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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (15)

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Post by julius mogyorossy on June 18, 2015

I didn't explain it as simply as I should have, about seeing where the solution set is going to be. You don't need a test point, it tells you right where it is going to be, it is obvious, it could not be more obvious. Look at example one, it say it said that y<x'2+4, that means that the y values for points in the solution set are going to be less than the y values on the line that make up the parabola, or in this case, the boundary. They are going to be below the parabola, what they would call in this case, outside the parabola, since the parabola is facing up. Since the vertex is at (0, 4) ever other x point at x=0 which is below 4, below the parabola is part of the solution set. Same holds true for every other infinite x point on the parabola, every other infinite x point at x=2 that has a corresponding y value that is less than the y value on the boundary, or on the parabola line is going to be part of the solution set, ie, x=8, y=7, x=8, y=-6, x=8, y=-infinity, any point, in this case, that is below the parabola, what they would call outside, is part of the solution set, I see the parabola, if it opens up or down, as being a horizontal line. It does not matter if the parabola opens up or down, if it says y is less than, that means below the parabola, as I see it, or course if y is greater than x'2 whatever, then above. I don't understand Dr. Carleen's logic, it is above my head, maybe when I have a PHD in mathematics. I am going to have to stop my Educator.com for a while, because I am a loser, because I am being blacklisted, it breaks my heart. Dr. Carleen, I love you, please forgive me, as the phony buddha would say, you really turned me on, to mathematics, when I am the most famous person on the planet, when I am a God, a trillionaire, call me if you need a favor, I shall make sure you have my number, I am your Godfather now, funny, they always compare me to mafia figures, which cracks me up since I am such a goofball, except in war, combat. Forgive my ADD, not time to proof. PS, same logic holds true for parabolas that open the the left or write, as they would say it, if x is less than y'2 whatever, the solution set is to the left of the parab, can you dig it.

1 answer

Last reply by: Dr Carleen Eaton
Sat Nov 1, 2014 4:11 PM

Post by Smriti Sharan on October 19, 2014

Q. Graph the inequality y < x2 − 6x + 12
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Step 1. Find the vertex x = − [b/2a]
Step 2. x = − [b/2a] = − [(( − 6))/2] = 3
Step 3. f(3) = (3)2 − 6(3) + 12 = 3

In this practice question, shouldn't x equal to -3?

1 answer

Last reply by: Dr Carleen Eaton
Thu Mar 27, 2014 6:54 PM

Post by Christopher Lee on January 22, 2014

At 11:13, you said that when x<=-3, y is positive, but 0 isn't considered positive, so it should be when x<-3, not equal to.

1 answer

Last reply by:
Sat Dec 14, 2013 7:58 AM

Post by julius mogyorossy on September 24, 2013

Funny to me, I used the midpoint formula for the x dimension, from the Geometry I have learned so far, to find the axis for two roots, to find the vertex, I am starting to put it all together. I have incredible instincts, I think I am going to discover something incredible in Geometry. I have already figured out a way to do my spinning kicks in the air in my apartment much less dangerously because of geometry. Once I ended up hanging from my heavy bag upside down hanging via one foot and my jump rope on my heavybag, that one commercial is making fun of me.

0 answers

Post by julius mogyorossy on June 10, 2013

Dr. Eaton, it seems you said that x'2+x-6 is y, but it seems to me that the 0 is y. This seems to be shown to be the case because you end of with x>2, and one of the roots is (2,0) where x=2 and y=0. I am also confused by why you say the solution set is x<-3 and x>2, when the quadratic equation clearly says it is x>-3, x>2. From the way mathematicians look at it it seems the negative solution is a lie, and the positive one the truth. The way I find the solution set is to look at the quadratic inequality when it is in the form where the 0, the y, is all alone on the right, then it is obvious where it is, it is telling you that 0, y values for the solution set, are less than x'2+x-6, which is the graph line that you will draw, since, a, is positive and it will face up, and y values for the solution set are less than the graph line they must be below that graph line, on the outside, since it faces up, it it faced downwards, on the inside. This method I call the Julius supergenius way of finding the solution set, maybe this what you were saying when you talked about the logical way to find the solution set, if so I could not understand so.

