INSTRUCTORS Carleen Eaton Grant Fraser

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Square Root Functions and Inequalities

• Exclude all values that make the radicand negative.
• The graph of a square root function is half of a parabola.
• The initial point of the graph can be translated horizontally or vertically from the origin, depending on the additive constants in the function.

Square Root Functions and Inequalities

Graph y = √{2x}
• Step 1: Determine the domain
• Since you cannot have a negative inside the radical sign, your domain is limited to values greater or equal to zero inside the radical.
• 2x > 0
• x > 0
• Your values of x can be greater or equal to zero
• Step 2 - Create a table of values
•  x y=√{2x} 2 8 18 32
•  x y=√{2x} 2 = √{2(2)} = √4 = 2 8 = √{2(8)} = √{16} = 4 18 = √{2(18)} = √{36} = 6 32 = √{2(32)} = √{64} = 8
• Step 3 - Draw a smooth curve
Graph y = [1/4]√{2x}
• Step 1: Determine the domain
• Since you cannot have a negative inside the radical sign, your domain is limited to values greater or equal to zero inside the radical.
• 2x > 0
• x > 0
• Your values of x can be greater or equal to zero
• Step 2 - Create a table of values
•  x y = [1/4]√{2x} 2 8 18 32
•  x y = [1/4]√{2x} 2 = [1/4]√{2(2)} = [1/4]√4 = [1/2] 8 = [1/4]√{2(8)} = [1/4]√{16} = 1 18 = [1/4]√{2(18)} = [1/4]√{36} = [6/4] = [3/2] 32 = [1/4]√{2(32)} = [1/4]√{64} = 2
• Step 3 - Draw a smooth curve
Graph y = 6√{3x}
• Step 1: Determine the domain
• Since you cannot have a negative inside the radical sign, your domain is limited to values greater or equal to zero inside the radical.
• 3x > 0
• x > 0
• Your values of x can be greater or equal to zero
• Step 2 - Create a table of values
•  x y = 6√{3x} 3 12 27 81
•  x y = 6√{3x} 3 = 6√{3(3)} = 6√9 = 6*3 = 18 12 = 6√{3x} = 6√{3*12} = 6√{36} = 6*6 = 36 27 = 6√{3*27} = 6√{81} = 6*9 = 54 81 = 6√{3*48} = 6√{144} = 6*12 = 72
• Step 3 - Draw a smooth curve
Graph y = √{x − 2} + 3
• Step 1: Determine the domain
• Since you cannot have a negative inside the radical sign, your domain is limited to values greater or equal to zero inside the radical.
• x − 2 ≥ 0
• x ≥ 2
• Your values of x can be greater or equal to 2
• Step 2 - Create a table of values
•  x y = √{x − 2} + 3 2 6 11 18
•  x y = √{x − 2} + 3 2 = √{2 − 2} + 3 = 0 + 3 = 3 6 = √{6 − 2} + 3 = √4 + 3 = 2 + 3 = 5 11 = √{11 − 2} + 3 = √9 + 3 = 3 + 3 = 6 18 = √{18 − 2} + 3 = √{16} + 3 = 4 + 3 = 7
• Step 3 - Draw a smooth curve
Graph y = √{2x − 6} + 5
• Step 1: Determine the domain
• Since you cannot have a negative inside the radical sign, your domain is limited to values greater or equal to zero inside the radical.
• 2x − 6 ≥ 0
• 2x ≥ 6
• x ≥ 3
• Your values of x can be greater or equal to 3
• Step 2 - Create a table of values
•  x y = √{2x − 6} + 5 3 5 11 21
•  x y = √{2x − 6} + 5 3 = √{2(3) − 6} + 5 = √{6 − 6} + 5 = 5 5 = √{2(5) − 6} + 5 = √{10 − 6} + 5 = √4 + 5 = 7 11 √{2(11) − 6} + 5 = √{22 − 6} + 5 = √{16} + 5 = 9 21 √{2(21) − 6} + 5 = √{42 − 6} + 5 = √{36} + 5 = 11
• Step 3 - Draw a smooth curve
Graph y ≥ √{x − 1} + 2
• In order to graph an a square root function, you must first determine the domain. You cannot have
• negatives inside the square root.
• x − 1 ≥ 0
• x ≥ 1
• Your domain is restricted to values greater or equal to 1.
• Step 1: Create a table of values to graph the square root function.
•  x y = √{x − 1} + 2 1 5 10 17 26
