•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
ObjectivesStudents will learn how to;
Quadratic Function
A function is a quadratic function if
2( ) ,x ax bx c f
where a, b, and c are real numbers, with a ≠ 0.
Simplest Quadratic
x (x)
– 2 4
– 1 1
0 0
1 1
2 4
2 3
2
– 2
3
– 2
4
– 3
– 4
– 3– 4
4
range[0, )
2x xfdomain (−, )
x
y
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.
Vertex
Vertex
Axis
Axis
Opens up
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
the graph of F(x) is the same as the graph of g(x) = ax2 translated hunits 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).
2( ) ( ) .F x a x h k
Example 1 GRAPHING QUADRATIC FUNCTIONS
Solution
a.
Graph the function. Give the domain and range.
2 4 2 (by plotting points)x x x fx (x)
– 1 3
0 – 2
1 – 5
2 – 6
3 – 5
4 – 2
5 3
2
3
– 2
– 6
Domain (−, )
Range[– 6, )
Example 1 GRAPHING QUADRATIC FUNCTIONS
Solution
b.
Graph the function. Give the domain and range.
212
x xg
Domain (−, )
Range(–, 0]
2
3
– 2
– 6
212
y x
2y x
212
x xg
Example 1 GRAPHING QUADRATIC FUNCTIONS
Solution
c.
Graph the function. Give the domain and range.
214 3
2F x x
Domain (−, )
Range(–, 3]
212
x xg
214 3
2x x F
(4, 3)
3
– 2
– 6 x = 4
Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution Express x2– 6x + 7 in the form (x– h)2 + k by completing the square.
Graph by completing the square and locating the vertex.
2 6 7x x x f
2 6 7x x x f Complete the square.
2
212
96 3
Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution Express x2 – 6x + 7 in the form (x– h)2 + k by completing the square.
Graph by completing the square and locating the vertex.
2 6 7x x x f
2 9 96 7x x x f Add and subtract 9.
2 6 9 79x xx f Regroup terms.
23 2x x f Factor; simplify.
This form shows that the vertex is (3, – 2)
Example 2 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution
Graph by completing the square and locating the vertex.
2 6 7x x x f
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
Solution To complete the square, the coefficient of x2 must be 1.
Graph by completing the square and locating the vertex.
23 2 1x x x f
2 23
3 1x x x
f Factor – 3 from the first two terms.
2 1 19
29
3 13
x x x
f
2 21 1 1
; add2 3 9
1and subtra
23
ct .9
Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution
Graph by completing the square and locating the vertex.
23 2 1x x x f
2 2 1 13 3 1
3 9 9x x x
f
Distributive property
2
1 43
3 3x x
f
Be careful here.
Factor; simplify.
Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution
Graph by completing the square and locating the vertex.
23 2 1x x x f
2
1 43
3 3x x
f Factor; simplify.
1 4The vertex is , .
3 3
Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution Intercepts are good additional points to find. Here is the y-intercept.
Graph by completing the square and locating the vertex.
23 2 1x x x f
Let x = 0. 23 2 10 0 1y
Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
Solution The x-intercepts are found by setting (x) equal to 0 in the original equation.
Graph by completing the square and locating the vertex.
23 2 1x x x f
Let (x) = 0.
Multiply by –1; rewrite.
23 2 10 x x 23 2 1 0x x
3 1 1 0x x Factor.
1or 1
3x x Zero-factor property
Example 3 GRAPHING A PARABOLA BY COMPLETING THE SQUARE
2
2
1 3,
2 4
1 4,
3 3
The y-intercept is 1
This x-intercept is – 1.
This x-intercept is 1/3.
The x intercepts or zeros are the solutions to the quadratic equations
Graph of a Quadratic FunctionThe quadratic function defined by (x) = ax2 + bx + c can be written as
where
2 , 0,y x a x h k a f
and .2b
h k ha
f
Vertex Formula
Graph of a Quadratic FunctionThe graph of a quadratic function has the following characteristics. Vertex form
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.
2( ) ( ) .F x a x h k
Example 4 FINDING THE AXIS,VERTEX, SOLUTIONS AND GRAPH OF A PARABOLA.