1 answer

Last reply by: Dr Carleen Eaton
Thu Jun 6, 2013 11:05 PM

Post by Kavita Agrawal on June 6, 2013

The last example was greater than, so the boundary should have been dotted, not solid.

0 answers

Post by julius mogyorossy on March 25, 2013

As soon as you see -x'2 + x + 2 >y, -x'2 + x + 2 < y, x'2 + x + 2 > y, or, x'2 + x + 2 < y, you should know instantly where the solutions are at, the equations tell you even without finding the roots, sometimes the roots are not the roots, have you noticed.

1 answer

Last reply by: Dr Carleen Eaton
Tue Feb 19, 2013 7:44 PM

Post by Kenneth Montfort on February 13, 2013

How do you know when to combine the solutions and when not to combine the solutions? I keep hearing "this 'and' that" or "this 'or' that".

1 answer

Last reply by: Dr Carleen Eaton
Mon Jul 16, 2012 7:53 PM

Post by George Welborn on July 14, 2012

At around 11:08 you said "at points when x is less than or equal to negative three or greater than or equal to negative two" but it should be just two, not negative two.

Graphing and Solving Quadratic Inequalities

  • Use the same techniques that were used to solve linear inequalities. Follow the conventions about how to draw the boundary curve (solid line or dotted line) and use a test point to determine the region that is the solution set. Sometimes, more than one test point must be used.