•  x y = √{x − 1} + 2 1 = √{1 − 1} + 2 = 0 + 2 = 2 5 = √{5 − 1} + 2 = √4 + 2 = 2 + 2 = 4 10 √{10 − 1} + 2 = √9 + 2 = 3 + 2 = 5 17 √{17 − 1} + 2 = √{16} + 2 = 4 + 2 = 6 26 √{26 − 1} + 2 = √{25} + 2 = 5 + 2 = 7
item Step 2 - Graph the square root function using a solid line for the boundary line.
• Step 3 - Using a test point, check to see which way to shade along the boundary line
•  Test (5,0) y ≥ √{x − 1} + 2 0 ≥ √{5 − 1} + 2 0 ≥ √4 + 2 0 ≥ 4 Not True
• Shade outside the boundary line not including (5,0). Be carefull when shading to the left of the imaginary line
• passing through x = 1, nothing should be shaded in that region.
Graph y√{x + 4} + 2
• In order to graph an a square root function, you must first determine the domain. You cannot have
• negatives inside the square root.
• x + 4 ≥ 0
• x ≥ − 4
• Your domain is restricted to values greater or equal to − 4.
• Step 1: Create a table of values to graph the square root function.
•  x y = √{x + 4} + 2 -4 0 5 12 21
•  x y = √{x + 4} + 2 -4 = √{ − 4 + 4} + 2 = 0 + 2 = 2 0 = √{0 + 4} + 2 = √4 + 2 = 2 + 2 = 4 5 √{5 + 4} + 2 = √9 + 2 = 3 + 2 = 5 12 √{12 + 4} + 2 = √{16} + 2 = 4 + 2 = 6 21 √{21 + 4} + 2 = √{25} + 2 = 5 + 2 = 7
• Step 2 - Graph the square root function using a solid line for the boundary line.
• Step 3 - Using a test point, check to see which way to shade along the boundary line
•  Test (5,0) y ≥ √{x + 4} + 2 0 ≥ √{0 + 4} + 2 0 ≥ √4 + 2 0 ≥ 4 Not True
• Shade outside the boundary line not including (0,0). Be carefull when shading to the left of the imaginary line
• passing through x = − 4, nothing should be shaded in that region.
Graph y <√{x + 4} − 6
• In order to graph an a square root function, you must first determine the domain. You cannot have
• negatives inside the square root.
• x + 4 ≥ 0
• x ≥ − 4
• Your domain is restricted to values greater or equal to − 4.
• Step 1: Create a table of values to graph the square root function.
•  x y = √{x + 4} − 6 -4 0 5 12 21
•  x y = √{x + 4} − 6 -4 = √{ − 4 + 4} − 6 = 0 − 6 = − 6 0 = √{0 + 4} − 6 = √4 − 6 = 2 − 6 = − 4 5 √{5 + 4} − 6 = √9 − 6 = 3 − 6 = − 3 12 √{12 + 4} − 6 = √{16} − 6 = 4 − 6 = − 2 21 √{21 + 4} − 6 = √{25} − 6 = 5 − 6 = − 1
• Step 2 - Graph the square root function using a dashed line for the boundary line.
• Step 3 - Using a test point, check to see which way to shade along the boundary line
•  Test (0,-6) y <√{x + 4} − 6 − 6 <√{0 + 4} − 6 − 6 <√4 − 6 − 6 <− 4 True
• Shade inside the boundary line including (0, − 6). Be carefull when shading to the left of the imaginary line
• passing through x = − 4, nothing should be shaded in that region.
Graph y <√x + 9
• In order to graph an a square root function, you must first determine the domain. You cannot have
• negatives inside the square root.
• x ≥ 0
• Your domain is restricted to values greater or equal to 0.
• Step 1: Create a table of values to graph the square root function.
•  x y = √x + 9 0 4 9 16 25
•  x y = √x + 9 0 = √0 + 9 = 0 + 9 = 9 4 = √4 + 9 = 2 + 9 = 11 9 √9 + 9 = 3 + 9 = 12 16 √{16} + 9 = 4 + 9 = 13 25 √{25} + 9 = 5 + 9 = 14
• Step 2 - Graph the square root function using a dashed line for the boundary line.
• Step 3 - Using a test point, check to see which way to shade along the boundary line
•  Test (4,2) y <√x + 9 2 <√4 + 9 2 < 2 + 9 2 < 11 True
• Shade inside the boundary line including (4,2). Be carefull when shading to the left of the imaginary line
• passing through x = 0, nothing should be shaded in that region.

