the domain and
range.
2
a. f  x   x  4 x  2 (by plotting points)
Solution
3
2
Domain
(−, )
Range
[–6, )
–2
–6
x
(x)
–1
3
0
–2
1
–5
2
–6
3
–5
4
–2
5
3
Example 1
GRAPHING QUADRATIC
FUNCTIONS
Graph the function. Give the domain and
range.
y  x2
1 2
b. g  x    x
2
3
1 2
Solution
y x
2
Domain
(−, )
2
–2
1
g  x    x2
2
Range
(–, 0]
–6
Example 1
GRAPHING QUADRATIC
FUNCTIONS
Graph the function. Give the domain and
1
2
range.
F  x     x  4  3
1
2
2
3
c. F  x     x  4   3
2
(4, 3)
Solution
Domain
(−, )
Range
(–, 3]
–2
1 2
g x   x
2
–6
x=4
Example 2
GRAPHING A PARABOLA BY
COMPLETING THE SQUARE
Graph f  x   x 2  6 x  7 by completing the
square and locating the vertex.
Solution Express x2– 6x + 7 in the form
(x–h)2 + k by completing the square.
f  x    x  6x
2
2
2
1


6


3

9




 2

7
Complete the square.
GRAPHING A PARABOLA BY
COMPLETING THE SQUARE
Example 2
Graph f  x  x 2  6 x  7 by completing the
square and locating the vertex.
Solution Express x2 – 6x + 7 in the form
(x–h)2 + k by completing the square.
f  x    x  6x  9  9  7
Add and subtract 9.
f  x    x  6x  9  9  7
Regroup terms.
2
2
f  x    x  3  2
2
Factor; simplify.
This form shows that the vertex is (3, – 2)
Example 2
GRAPHING A PARABOLA BY
COMPLETING THE SQUARE
Graph f  x  x 2  6 x  7 by completing the
square and locating the vertex.
Solution
Find additional ordered pairs
that satisfy the equation.
Use symmetry about the
axis of the parabola to find
other ordered pairs.
Connect to obtain the graph.
Domain is (−, ) Range is [–2, )
Example 3
GRAPHING A PARABOLA BY
COMPLETING THE SQUARE
Graph f  x   3 x 2  2x  1 by completing the
square and locating the vertex.
Solution To complete the square, the
coefficient of x2 must be 1.
 2 2

f  x   3  x  x   1 Factor –3 from the
first two terms.
3


2
1 1
 2 2
f  x   3  x  x     1
3
9 9

2
1
 1  2   1 


 2  3    3  9 ; add
1
and subtract .
9
Example 3
GRAPHING A PARABOLA BY
COMPLETING THE SQUARE
Graph f  x   3 x 2  2x  1 by completing the
square and locating the vertex.
Solution
1
 2 2
 1
f  x   3  x  x    3     1
3
9

 9
Distributive
property
2
Be careful
here.
1 4

f  x   3  x   
3 3

Factor; simplify.
Example 3
GRAPHING A PARABOLA BY
COMPLETING THE SQUARE
Graph f  x   3 x 2  2x  1 by completing the
square and locating the vertex.
Solution
2
1 4