Graphing and Solving Quadratic Inequalities

Graph the inequality y ≤ x2 + 8x + 19
  • Find the vertex x = − [b/2a]
  • x = − [b/2a] = − [8/2] = − 4
  • f( − 4) = ( − 4)2 + 8( − 4) + 19 = 3
  • Vertex = ( − 4,3)
  • Locate three points on the parabola to the right of the vertex. Use property of symmetry to sketch the other half of the parabola
  • x y=x2+8x+19
    -3 4
    -2 7
    -1 12
  • Graph the parabola using the vertex and the three additional points, as well as the other points you found using property of symmetry. The line should be a solid line.
  • Test which way to shade using the point (0,0) since it is not on the parabola.
  •   Test Using (0,0)
    y ≤ x2+8x+19  
  •   Test Using (0,0)
    y ≤ x2+8x+19 0 ≤ (0)2 +8(0)+19
      0 ≤ 19
      True, Shade Towards (0,0)
  • Shade from the line of the parabola outwards towards and including (0,0)
Graph the inequality yx2 − 8x + 19
  • Find the vertex x = − [b/2a]
  • x = − [b/2a] = − [(( − 8))/2] = 4
  • f(4) = (4)2 − 8(4) + 19 = 3
  • Vertex = (4,3)
  • Locate three points on the parabola to the right of the vertex. Use property of symmetry to sketch the other half of the parabola
  • x y=x2−8x+19
    5 4
    6 7
    7 12
  • Graph the parabola using the vertex and the three additional points, as well as the other points you found using property of symmetry. The line should be a solid line.
  • Test which way to shade using the point (0,0) since it is not on the parabola.
  • Test using (0,0)
  • yx2 − 8x + 19
  • 0 ≥ (0)2 − 8(0) + 19
  • 0 ≥ 19
  • Not True, shade inside the parabola only
  • Shade from the line of the parabola inwards, excluding everything outside the line of the parabola including (0,0)
Graph the inequality y < x2 − 6x + 12
  • Find the vertex x = − [b/2a]
  • x = − [b/2a] = − [(( − 6))/2] = 3
  • f(3) = (3)2 − 6(3) + 12 = 3
  • Vertex = (3,3)
  • Locate three points on the parabola to the right of the vertex. Use property of symmetry to sketch the other half of the parabola
  • x y=x2−6x+12
    4 4
    5 7
    6 12
  • Graph the parabola using the vertex and the three additional points, as well as the other points you found using property of symmetry. The line should be a dashed lined because of the < sign.
  • Test which way to shade using the point (0,0) since it is not on the parabola.
  • Test using (0,0)
  • y < x2 − 6x + 12
  • 0 < (0)2 − 6(0) + 12
  • 0 < 12
  • True, shade outside the parabola
  • Shade everything outside the parabola.
Graph the inequality y < x2 + 4x + 6
  • Find the vertex x = − [b/2a]
  • x = − [b/2a] = − [(4)/2] = − 2
  • f( − 2) = ( − 2)2 + 4( − 2) + 6 = 2
  • Vertex = ( − 2,2)
  • Locate three points on the parabola to the right of the vertex. Use property of symmetry to sketch the other half of the parabola
  • x y=x2−6x+12
    -1 3
    0 6
    1 11
  • Graph the parabola using the vertex and the three additional points, as well as the other points you found using property of symmetry. The line should be a dashed lined because of the < sign.
  • Test which way to shade using the point (0,0) since it is not on the parabola.
  • Test using (0,0)
  • y < x2 + 4x + 6
  • 0 < (0)2 + 4(0) + 6
  • 0 < 6
  • True, shade outside the parabola
  • Shade everything outside the parabola.
Graph the inequality y > x2 + 2x + 4
  • Find the vertex x = − [b/2a]
  • x = − [b/2a] = − [(2)/2] = − 1
  • f( − 1) = ( − 1)2 + 2( − 1) + 4 = 3
  • Vertex = ( − 1,3)
  • Locate three points on the parabola to the right of the vertex. Use property of symmetry to sketch the other half of the parabola
  • x y=x2+2x+4
    0 4
    1 7
    2 12
  • Graph the parabola using the vertex and the three additional points, as well as the other points you found using property of symmetry. The line should be a dashed lined because of the > sign.
  • Test which way to shade using the point (0,0) since it is not on the parabola.
  • Test using (0,0)
  • y > x2 + 2x + 4
  • 0 > (0)2 + 2(0) + 4
  • 0 > 4
  • Not True, shade inside the parabola
  • Shade everything inside the parabola.
Solve x2 − x − 6 > 0
  • Factor out the quadratic.
  • x2 − x − 6 = (x − 3)(x + 2) = 0
  • Solve using Zero Product Property
  • x − 3 = 0 and x + 2 = 0
  • x = 3 and x = − 2
  • Plot the roots and sketch the graph of the parabola given that a is > 0.
  • Test the Inequality using a point not on the graph (0, 0)
  • x2 − x − 6 > 0
  • (0)2 − (0) − 6 > 0
  • − 6 > 0
  • Not True, which means that
  • x < − 2
  • x > 3
Solve x2 + 5x + 4 < 0
  • Factor out the quadratic.
  • x2 + 5x + 4 = (x + 4)(x + 1) = 0
  • Solve using Zero Product Property
  • x + 4 = 0 and x + 1 = 0
  • x = − 4 and x = − 1
  • Plot the roots and sketch the graph of the parabola given that a is > 0.
  • Test the Inequality using a point not on the graph (0, 0)
  • x2 + 5x + 4 < 0
  • (0)2 + 5(0) + 6 < 0
  • 6 < 0
  • Not True, which means that
  • x > − 4 and
  • x < − 1
  • Can also be written as − 4 < x < − 3
Solve x2 − 8x + 7 > 0
  • Factor out the quadratic.
  • x2 − 8x + 7 = (x − 1)(x − 7) = 0
  • Solve using Zero Product Property
  • x − 1 = 0 and x − 7 = 0
  • x = 1 and x = 7
  • Plot the roots and sketch the graph of the parabola given that a is > 0.
  • Test the Inequality using a point not on the graph (0, 0)
  • x2 − 8x + 7 > 0
  • (0)2 − 8(0) + 7 > 0
  • 7 > 0
  • True, which means that
  • x < 1 or
  • x > 7
Solve x2 − 4x + 3 > 0
  • Factor out the quadratic.
  • x2 − 4x + 3 = (x − 1)(x − 3) = 0
  • Solve using Zero Product Property
  • x − 1 = 0 and x − 3 = 0
  • x = 1 and x = 3
  • Plot the roots and sketch the graph of the parabola given that a is > 0.
  • Test the Inequality using a point not on the graph (0, 0)
  • x2 − 4x + 3 > 0
  • (0)2 − 4(0) + 3 > 0
  • 3 > 0
  • True, which means that
  • x < 1 or
  • x > 3
Solve x2 − 2x − 15 < 0
  • Factor out the quadratic.
  • x2 − 2x − 15 = (x + 3)(x − 5) = 0
  • Solve using Zero Product Property
  • x + 3 = 0 and x − 5 = 0
  • x = − 3 and x = 5
  • Plot the roots and sketch the graph of the parabola given that a is < 0.
  • Test the Inequality using a point not on the graph (0, 0)
  • x2 − 2x − 15 < 0
  • (0)2 − 2(0) − 15 < 0
  • − 15 < 0
  • True, which means that
  • x > − 3 or
  • x < 5
  • Can also be written as − 3 < x < 5

