INEQUALITIES

QUADRATIC

INVOLVING

FUNCTIONS

We are going to use what we’ve

learned about the graphs of quadratic

functions to solve quadratic

inequalities.

You walk directly east from your house one block. How far from your house are you?

1 block1 block

You walk directly west from your house one block. How far from your house are you?

It didn't matter which direction you walked, you were still 1 block from your house.

This is like absolute value. It is the distance from zero. It doesn't matter whether we are in the positive direction or the negative direction, we just care about how far away we are.

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8

44 4 units away from 044 4 units away from 0

6x What we are after here are values of x such that they are 6 away from 0.

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8

6 and -6 are both 6 units away from 0

6or 6 xx

104 xThe "stuff" inside the absolute value signs could = 10 (the positive direction) or the "stuff" inside the absolute value signs could = -10 (the negative direction)

104 x 104 x 14or 6 xx

Let's check it: 1046

1010

10414

1010

Let's look at absolute value with an inequality. 5x

This is asking, "For what numbers is the distance from 0 less than 5 units?"

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8 Everything

inbetween these lines is less than 5 units away from 0

55 x

Inequality notation

)5,5(

Interval notation

So if we have it is equivalent to ax axa

This means x is greater than -a and x is less than a(or x is inbetween -a and a)

What if the inequality is greater than? 5x

This is asking, "When is our distance from 0 more than 5 units away?"

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8 Everything outside these

lines is more than 5 units away from 0

5x 5x

So if we have it is equivalent to ax axax or

Everything outside these lines is more than 5 units away from 0We'll have to express

this with two difference pieces

OR

In interval notation: ),5(or )5,(

3321 xThis means if there are other things on the left hand side of the inequality that are outside of the absolute value signs, we must get rid of them first.

We must first isolate the absolute value.

+3 +3

6

621 xFrom what we saw previously, the "stuff" inside the absolute value is either less than or equal to -6 or greater than or equal to 6

621or 621 xxIsolate x, remembering that if you multiply or divide by a negative you must turn the sign.

72 x 52 x- 2 - 2 - 2 - 2

We are dividing by a negative so turn the signs!

2

7x

2

5xor

2 -7 -6 -5 -4 -3 -2 -1 1 5 7 3 0 4 6 8 []

025 xxConsider the inequality:

Let’s look at the graph 25 xxxf

It is a parabola with x intercepts (5, 0) and (-2, 0) and y intercept (0, -10).

We could also find the vertex by FOILing and then using –b/2a but we will see that this will not be necessary to solve the inequality.

025 xx is still the problem we want to solve

We have a graph of: 25 xxxf

Since the left hand side of the inequality is f(x), the inequality is asking, “Where is f(x) > 0”?

Look at the graph to answer this question. f(x) is the y value so where is the function value or y value above the x axis?

(-,-2) or (5, )

Since the left hand side of the inequality is f(x), the inequality is asking, “Where is f(x) > 0”?

0xf

1272 xxLet’s try another one:

First of all we want the right hand side to be zero so that when we look at the graph we are looking where the function is either above or below the x axis depending on the inequality.

01272 xx Factor

034 xx Graph f(x) = left hand side, by finding intercepts

34 xxxf

Where is this graph less than or equal to 0?

[-4, -3]Using interval notation:

Using inequality notation: 34 xx

322 xxLet’s try another one:

We want the right hand side to be 0 and then factor

Graph f(x) = left hand side, by finding intercepts and knowing it is a parabola opening up. 13 xxxf

Where is this graph (the y value) greater than 0?

(- , -3) or (1, )Using interval notation:

Using inequality notation:

1or 03 xxx

013 xx0322 xx