Solution Here a = 2, b = 4, and c = 5. The axis of the parabola is the vertical line
Find the axis, vertex, solutions and graph of the parabola having equation (x) = 2x2 +4x + 5 using the vertex formula.
4
2 2 21
bx h
a
The vertex is (– 1, (– 1)). Since (– 1) = 2(– 1)2 + 4 (– 1) +5 = 3, the vertex is (– 1, 3).
Axis is x = -1
Example 4 FINDING THE AXIS,VERTEX, SOLUTIONS AND GRAPH OF A PARABOLA.
X Y
-3 11
-2 5
-1 3
0 5
1 11
Axis; x = -1
No real solutions because the graph does not have a value at y = 0; the graph does not cross the x-axis
This function has solutions at x=3 and x=6. These values are also called zeros because the y value is zero for x=3 and x=6
Zero-Factor Property
If a and b are numbers with ab = 0, then a = 0 or b = 0 or both.
Example 1 USING THE ZERO-FACTOR PROPERTY
Solve
Solution:
26 7 3x x
26 7 3x x
26 7 3 0X X Standard form
(3 1)(2 3) 0x x Factor.
3 1 0 or 2 3 0x x Zero-factor property.
Example 1 USING THE ZERO-FACTOR PROPERTY
Solve
Solution:
26 7 3x x
3 1 0 or 2 3 0x x Zero-factor property.
3 1x or 2 3x
13
x 32
x or
Solve each equation.
Square-Root Property
A quadratic equation of the form x2 = k can also be solved by factoring
2x k2 0x k
0x k x k
or0x k 0x k
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 of2x kis
,k k
or
k
If k = 0, then this is sometimes called a double solution.
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
Example 2 USING THE SQUARE ROOT PROPERTY
a.
Solution:
2 17x
By the square root property, the solution set is 17
Solve each quadratic equation.
Example 2 USING THE SQUARE ROOT PROPERTY
b.
Solution:
2 25x
Since
5 .i
1 ,i
the solution set of x2 = − 25
is
Solve each quadratic equation.
Example 2 USING THE SQUARE ROOT PROPERTY
c.
Solution:
2( 4) 12x
Use a generalization of the square root property. 2( 4) 12x
4 12x Generalized square root property.
124x Add 4.
2 34x 12 4 3 2 3
Solve each quadratic equation.
Solving A Quadratic Equation By Completing The SquareTo 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 2 144x x Add 14 to both sides.
Step 3 2 44 14 4x x
add 4 to both sides.
Step 4 2( 2) 18x Factor; combine terms.
21
( )4 4;2
Example 3 USING THE METHOD OF COMPLETING THE SQUARE a = 1
Solve x2 – 4x –14 = 0 by completing the square.Solution
Step 4 2( 2) 18x Factor; combine terms.
Step 5 2 18x Square root property.
Take both roots. 2 18x Add 2.
2 3 2x Simplify the radical.
The solution set is 2 3 2 .
Example 4 USING THE METHOD OF COMPLETING THE SQUARE a ≠ 1
Solve 9x2 – 12x + 9 = 0 by completing the square.Solution 29 12 9 0x x
2 41 0
3x x Divide by 9. (Step 1)
2 41
3x x Add – 1. (Step 2)
2 4 49
13 94
x x 2
1 4 4; add
432 9 9
Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the square.Solution 2 4 4
91
3 94
x x 2
1 4 4; add
432 9 9
22 53 9
x
Factor, combine terms. (Step 4)
2 53 9
x Square root property
Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1
Solve 9x2 – 12x + 9 = 0 by completing the square.Solution
2 53 9
x
Quotient rule for radicals
2 53 3
x i a i a
Square root property
2 53 3
x i Add ⅔.
Example 4 USING THE METHOD OF COMPLETING THE SQUARE a = 1
The solution set is
Solution 2 5
3 3x i Add ⅔.
2 5.
3 3i
Solve 9x2 – 12x + 9 = 0 by completing the square.
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
2 4.
2b b ac
xa
Caution Notice that the fraction bar in the quadratic formula extends under the – b term in the numerator.
2 4.