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Graphing and Solving Quadratic Inequalities

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Graphing Quadratic Inequalities 0:11
    • Test Point
    • Example: Quadratic Inequality
  • Solving Quadratic Inequalities 3:57
    • Example: Parameter
  • Example 1: Graph Inequality 11:16
  • Example 2: Solve Inequality 14:27
  • Example 3: Graph Inequality 19:14
  • Example 4: Solve Inequality 23:48

Transcription: Graphing and Solving Quadratic Inequalities

Welcome to Educator.com.0000

Today, we are going to be talking about graphing and solving quadratic inequalities.0002

And we are going to apply some techniques learned from linear inequalities in our work.0006

So, in order to graph quadratic inequalities, use the techniques learned for graphing linear inequalities.0012

And recall: that involves graphing the related equation and using a test point,0018

except this time, we are going to be using a quadratic equation.0024

For example, just starting out with something simple: if you had an inequality y > x2 - 2,0029

the first step is to graph the related equation; and in this case, the related equation is going to be y = x2 - 2.0038

And that is going to give the boundary line for the solution set.0048

So, looking at this form, this is actually in vertex form; and here, a is 1; h is 0; and k is -2.0052

So, what this tells me is that the vertex of my graph is at (0,-2).0064

So, I have the vertex at (0,-2), right here.0071

And I also know that a is positive; a is greater than 0; so I know that the parabola opens upward.0074

So, I have a general idea of the shape of this graph before I even get started.0085

Now, to make a more exact graph, I am going to find some points.0089

And the first point is going to be -2, squared is 4, minus 2 is 2.0094

-1 squared is 1, minus 2 is -1; and then, I already have (0,-2); this is my vertex.0100

1 squared is 1, minus 2 is -1; and now, 2 squared is 4, minus 2 is 2.0110

This is enough for me to go ahead and graph.0119

(-2,2), (-1,-1), the vertex I already have graphed...actually, this goes -1, -1 is right there; and then (1,-1); (2,2).0122

So, I can see the shape of the graph.0146

Now, since this is a strict inequality, the boundary line is going to be dashed; it is not going to be part of the solution set.0148

And as expected, this parabola faces upward.0159

So, just as with linear inequalities, my first step is to graph the related equation.0163

My second step is to use a test point; and recall that a good test point to use is...I am going to use the test point (0,0),0167

because it is not near the boundary line, and it is a very easy point to use.0178

So, I am going to substitute (0,0) into this inequality, y > x2 - 2; so 0 is greater than 0 squared minus 2.0183

So, 0 is greater than -2; and that is true.0193

Since this is true, this test point is part of the solution set.0199

Now, the boundary line divides this into two sections: the area inside the parabola here, and the area outside the parabola.0205

And what the test point tells me is that, since the test point is inside the parabola, the solution set for the inequality is also inside the parabola.0214

If this had turned out to be not true, and the test point wasn't part of the solution set, then the solution set would have been out here.0223

OK, so that was graphing quadratic inequalities.0233

Now, we are going to talk about actually solving quadratic inequalities.0237

In this case, we are going to again graph the related quadratic function.0241

But then, we are going to find the roots of the related quadratic equation.0245

So, we are looking for something slightly different: remember, what I just looked at was this,0249

where I just had that y is greater than some function.0256

Now, this time, I am actually going to say something like x2 + x - 6 ≥ 0.0262

So, I actually have a number here; I have a parameter; I am saying that this function value is greater than or equal to a specific value.0270

