Aim: How do we solve quadratic inequalities?

Aim: Quadratic Inequalities Course: Adv. Alg. & Trig.

Aim: How do we solve quadratic inequalities?

Do Now:

What are the roots for

y = x2 - 2x - 3?


Graph y = x2 - 2x - 3

wherethe parabola

crossesthe x-axis

Finding the roots/zeroes:

Graphically:

-1,0 3,0x-axis

y = 0 represents the x-axis and the solution to

quadratic x2 - 2x - 3 = 0 is found at the intersection of

the parabola and x-axis

0 = x2 - 2x - 30 = (x - 3)(x + 1)x = 3 and x = -1

factorand solve

for x.

Algebraically:

x-intercepts


Graphing a Linear Inequality

Graph the inequality y - 2x > 2 y - 2x > 21. Convert to standard form +2x +2x

y > 2 + 2x2. Create Table of Values

y =

2 +

2x

3. Shade the region above the line.

x y2 + 2x

0

1

2

2

4

6

2 + 2(0)

2 + 2(1)

2 + 2(2)

4. Check the solution by choosing a point in the shaded region to see if it satisfies the inequality

(-2,4)

4 - 2(-2) > 24 - (-4) > 2 8 > 2

y - 2x > 2

Note: the line is now

solid.


Graphing a Linear Inequality

An inequality may contain one of these four symbols:

<, >, >, or <.

y < mx + b y > mx + b

The boundary line is part of the solution. It is drawn as a

SOLID line.

The boundary line is not part of the solution. It is drawn as a

DASHED line.

y < mx + b y > mx + b



Do Now:

Graph: y – 3x < 3

Solve: y – 3x2 = 9x – 12


y > x2 - 2x - 3 y < x2 - 2x - 3

Quadratic Inequalities - Graphically y > x2 - 2x - 3 & y < x2 - 2x -

3

-1,0 3,0 -1,0 3,0

-1 < x < 3 x < -1 or x > 3

> Shaded inside the curve < Shaded outside the curve

The values of x found within the shaded regions.0 > x2 - 2x - 3 0 < x2 - 2x - 3

What values of x satisfy these inequalities when y = 0

(x, y)

(x, y)


Graph y ≥ x2 - 1 or x2 - 1 ≤ y

Because y is greater than or

equal to (≥) x2 - 1, the parabola is

shaded inside the curve and

includes the curve itself

Graphically:

-1 ≤ x ≤ 1

What values of x satisfy the quadratic inequality when y = 0?

x-axis

(x2 – 1 = 0)

-1,0 1,0

roots


Exceptions - 1

What values of x satisfy the quadratic inequality

0 > x2 - 4x + 4?

y < x2 - 4x + 4 Solution: {x| x = 2}

y > x2 - 4x + 4x = 2 root/zero


0 < x2 - 4x + 4?

y > x2 - 4x + 4

Solution:

= (x - 2)(x - 2)0 =

Quadratic Inequalities that have roots that are equal

(2,0)

(2,0)

0

y > x2 - 4x + 4

0


Exceptions - 2

Quadratic Inequalities that have no roots.


0 > x2 + 1?

Solution: {x| x = }


0 < x2 + 1?

y > x2 + 1

Solution: {x| x = }(0, 1)

y < x2 + 1

(0, 1)


General Solutions

of Quadratic Inequalities where a > 0 and r1 < r2

(r1 and r2 are the unequal roots)

Quadratic Inequality

Solution Interval

Graph of Solution

ax2 + bx + c < 0 r1 < x < r2

ax2 + bx + c < 0 r1 < x < r2

ax2 + bx + c > 0 r1 < x or

x > r2

ax2 + bx + c > 0 r1 < x or

x > r2

r1 r2

r1 r2

r1 r2

r1 r2


4

2

-2

-4

-5 5

f x = x2-2x-3

Critical Numbers & Test Intervals

x2 – 2x – 3 < 0

(-1,0) (3,0)

roots, orzeros

Critical Numbers fortesting the inequality

(x + 1)(x – 3) = 0

x = -1 and x = 3 are the roots or the zeros that create 3 test

intervals

(-, -1)

(-1, 3)

(3, )

Test Interval

Representative x-value

Value of Polynomial

x = -3 (-3)2 – 2(-3) – 3 = 12

x = 0 (0)2 – 2(0) – 3 = -3

(-, -1)

(-1, 3)

(3, ) x = 5 (5)2 – 2(5) – 3 = 12

4

2

-2

-4

-5 5

f x = x2-2x-3

(-1, 3)

Is this value < 0?

NoYes

No


Model Problems

Test Interval

Representative x-value

Value of Polynomial

Solve algebraically and Graph:

y < x2 – 12x + 27

0 < x2 – 12x + 27


Model Problems

Graph the solution set for

x2 – 2 > -x – 3


Regents Question

Which graph best represents the inequality y + 6 > x2 – x?

6

4

2

-2

-4

-6

-8

-5 5

6

4

2

-2

-4

-6

-8

-5 5

-5 5

6

4

2

-2

-4

-6

-8

-5 5

6

4

2

-2

-4

-6

-8

1) 2)

3) 4)


Model Problems

Graph the solution set for

2(x – 2)(x + 3) < (x – 2)(x + 3)


Because 0 ≥ x2 - 1, • x2 - 1 must be a negative number or 0

Solve 0 ≥ x2 - 1 algebraically

-1 ≤ x ≤ 1

Algebraically:

0 ≥ x2 - 1

0 ≥ (x - 1)(x + 1)

x ≥ 1 and x ≥ -1

?0 ≥ (x - 1) 0 ≥ (x + 1)

• a negative number is the product of a positive & negative #

one of the factors must be positive and the other negative

If ab < 0, then a < 0 and b > 0, or a > 0 and b < 0.

-1,0 1,0

roots


Solve 0 ≥ x2 - 1 algebraically (con’t)x2 - 1 ≤ 0

(x - 1)(x + 1) ≤ 0

(x - 1) ≥ 0 (x + 1) ≤ 0 (x - 1) ≤ 0 (x + 1) ≥ 0

x ≥ 1 x ≤ -1 x ≤ 1 x ≥ -1

and

and

and

and

x CANNOT be a number less than or

equal to -1 and greater than or equal to 1.

EXTRANEOUS

Set quadratic = 0

Factor

What values of x can satisfy both inequalities for each set?

“What values of x are less than or equal to 1

and greater than or equal to -1?”

-1 ≤ x ≤ 1

KEY WORD - “and”



Do Now:

Graph the inequality

y ≥ x2 – 1

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Aim: How do we solve quadratic inequalities?