f  x   3  x   
3 3

Factor; simplify.
 1 4
The vertex is   ,  .
 3 3
Example 3
GRAPHING A PARABOLA BY
COMPLETING THE SQUARE
Graph f  x   3 x 2  2x  1 by completing the
square and locating the vertex.
Solution Intercepts are good additional points
to find. Here is the y-intercept.
y  3  0   2  0   1  1
2
Let x = 0.
Example 3
GRAPHING A PARABOLA BY
COMPLETING THE SQUARE
Graph f  x   3 x 2  2x  1 by completing the
square and locating the vertex.
Solution The x-intercepts are found by
setting (x) equal to 0 in the original equation.
0  3 x  2 x  1
2
3 x  2x  1  0
2
 3 x  1 x  1  0
1
x
3
or
Let (x) = 0.
Multiply by –1; rewrite.
Factor.
x  1
Zero-factor property
Example 3
GRAPHING A PARABOLA BY
COMPLETING THE SQUARE
The x intercepts or
zeros are the solutions
to the quadratic
equations
 1 4
2
The y-intercept is 1
 , 
 3 3
This x-intercept
is – 1.
This x-intercept
is 1/3.
1 3
 , 
2 4
2
Vertex Formula
Graph of a Quadratic
Function
The quadratic function defined by
(x) = ax2 + bx + c can be written as
y  f  x   a  x  h   k, a  0,
2
where
b
h
and k  f  h  .
2a
Graph of a Quadratic
Function
The graph of a quadratic function has the following
characteristics. Vertex form F ( x )  a( x  h )2  k .
1.It is a parabola with vertex (h, k) and the vertical line
x = h as axis of symmetry.
2.It opens up if a > 0 and down is a < 0.
3.It is broader than the graph of y = x2 if a< 1 and
narrower if a> 1.
4.The y-intercept is (0) = c.
Example 4
FINDING THE AXIS,VERTEX, SOLUTIONS AND
GRAPH OF A PARABOLA.
Find the axis, vertex, solutions and graph of the
parabola having equation (x) = 2x2 +4x + 5
using the vertex formula.
Solution Here a = 2, b = 4, and c = 5. The
axis of the parabola is the vertical line
b
4
Axis is x = -1
xh

 1
2a
2  2
The vertex is (–1, (–1)).
Since (–1) = 2(–1)2 + 4 (– 1) +5 = 3,
the vertex is (–1, 3).
Example 4
FINDING THE AXIS,VERTEX, SOLUTIONS AND
GRAPH OF A PARABOLA.
No real solutions because
the graph does not have a
value at y = 0; the graph
does not cross the x-axis
X
-3
-2
-1
0
1
Y
11
5
3
5
11
Axis; x = -1
Zero-Factor Property
If a and b are numbers with ab = 0,
then a = 0 or b = 0 or both.
USING THE ZERO-FACTOR
PROPERTY
Example 1
Solve 6 x 2  7 x  3
Solution:
6x 2  7x  3
6X  7X  3  0
2
Standard form
(3 x  1)(2 x  3)  0
Factor.
3x  1  0
Zero-factor
property.
or
2x  3  0
Example 1
USING THE ZERO-FACTOR
PROPERTY
Solve 6 x 2  7 x  3
Solution:
3x  1  0
or
3x  1
or
2 x  3
1
or
x
3
3
x
2
2x  3  0
Zero-factor
property.
Solve each
equation.
Square-Root Property
A quadratic equation of the form x2 = k can
also be solved by factoring
x2  k
2
x k 0
 x  k  x  k   0
x  k  0 or x  k  0
x k
or x   k
Subtract k.
Factor.
Zero-factor property.
Solve each equation.
Square Root Property
If x2 = k, then
x k
or
x k
Square-Root Property
That is, the solution of
Both solutions
are real if k > 0,
and both are
imaginary if k < 0
If k < 0, we write
the solution set
as
i k 
x2  k

is
k,  k
or
 k 

If k = 0, then this
is sometimes
called a double
solution.
Example 2
USING THE SQUARE ROOT
PROPERTY
Solve each quadratic equation.
2
a. x  17
Solution:
By the square root property, the solution set
is
 17


USING THE SQUARE ROOT
PROPERTY
Example 2
Solve each quadratic equation.
b. x 2  25
Solution:
Since
1  i ,
the solution set of x2 = − 25
is
5i .
Example 2
USING THE SQUARE ROOT
PROPERTY
Solve each quadratic equation.
2
c. ( x  4)  12
Solution:
Use a generalization of the square root
property.
( x  4)2  12
x  4   12
x  4  12
x 42 3
Generalized square
root property.
Add 4.
12  4 3  2 3
Solving A Quadratic Equation
By Completing The Square
To solve ax2 + bx + c = 0, by completing the square:
Step 1 If a ≠ 1, divide both sides of the equation by a.
Step 2 Rewrite the equation so that the constant term is
alone on one side of the equality symbol.
Step 3 Square half the coefficient of x, and add this square
to both sides of the equation.
Step 4 Factor the resulting trinomial as a perfect square
and combine like terms on the other side.
Step 5 Use the square root property to complete the
solution.
Example 3
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve x2 – 4x –14 = 0 by completing the
square.
Solution
Step 1 This step is not necessary since a = 1.
Step 2
x 2  4 x  14
Step 3
x  4 x  4  14  4
Step 4
( x  2)  18
2
2
Add 14 to both
sides.
2
1