So, I am going to start out by finding the roots, because by finding the roots, my graphing will be easier,0279

because recall that the roots give me the x-intercept.0289

Let's write the related quadratic--let's go ahead and find the roots by writing this as a quadratic equation.0300

And looking at this, as you know, there are many different ways to solve quadratic equations.0312

In this case, factoring would be an easy way; so I can see that I have a negative here, so I have x + something and x - something.0317

And I also know that my middle term is x; so I have to look at factors of 6, which are 1 and 6, and 2 and 3,0326

and look for a combination here that will add up to 1x.0338

And it actually turns out to be (x + 3) (x - 2), because that would give me -2x + 3x, to give me x for the middle term.0343

So, I did my factoring; and then, using the zero product property, I know that x + 3 = 0 and x - 2 = 0.0354

So, this gives me x = -3 and x = 2, so those are my roots: x = -3, and it can also equal 2.0363

That means that my x-intercepts are at -3 (this is -3) and 2; those are the roots.0380

OK, the next thing I am going to do in order to have my graph is find the vertex.0400

And recall that the vertex equals -b/2a; the x-value for the vertex equals -b/2a.0406

Here, I have that b is 1, so this is -1; a is also 1; so this is going to give me -1/2.0420

At the vertex, x equals -1/2; in order to find the y-value, I am going to substitute in.0430

And I am going to find out that I am going to get (-1/2)2 + 1/2 - 6.0441

So, I need to solve this to get my y-value; and this is going to give me 1/2...0455

this is actually -1/2 right here...(1/2)2 is going to give me 1/4, minus 1/2,0468

minus 6, which equals -1/4 - 6, which equals -6 and 1/4.0476

OK, so now I have my vertex; my vertex is at (-1/2, -6 1/4).0484

So, I can't graph that exactly; but 1, 2, 3, 4, 5, 6...-6 will be right here; and -1/2, -6 1/4 places it about there.0492

Now, I also know that, since a is positive (a is greater than 0), this is going to open upward.0509

So, I have my vertex here; and I know that this is going to open upward.0518

And this is greater than or equal to, so I can use a solid line; OK.0529

Now, what we are looking for, if we go back and think about this, is times when f(x), or y, is greater than or equal to 0.0536

I can use a test point, or I can just use logic.0546

I can look and see that here, at these x-intercepts, this is where y equals 0.0550

Beyond that, up here...it was when y equals 0; y equals 0 right here at x = -3, and y equals 0 at x = -2.0558

Beyond this point, for x-values over here, and for x-values over here, y is positive: y is greater than 0.0568

So, I can actually just look at this graph and say, "OK, when x is less than or equal to -3, or when x is greater than or equal to 2, this inequality is satisfied."0577

Now, you can always look at it and do it by logic; or you can use a test point.0597

So, if I use a test point of (0,0), what I would do is go back here and say,0603

"OK, I am going to let x equal 0; and I am going to see if that satisfies this inequality."0611

So, I am using my test point, and then I put it in; now I have 02 + 0 - 6 ≥ 0.0618

This gives me 0 - 6 ≥ 0; -6 ≥ 0.0628

That is not true--no; therefore, this is not part of the solution set.0633

These values in here are not part of the solution set; so that tells me that these are the correct values.0639

And I knew that by logic; but you can always use a test point or verify using a test point.0645

Therefore, my solutions for this inequality, for times when the function (or y) ends up0651

being greater than or equal to 0, are at points when x is less than or equal to -3, or greater than or equal to -2.0657

And you can see on the graph that that is true; when x is big over here, y is positive; when x is less than or equal to -3, y is also positive.0666

OK, first we are just asked to graph this inequality.0677

And we are going to begin by graphing the corresponding quadratic function in order to find the boundary line for the solution.0681

The corresponding function is y = x2 + 4.0692

This gives me an equation with a vertex at (0,4).0701

Now, let's find some points and then do the graphing.0710

When x is -2, this gives me 4 plus 4, or 8; when x is -1, this is 1, plus 4 is 5.0715

