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Linear and quadratic inequalities

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Chapter 9

Chapter 9

We’ll be looking at linear inequalities in the following form where A, B, and C are real numbers:

Ax + By < C

Ax + By ≤ C

Ax + By > C

Ax + By ≥ C

The set of points that satisfy a linear inequality can be called the solution set or the solution region.

- A boundary is a line or curve that separates the Cartesian plane into two regions.
- It may or may not be part of the solution region.
- Drawn as a solid line and included in the solution region if the inequality involves ≤ or ≥.
- Drawn as a dashed line and not included in the solution region if the inequality involves < or >.

a) Graph 2x + 3y ≤ 6.

b) Determine if the point (–2, 4) is part of the solution.

Solve for y.

- b) Try putting the point into the inequality and seeing if it holds true.
- 2x + 3y ≤ 6
- 2(–2) + 3(4) ≤ 6
- –4 + 12 ≤ 6
- 8 ≤ 6
- This is a false statement, so (–2, 4) is not part of the solution region.

Try graphing 10x – 5x > 0.

Write an inequality to represent the graph.

- Recall slope-intercept form:
- y = mx + b
- b is y-intercept
- b = 1
- m is the slope
- How do you calculate slope?

- Rise over run!
- Choose two points.
- (0, 1) and (1, 3)
- Rise: 3 – 1 = 2
- Run: 1 – 0 = 1
- So: Rise/Run = 2/1 = 2
- m = 2

- Is the graph less than or greater than?
- Is it greater than or greater than or equal to?

y > 2x + 1

Suppose that you are constructing a tabletop using aluminum and glass. The most that you can spend on materials is $50. Laminated safety glass costs $60/m2, and aluminum costs $1.75/ft. You can choose the dimensions of the table and the amount of each material used. Find all possible combinations of materials sufficient to make the tabletop.

Independent Practice

Chapter 9

Solve x2 – 2x – 3 ≤ 0.

Method One:

Graph the function:

For which values of x is the function below the x-axis (below 0 on the y-axis)?

The solution set is all real values of x between –1 and 3, inclusive.

{x | –1 ≤ x ≤ 3, x E R}

Solve x2 – 2x – 3 ≤ 0.

- The x-axis is being divided into three intervals. We need to test points from each interval to see which works.
- Interval One: Try x = –2
- (–2)2 – 2(–2) – 3 ≤ 0
- 5 ≤ 0
- Doesn’t work
- Interval Two: Try x = 0
- 02 – 2(0) – 3 ≤ 0
- –3 ≤ 0
- Works!
- Interval Three: Try x = 4
- (4)2 – 2(4) – 3 ≤ 0
- 5 ≤ 0
- Doesn’t work

Method Two:

Consider x2 – 2x – 3 = 0. Factor it.

(x – 3)(x + 1) = 0

The roots are 3 and –1.

Put these values on a number line, with closed circles since it’s “less than or equal to.”

{x | –1 ≤ x ≤ 3, x E R}

Solve –x2 + x + 12 < 0.

Case 1:

-3

4

Method 3: Case Analysis

- Factor:
- –(x2 – x – 12) < 0
- –(x – 4)(x + 3) < 0
- Either both factors need to positive, or they are both negative.
- Case 1: x – 4 < 0 and x + 3 < 0
- x < 4 and x < –3
- Case 2: x – 4 > 0 and x + 3 > 0
- x > 4 and x > –3

-3

4

Case 2:

-3

4

-3

4

The solution set is values for x for each case that satisfy both inequalities. So, here it is where x < –3 or x > 4.

{x | x < –3 or x > 4, x E R}

Solve 2x2 – 7x > 12.

Rewrite the inequality: 2x2 – 7x – 12 > 0

Find the roots using the

quadratic formula:

Consider the graph:

If a baseball is thrown at an initial speed of 15 m/s from a height of 2 m above the ground, the inequality –4.9t2 + 15t + 2 >0 models the time, t, in seconds, that the baseball is in flight. During what time interval is the baseball in flight?

Independent Practice

Chapter 9

Graph y < –2(x – 3)2 + 1

Determine if the point (2, –4) is a solution to the inequality.

- Graph the parabola y = –2(x – 3)2 + 1
- Draw it with a dotted line, since it’s “<“.
- Choose two test points.
- (0, 0) and (3, –3)
- First point:
- y < –2(x – 3)2 + 1
- 0 < –2(0 – 3)2 + 1
- 0 < –17
- Doesn’t work.
- Second point:
- –3 < –2(3 – 3)2 + 1
- –3 < 1
- Works.

The portion of the graph to be shaded is the portion that includes (3, –3).

Graph y < –2(x – 3)2 + 1

Determine if the point (2, –4) is a solution to the inequality.

- b) We can see from the graph that the point is a solution to the inequality.
- We should check algebraically:
- y < –2(x – 3)2 + 1
- –4 < –2(2 – 3)2 + 1
- –4 < –2 + 1
- –4 < –1
- This is true, so the point is a solution to the inequality.

Graph y ≥ x2 – 4x – 5. Check if the point (–2, 4) is a solution to the inequality, both graphically and algebraically.

You can use a parabolic reflector to focus sound, light, or radio waves to a single point. A parabolic microphone has a parabolic reflector attached that directs incoming sounds to the microphone. René, a journalist, is using a parabolic microphone as he covers the Francophone Summer Festival of Vancouver. Describe the region that René can cover with his microphone if the reflector has a width of 50 cm and a max depth of 15 cm.

Samia and Jerrod want to learn the exhilarating sport of alpine rock climbing. They have enrolled in one of the summer camps at the Cascade Mountains in southern British Columbia. In the brochure, they come across an interesting fact about the manila rope that is used for rappelling down a cliff. It states that the rope can safely support a mass, M, in pounds, modelled by the inequality M ≤ 1450d2, where d is the diameter of the rope, in inches. Graph the inequality to examine how the mass that the rope supports is related to the diameter of the rope.

Independent Practice