(

4
)
 2
  4;
add 4 to both sides.
Factor; combine
terms.
Example 3
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve x2 – 4x –14 = 0 by completing the
square.
Solution
Step 4
( x  2)  18
Factor; combine terms.
Step 5
x  2   18
Square root property.
Take both
roots.
2
x  2  18
Add 2.
x  2  3 2 Simplify the radical.
The solution set is 2  3 2 .


USING THE METHOD OF
COMPLETING THE SQUARE a ≠ 1
Example 4
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
2
9 x  12 x  9  0
4
x  x 1 0
3
4
2
x  x  1
3
4
4
4
2
x  x   1 
3
9
9
2
Divide by 9. (Step 1)
Add – 1. (Step 2)
2
4
4
 1  4 


;
add
 2  3  
9
9
Example 4
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
4
4
4  1  4  4
2
4


;
add
x  x   1 
 2  3  
9
9
3
9
9
2
2
2
5

x    
3
9

2
5
x  
3
9
Factor, combine
terms. (Step 4)
Square root property
Example 4
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
2
5 Square root property
Solution
x  
3
9
2
5
x 
i
3
3
2
5
x 
i
3 3
a  i a
Quotient rule for
radicals
Add ⅔.
Example 4
USING THE METHOD OF
COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the
square.
Solution
2
5
x 
i Add ⅔.
3 3
2
5 
i .
The solution set is  
3 3 
The Quadratic Formula
The method of completing the square can
be used to solve any quadratic equation. If
we start with the general quadratic equation,
ax2 + bx + c = 0, a ≠ 0, and complete the
square to solve this equation for x in terms
of the constants a, b, and c, the result is a
general formula for solving any quadratic
equation. We assume that a > 0.
Quadratic Formula
The solutions of the quadratic equation
ax2 + bx + c = 0, where a ≠ 0, are
b  b 2  4ac
x
.
2a
Caution Notice that the fraction bar
in the quadratic formula extends under
the –b term in the numerator.
b  b  4ac
x
.
2a
2
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Example 5
Solve x2 – 4x = –2
Solution:
x  4x  2  0
2
Write in standard
form.
Here a = 1, b = –4, c = 2
b  b 2  4ac
x
2a
Quadratic formula.
Example 5
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = –2
Solution:
b  b  4ac
x
2a
2
The fraction
bar extends
under – b.
Quadratic formula.
(  4)  (  4)2  4(1)( 2)

2(1)
Example 5
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = –2
Solution:
(  4)  (  4)  4(1)( 2)

2(1)
2
The fraction
bar extends
under – b.
4  16  8

2
42 2

2
16  8  8  4 2  2 2
USING THE QUADRATIC
FORMULA (REAL SOLUTIONS)
Example 5
Solve x2 – 4x = –2
Solution:
42 2

2
2 2 2

2

Factor first,
then divide.
 2 2
16  8  8  4 2  2 2

Factor out 2 in the numerator.
Lowest terms.