I already know that, when x is 0, y is 4, because that is the vertex.0726

And I know that this is a minimum, because the parabola opens upward, because a is positive.0731

So, a is greater than 0, so this opens upward.0736

And when x is 1, this is 1 plus 4; that gives me 5.0746

And when x is 2, that is 4 plus 4; that gives me 8.0751

So, 2, 4, 6, 8, 2, 4, 6, 8; let's start graphing, starting with the vertex at (0,4).0755

And I know this opens upward; and then, when x is -1, y is 5; when x is -2, y is 8.0768

When x is 1, y is 5; and when x is 2, y is 8.0781

So, because this is less than or equal to, I am going to use a solid line for this parabola.0788

And it is opening upward, as I expected.0794

Now, I graphed my corresponding quadratic function; I found this parabola; I graphed it;0798

and now, I need to use a test point--I am going to use the test point of (0,0) once again, because that is an easy point to use.0807

And I am going to go back here, and I am going to say y < x2 + 4.0818

0 is less than or equal to 0 squared plus 4; 0 is less than or equal to 0 plus 4; 0 ≤ 4.0824

And this is true: since this is true, the test point is part of the solution set;0832

and this parabola divides this graph into two sections--the area within the parabola and the area outside the parabola.0839

And since this is true, then my solution set lies out here.0847

So again, this is very similar to graphing linear inequalities, where you find the boundary line0852

by graphing the corresponding equation, and then use a test point to determine where the solution set is.0858

Now, I am asked to solve this quadratic inequality.0868

I am going to start out by finding the roots of the corresponding quadratic equation, x2 - x - 6 = 0.0872

This is also one that is most easily solved through factoring.0880

And since this term is negative, I am going to have x + something, and then x - something.0884

And I know that my factors of 6 are 1 and 6, and 2 and 3, and that I want these to sum up to -1.0892

1 and 6 are too far apart, so I look at 2 and 3.0901

And if I added -3 + 2, I would get -1; therefore, this is (x + 2) (x - 3).0905

And I can check that by saying, "OK, that would give me -3x + 2x, so my middle term would come out to -x."0913

OK, and this equals 0; using the zero product property, I get x + 2 = 0, and x - 3 = 0.0922

So, I solve and get x = -2 and x = 3.0934

So, my roots are right here: -2 and 3.0941

Something else I know is that this is going to open upward, because a is positive.0948

And I am going to go ahead and find the vertex, and that is -b/2a (that is the x-value for the vertex).0953

And that is going to give me negative...and b is -1/2, and a is 1.0964

So, that equals 1/2; so the vertex is at 1/2.0971

Now, to find the y-value, the x-value for the vertex (x = 1/2)--substitute that in to find y (the y point for the vertex).0977

That gives me (1/2)2 - 1/2 - 6.0994

This equals 1/4 - 1/2 - 6, which equals -1/2 - 6...actually, -1/4...that would be minus 2/4, and that would give me -1/4, minus 6 equals -6 1/4.1000

Therefore, the vertex lies at (1/2,-6 1/4).1022

Now I have the vertex, which is going to be right about down here.1033

And I know that this opens upward, and I know that it is going to be a solid line, because this is less than or equal to.1041

Now, what I want to know is, "What are the situations when y, or this function, is less than or equal to 0?"1069

And I can actually just look on here and say, "OK, y is less than or equal to 0 at all these values for x in here."1081

So, when x is less than or equal to 3, I see that the y becomes negative--dips below 0.1095

I could always use a test point of (0,0): substitute in (0,0) here; that would give me...1105

if my test point is (0,0), I am going to get 02 - 0 - 6 ≤ 0, or -6 ≤ 0.1113

And that is true; since that is true, this is part of the solution set.1122

And the solution set is comprised of x ≥ -2 and x < 3.1126

So, that is going to give me x ≥ -2 and ≤ 3.1137

So, that is my solution for this quadratic inequality.1150

OK, now graph the inequality: first graph the corresponding quadratic equation.1155