*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.

Square Root Functions and 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
• Square Root Functions 0:07
• Examples: Square Root Function
• Example: Not Square Root Function
• Example: Restriction
• Graphing Square Root Functions 3:42
• Example: Graphing
• Square Root Inequalities 8:47
• Same Technique
• Example: Square Root Inequality
• Example 1: Graph Square Root Function 15:19
• Example 2: Graph Square Root Function 18:03
• Example 3: Graph Square Root Function 22:41
• Example 4: Square Root Inequalities 25:37

Transcription: Square Root Functions and Inequalities

Welcome to Educator.com.0000

Today, we will be talking about square root functions and inequalities.0002

First, defining what a square root function is: a square root function contains a square root involving a variable.0007

So, let's look at some examples before we go on to talk more about these.0016

A square root function could be something like f(x) = √(2x + 1), or g(x) = √(x2 + 4) - 2x.0022

Notice that it says "a square root involving a variable"; so there is a variable under the square root sign.0039

If I had a function--say h(x) =...let's say, instead of x2 + 4, I said √3, minus 2x.0046

This is not a square root function; and it is not a square root function because there is no variable in the radicand.0057

So, recall that whatever is under the square root sign here is the radicand.0072

And recall that terminology, because here it says that the radicand must be non-negative.0080

So, the domain is restricted to values that make the radicand non-negative.0085

Let's look at this first one, f(x) = √(2x + 1): if this expression were to become negative,0092

then what I would end up with here is...let's say I ended up with something like...this was a value like -4:0110

So, let's let x equal -4; then, what I would end up with is 2 times -4, plus 1; and that would give me -8 + 1, and that would be √-7.0119

The problem is that that is not a real number; and although we have talked about complex numbers0133

and imaginary numbers, we are going to restrict our discussion of square root functions to only functions that result in real numbers.0142

We are not going to allow values such as this.0151

Instead, when we look at a square root function, the first thing we are going to do is determine what the domain will be.0154

And we can do that by saying that this domain, this radicand, needs to be greater than or equal to 0.0160

So, we need to find values of x that will allow the radicand to be greater than or equal to 0, and our domain will be restricted to those values.0167

If I have f(x) = √(2x + 1), I am saying that I want 2x + 1 to be greater than or equal to 0.0177

Therefore, subtracting 1 from both sides gives me that 2x must be greater than or equal to -1.0185

Dividing both sides by 2 gives me x ≥ -1/2.0193

So, the domain is restricted to values of x that are greater than or equal to -1/2, for this function.0199

This function is not defined; we are not defining the function for values of x that are less than -1/2.0211

OK, in graphing a square root function, we are going to exclude values of x that make the radicand negative.0222

So, I just discussed how to find what the domain is; and values outside the domain are restricted.0227

And we won't even include those values on the graph.0233

For example, looking at the function f(x) = √(x + 3) - 2: what I want to do is make sure that x + 3 ≥ 0,0239

so that I don't end up with a negative radicand, and then an imaginary or complex number.0259

