Graph quadratic equations. Complete the square to graph quadratic equations. Use the Vertex Formula...

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•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

– 2 4

– 1 1

– 3– 4

range[0, )

2x xfdomain (−, )

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

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

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

Domain (−, )

Range[– 6, )

Solution

Domain (−, )

Range(–, 0]

Solution

2F x x

Domain (−, )

Range(–, 3]

2x x F

(4, 3)

– 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.

Solution Express x2 – 6x + 7 in the form (x– h)2 + k by completing the square.

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)

Solution

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, )

Solution To complete the square, the coefficient of x2 must be 1.

23 2 1x x x f

3 1x x x

f Factor – 3 from the first two terms.

2 1 19

2 21 1 1

; add2 3 9

1and subtra

Solution

23 2 1x x x f

2 2 1 13 3 1

3 9 9x x x

Distributive property

3 3x x

Be careful here.

Factor; simplify.

Solution

23 2 1x x x f

3 3x x

f Factor; simplify.

1 4The vertex is , .

Solution Intercepts are good additional points to find. Here is the y-intercept.

23 2 1x x x f

Let x = 0. 23 2 10 0 1y

Solution The x-intercepts are found by setting (x) equal to 0 in the original equation.

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.

3x x Zero-factor property

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

2 , 0,y x a x h k a f

and .2b

h k ha

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.

2 2 21

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.

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

Solution:

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

Solution:

26 7 3x x

3 1 0 or 2 3 0x x Zero-factor property.

3 1x or 2 3x

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

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

Solution:

By the square root property, the solution set is 17

Solve each quadratic equation.

Solution:

the solution set of x2 = − 25

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

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.

( )4 4;2

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)

3x x Add – 1. (Step 2)

2 4 49

1 4 4; add

432 9 9

Solve 9x2 – 12x + 9 = 0 by completing the square.Solution 2 4 4

1 4 4; add

432 9 9

22 53 9

Factor, combine terms. (Step 4)

2 53 9

x Square root property

Solve 9x2 – 12x + 9 = 0 by completing the square.Solution

2 53 9

Quotient rule for radicals

2 53 3

x i a i a

Square root property

2 53 3

x i Add ⅔.

The solution set is

Solution 2 5

3 3x i Add ⅔.

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

2b b ac

Caution Notice that the fraction bar in the quadratic formula extends under the – b term in the numerator.

2b b ac

Example 5 USING THE QUADRATIC FORMULA (REAL SOLUTIONS)

Solve x2 – 4x = – 2

Solution: 2 4 2 0x x Write in standard

Here a = 1, b = – 4, c = 2

b b acx

a Quadratic formula.

Solve x2 – 4x = – 2

Solution: 2 4

2b b ac

Quadratic formula.

2( ) ( ) 4( )( )2(

4 24 11)

The fraction bar extends under – b.

Solve x2 – 4x = – 2

Solution: 2( ) ( ) 4( )( )

2(4 24 1

The fraction bar extends under – b. 4 16 8

4 2 22

16 8 8 4 2 2 2

Solve x2 – 4x = – 2

Solution:

Factor first, then divide.

4 2 22

16 8 8 4 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

The solution set is1 31

Example 8 SOLVING FOR A QUADRATIC VARIABLE IN A FORMULA

Solve for the specified variable. Use when taking square roots.

Solution

, for 4d

d Goal: Isolate d, the specified

variable.24A d Multiply by 4.

Divide by .

Solve the specified variable. Use when taking square roots.

Solution

, for 4d

See the Note following this

example.

Divide by .

Solution

, for 4d

Rationalize the denominator.

Solution

, for 4d

Rationalize the denominator.

Multiply numerators; multiply denominators.

Solution

, for 4d

Multiply numerators; multiply denominators.

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.

Solution

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.

tb b ac

Solution

a = r, b = – s, and c = – k

2 ( 0), for rt st k r t 2 4

b b aca

2( ) ( ) 4( )( )2

ts s r k

Solution

a = r, b = – s, and c = – k

2 ( 0), for rt st k r t

2( ) ( ) 4( )( )2

ts s r k

2 42s k

s r Simplify.

The Discriminant

The Discriminant The quantity under the radical in the quadratic formula, b2 – 4ac, is called the discriminant.

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

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.

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.”

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.)

Graph quadratic equations. Complete the square to graph quadratic equations. Use the Vertex Formula...

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