And I have y = x2 - x - 12, and I am going to start out by finding the vertex at -b/2a, which equals...1163

b is -1, so it is negative -1, over 2...a is 1; that is going to give me 1/2.1181

The x-value for the vertex is going to be 1/2; what is y going to be?1195

y is going to be (1/2)2 - 1/2 - 12, so y = -12 1/4.1202

So, 2, 4, 6, 8, 10, 12; the vertex is going to be here at 1/2, which is going to be about here, all the way down here.1222

And this parabola does open upward, because a is positive.1238

Now, I am going to go ahead and find some more points (in addition to the vertex) to graph.1243

So, when x is -2, if you figure this out, y ends up being -6.1248

When x is -1, this is going to give me 1 + 1; that is 2, so this will come out to -10.1255

When x is 0, that is going to give me -12.1263

When x is 1, that is going to be 1 - 1 is 0, minus 12; so that is going to be -12.1268

When x is 2, that is going to be 4 - 2 is 2, minus 12 is -10.1275

OK, graphing these: this is -2 right here; when x is -2, y is -6.1281

When x is -1, which is right about here, y is all the way down here at -10.1290

When x is 1, y is down here at -12, right about here; when x is 2, we get -10 right here.1297

So, I have a graph right here, and it is actually going to be a dashed line; it is going to give me a parabola that is a dashed line,1315

because I have greater than; so, my boundary line is not part of the solution set.1335

So now, I have this parabola; and I need to use a test point right here.1342

Test point (0,0): substituting in here is going to give me 0 > 02 - 0 - 12.1346

0 is greater than 0 - 12; 0 is greater than -12.1357

OK, so is 0 greater than -12? Yes, therefore, since 0 is greater than -12, this point would be part of the solution set.1366

So, the solution set is going to be in here.1386

Again, we handled this much the way we would handle a linear inequality.1392

And that is by graphing out the corresponding equation, y = x2 - x - 12.1398

And it gave me this parabola, with a vertex way down here that is a minimum that opens upward.1405

Then, I went and found a test point, (0,0), and I substituted this in and came up with 0 > -12.1410

And 0 actually is greater than -12, so this test point is part of the solution set, which would be right here within the parabola.1420

Now, solve this quadratic inequality; first I am going to go ahead and find the roots,1429

which are of the corresponding quadratic equation.1435

And this is another one that I can do by factoring; I am going to have x + something, and I am going to have x - something.1443

And I know that, because I have a negative 10; so I am going to have a positive and a negative.1451

Now, the factors of 10 are 1 and 10, and 2 and 5; and I need something that is going to add up to 3x.1456

And what would give me that is if I had a positive 5 and a negative 2, because then I am going to get -2x + 5x, to get 3x.1464

Using the zero product property, I will have x + 5 = 0, or x = -5; x - 2 = 0 gives me x = 2.1475

So, I have two roots that I found; and these are x = -5 and x = 2: right here and right here.1486

Now, I could find the vertex, and I have been finding it; but you actually don't need it for these.1498

All you need to know is that the zeroes are here, and this graph opens upward.1505

So, I am just going to sketch this out--that this is a parabola that opens upward,1510

and I don't know exactly where the vertex is; that is OK, because I am honestly not concerned with that.1526

What I am concerned with are these points right here.1532

And I am looking for values of x where this function would end up being greater than 0.1534

So, I am going to look over here and see that, whenever x is less than -5, y is greater than 0.1542

Whenever x is greater than 2, y is also positive.1550

I could also use my test point; and again, I use the origin as a test point, as long as it is not too close to the boundary.1559

So, my test point is (0,0); and now I am going to use the x-value and put it in here and find that 02 + 3 times 0 - 10 > 0.1568

Let's see if that is true: 0 + 0 - 10 > 0--is -10 greater than 0? No.1579

So, this test point is not part of the solution set; the solution set is not in here--it is out here, just as I thought.1587

So, what this gives me is a solution set where x is less than -5, or x is greater than 2.1593

So, my solution set are values of x that are less than -5 or greater than 2.1609

That concludes this session of Educator.com on quadratic inequalities.1619

And I will see you next lecture!1624