I am going to subtract 3 from both sides; and this is telling me that I must restrict my domain to x ≥ -3.0265

That is going to be the domain; I am not going to define this function for values of x less than -3; those are excluded values.0274

Values of x that are less than -3 are excluded from the domain.0282

OK, so with that in mind, we can pick some values for x that are part of the domain, and then evaluate the function for those values.0302

So, if I let x be -3 (because it says "greater than or equal to," so -3 is included), then I am going to get a 0 under here.0313

The square root of that is 0; minus 2 is going to give me -2 for y.0321

OK, when x is -2, the radicand will be 1; the square root of that is 1; minus 2 gives me -1.0329

Let's let x be -1: when x is -1, what I am going to end up with is the square root of 2.0341

Well, the square root of 2 is approximately 1.4; so that is going to give me 1.4 - 2, which is going to give me -0.6.0348

OK, when x is 1, this becomes 4; the square root of that is 2; minus 2 is 0.0362

Now, it is good to pick values that are easy to work with, so I am going to think about perfect squares.0370

How can I get perfect squares under here (like 1 + 3 gave me 4--that was a perfect square; -3 gave me 0--I can get a square root of that easily)?0376

Another nice number to work with is 9: pick values for x strategically--if I make x 6, this becomes 9, and the square root of that is 3; minus 2 gives me 1.0386

Another perfect square is 16: I just think, "OK, 16 - 3 is 13; 13 would be a good value to work with."0399

13 + 3 is 16; the square root of that is 4; subtract 2 from that, and I have 2.0408

Now, you will notice, I picked a large value here; and it was a convenient value;0416

but also, if you look at the coordinate axis I am using here, it is short but long,0421

because, to get a good graph, you are going to need to pick some big x-values,0427

because often, these graphs are not very tall, but they are wide--the slope is not very great.0433

For each change in x, I don't get a big change in y.0440

And then, if I don't pick enough values, I am not going to be able to find the shape of my graph.0444

Here, when x is -1, 2...-3 is right here...y is -2; that is (-3,-2), right here.0449

When x is -2, y is -1; -1 puts me right up here; -.6 is going to be about there; (1,0)--when x is 1, y is 0.0460

When x is 6, y is 1--not much change in the graph there--not much of a slope.0476

When x is 13, y is 2; and that is why I picked some values out here--so I could get a better sense of what this graph looks like.0484

So, just keep that in mind.0493

Now, one thing to also keep in mind is that the graph starts here, or ends here, because this function is not defined for values of x beyond this.0496

Beyond this is not defined, so I can't graph the function past x = -3, for values of x less than -3.0515

This is graphing square root functions: now, talking about graphing square root inequalities:0528

when we are graphing an inequality involving square roots, we are going to use0533

the same techniques that we used to graph linear and quadratic inequalities.0536

And recall that, when we handled these, what we did is...let's say quadratic inequalities:0540

the first thing we would do is graph the corresponding quadratic equation.0545

That equation formed the boundary line for our solution set; and then we used a test point to find on which side of the boundary the solution set lay.0550

And we are going to use those same techniques here, except now we are working with square root inequalities.0561

For example, if I am given the inequality y ≥ √2x, the one extra thing I do have to do is find the excluded values that we talked about.0566

I am going to need to graph this corresponding equation, which is going to be y = √2x.0585

But I have to make sure that I don't end up with some excluded value.0592

So, I know that, when I look at the radicand, I need 2x (the radicand) to be greater than or equal to 0.0597

So, that means that, if I divide both sides by 2, I am going to get that x must be greater than or equal to 0.0609

So, this is the domain; for this inequality, I am only going to graph the corresponding equation for a domain where x is greater than or equal to 0.0616

First, graph the corresponding equation; and in this case, the corresponding equation here is y = √2x.0627

And I can plot points, now that I have figured out what my domain is.0643

And the smallest value for my domain is 0; so when x is 0, y is 0.0647

When x is 1, that gives me the square root of 2, which is about 1.4.0654

When x is 2, that gives me a perfect square, 4; the square root of that is 2.0660