2b b ac
xa
Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2
Solution: 2 4 2 0x x Write in standard
form.
Here a = 1, b = – 4, c = 2
2 42
b b acx
a Quadratic formula.
Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2
Solution: 2 4
2b b ac
xa
Quadratic formula.
2( ) ( ) 4( )( )2(
4 24 11)
The fraction bar extends under – b.
Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2
Solution: 2( ) ( ) 4( )( )
2(4 24 1
1)
The fraction bar extends under – b. 4 16 8
2
4 2 22
16 8 8 4 2 2 2
Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS)
Solve x2 – 4x = – 2
Solution:
Factor first, then divide.
4 2 22
16 8 8 4 2 2 2
2
2
2 2 Factor out 2 in the numerator.
2 2 Lowest terms.
The solution set is 2 2 .
Example 6 USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS)
Solve 2x2 = x – 4.
Solution: 22 4 0x x Write in standard form.
2( 1) ( 1) 4(2)(4)2(2)
x Quadratic formula;
a = 2, b = – 1, c = 4
1 1 324
Use parentheses and substitute carefully to
avoid errors.
Example 6 USING THE QUADRATIC FORMULA (NONREAL COMPLEX SOLUTIONS)
Solve 2x2 = x – 4.
Solution:
1 1 324
411 3
x
1 i
The solution set is1 31
.4 4
i
Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA
Solve for the specified variable. Use when taking square roots.
Solution
a.2
, for 4d
A d
2
4A
d Goal: Isolate d, the specified
variable.24A d Multiply by 4.
24Ad
Divide by .
Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA
Solve the specified variable. Use when taking square roots.
Solution
a.2
, for 4d
A d
See the Note following this
example.
Square root property
24Ad
Divide by .
4d
A
Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA
Solve the specified variable. Use when taking square roots.
Solution
a.2
, for 4d
A d
Square root property
4d
A
4Ad
Rationalize the denominator.
Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA
Solve the specified variable. Use when taking square roots.
Solution
a.2
, for 4d
A d
4Ad
Rationalize the denominator.
4d
A
Multiply numerators; multiply denominators.
Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA
Solve the specified variable. Use when taking square roots.
Solution
a.2
, for 4d
A d
4d
A
Multiply numerators; multiply denominators.
2d
A
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.
Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA
Solve the specified variable. Use when taking square roots.
Solution
b.
Write in standard form.
2 ( 0), for rt st k r t 2 0t tr s k
Now use the quadratic formula to find t.
2 42
tb b ac
a
Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA
Solve the specified variable. Use when taking square roots.
Solution
b.
a = r, b = – s, and c = – k
2 ( 0), for rt st k r t 2 4
2t
b b aca
2( ) ( ) 4( )( )2
ts s r k
r
Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA
Solve the specified variable. Use when taking square roots.
Solution
b.
a = r, b = – s, and c = – k
2 ( 0), for rt st k r t
2( ) ( ) 4( )( )2
ts s r k
r
2 42s k
rt
s r Simplify.
The Discriminant
The Discriminant The quantity under the radical in the quadratic formula, b2 – 4ac, is called the discriminant.
2
24ab
ab
xc Discriminant
The Discriminant
DiscriminantNumber of Solutions
Type of Solution
Positive, perfect square
Two Rational
Positive, but not a perfect square
Two Irrational
ZeroOne
(a double solution)Rational
Negative TwoNonreal complex
Caution The restriction on a, b, and c is important. For example, for the equation
2 5 1 0x x
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
2x
Example 9 USING THE DISCRIMINANT
Determine the number of distinct solutions, and tell whether they are rational, irrational, or nonreal complex numbers.
Solution
a. 25 2 4 0x x
For 5x2 + 2x – 4 = 0, a = 5, b = 2, and c = – 4. The discriminant is
b2 – 4ac = 22 – 4(5)(– 4) = 84The 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.
Solution
b. 2 10 25x x
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) = 0There 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.
Solution
c. 22 1 0x x
For 2x2 – x + 1 = 0, a = 2, b = –1, and c = 1, sob2 – 4ac = (–1)2 – 4(2)(1) = –7.
There are two distinct nonreal complex solutions. (They are complex conjugates.)