The solution set is 2  2 .
Example 6
USING THE QUADRATIC FORMULA
(NONREAL COMPLEX SOLUTIONS)
Solve 2x2 = x – 4.
Solution:
2x 2  x  4  0
Write in standard form.
( 1)  ( 1)  4(2)(4)
x
2(2)
2
1  1  32

4
Quadratic formula;
a = 2, b = – 1, c = 4
Use parentheses and
substitute carefully to
avoid errors.
Example 6
USING THE QUADRATIC FORMULA
(NONREAL COMPLEX SOLUTIONS)
Solve 2x2 = x – 4.
Solution:
1  1  32

4
1  31
x
1  i
4
1
31 
i .
The solution set is  
4 
4
Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve for the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution
d 2
Goal: Isolate d,
A
the specified
4
variable.
4A  d
4A
2
d

2
Multiply by 4.
Divide by .
Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution 4A
2
Divide by .
d

See the Note
following this
example.
4A
d 

Square root
property
Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution
4A
Square root
d 
property

 4A
d



Rationalize the
denominator.
Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution
 4A 
Rationalize the
d
denominator.


 4 A
d

Multiply numerators;
multiply
denominators.
Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve the specified variable. Use  when
taking square roots.
2

d
a. A 
, for d
4
Solution
 4 A Multiply numerators;
multiply
d

denominators.
2 A
d

Simplify.
Solving for a Specified Variable
Note In Example 8, we took both
positive and negative square roots.
However, if the variable represents a
distance or length in an application, we
would consider only the positive square
root.
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Example 8
Solve the specified variable. Use  when
taking square roots.
b. rt 2  st  k
Solution
(r  0), for t
r t  st  k  0 Write in standard form.
2
Now use the quadratic formula to find t.
b  b 2  4ac
t
2a
Example 8
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Solve the specified variable. Use  when
taking square roots.
b. rt 2  st  k
(r  0), for t
Solution t  b  b  4ac
2a
2
( s )  ( s )  4(r )( k )
t
2r
2
a = r, b = – s,
and c = – k
SOLVING FOR A QUADRATIC
VARIABLE IN A FORMULA
Example 8
Solve the specified variable. Use  when
taking square roots.
b. rt 2  st  k
(r  0), for t
Solution
2
( s )  ( s )  4(r )( k )
t
2r
s  s  4rk
t
2r
a = r, b = – s,
and
c = –k
2
Simplify.
The Discriminant
The Discriminant The quantity under the
radical in the quadratic formula, b2 – 4ac, is
called the discriminant.
b  b  4ac
x
2a
2
Discriminant
The Discriminant
Discriminant
Positive, perfect
square
Positive, but not
a perfect
square
Zero
Negative
Number of
Solutions
Type of
Solution
Two
Rational
Two
Irrational
One
(a double solution)
Rational
Two
Nonreal
complex
Caution The restriction on a, b, and c
is important. For example, for the equation
x  5x  1  0
2
the discriminant is b2 – 4ac = 5 + 4 = 9,
which would indicate two rational solutions
if the coefficients were integers. By the
quadratic formula, however, the two
solutions are irrational numbers,
53
x
2
Example 9
USING THE DISCRIMINANT
Determine the number of distinct solutions,
and tell whether they are rational, irrational, or
nonreal complex numbers.
a. 5 x 2  2 x  4  0
Solution
For 5x2 + 2x – 4 = 0, a = 5, b = 2, and c = –4.
The discriminant is
b2 – 4ac = 22 – 4(5)(–4) = 84
The discriminant is positive and not a perfect
square, so there are two distinct irrational
solutions.
Example 9
USING THE DISCRIMINANT
Determine the number of distinct solutions,
and tell whether they are rational, irrational, or
nonreal complex numbers.
b. x 2  10 x  25
Solution
First write the equation in standard form as
x2 – 10x + 25 = 0. Thus, a = 1, b = –10, and
c = 25, and b2 – 4ac =(–10 )2 – 4(1)(25) = 0
There is one distinct rational solution, a “double
solution.”
Example 9
USING THE DISCRIMINANT
Determine the number of distinct solutions,
and tell whether they are rational, irrational, or
nonreal complex numbers.
c. 2 x 2  x  1  0
Solution
For 2x2 – x + 1 = 0, a = 2, b = –1, and c = 1, so
b2 – 4ac = (–1)2 – 4(2)(1) = –7.
There are two distinct nonreal complex
solutions. (They are complex conjugates.)