I like to work with perfect squares; I know that (even though it seems like a strange number to pick)--if I pick 4.5 and multiply it by 2, I will get 9.0665

And the square root of 9 is 3, so that makes it easy on that end of things.0676

And I will pick one more value; and again, I want to look for perfect squares, if I can, to make my life easy.0681

So, we used 4; we ended up with 4 when we made x 2.0688

When we made x 4.5, we ended up with 9; if I make x 8, I am going to get 8, times 2 is 16;0692

that is a perfect square, and the square root of that is 4.0699

I have enough points to plot; and when x is 0, y is 0; when x is 1, y is 1.4--about there; when x is 2, y is 2.0703

When x is 4.5, y is going to be 3; and then, when x is 8, y will be 4, right here.0715

Now, this is where the graph starts--or ends, depending on how you want to look at it.0733

I cannot define values of this function--evaluate the function--for any value smaller than x; the domain is restricted.0740

So, this is what the graph is going to look like.0751

Now, something else to point out: we are working with an inequality, and I did make this a solid line.0754

As previously, when we are working with inequalities, if it is ≥ or ≤, that means that the boundary line is part of the solution set.0761

If this had been a strict inequality, if it was greater than or just less than, I would have made this a dashed line0771

to indicate that the boundary line is not part of the solution set.0779

So, I graphed the corresponding equation; my second step is to use a test point to determine which side of the boundary the solution set is on.0782

Now, often, we use (0,0), the origin, as the test point.0794

But in this case, that is on the boundary line; and you want to find a test point that is away from the boundary line.0799

So, I am going to choose (right here) (3,0) as a test point.0805

And recall that, to use a test point, we go back to the inequality, y ≥ √2x, and we insert these values.0811

We substitute these values: so, I am going to let y be 0, and I am going to let x be 3.0824

That is going to give me 2 times 3; and this is going to end up being 0 ≥ √6.0833

And even if you don't know the exact value of the square root of 6 (it is about 2.5, but), we know that it is a positive number.0844

And we know that the square root of 4 is 2, and we know that the square root of 6 is going to be greater than that.0851

So, I know that this is not true: so this is not valid; therefore, the test point is not part of the solution set.0856

When I inserted these values, I came up with something not valid; so it is not part of the solution set.0872

So, the solution set is actually up here, not down here.0876

And I am going to shade in this region, and the boundary line is actually going to be included in my solution set.0880

But I need to keep in mind that this inequality is not going to be defined for any values of x smaller than 0.0886

So, I am not going to shade over here: the graph ends here; my shading ends there.0895

OK, so that was inequalities with square roots.0899

And with square root inequalities, you graph the corresponding equation, find the boundary line,0902

look at the inequality to determine if you should use a dashed line or a solid line,0909

and then use a test point to find if the solution set is above or below the boundary line.0914

The first example is asking me to graph y = √4x.0921

First, we are going to define the domain; and the domain has to be such that 4x is greater than or equal to 0.0928

I am dividing both sides by 4, so x has to be greater than or equal to 0.0935

All values for x that are less than 0 are excluded.0940

Now, I can plot points, starting with 0: 0 times 4 is 0; the square root of that is 0.0948

1 times 4 is 4; the square root of that is 2; 2 times 4 is 8; the square root of 8--you can use a calculator,0957

and figure out that it is approximately 2.8; or you can just estimate by knowing that0969

it is going to be greater than a square root you know of, such as the square root of 4,0975

but less than another square root that you are familiar with.0980

4 times 4 is 16; and so, the square root of that is going to be 4; and then the square root of 6 times 4...0985

that is 24, and the square root of 24 is approximately equal to 5--a little bit less than 5, actually...4.9.0995

We know that the square root of 25 is 5, so the square root of 24 is going to be slightly less than that; we will say about 4.9--close enough for this graph.1013

When x is 0, y is 0; when x is 1, y is 2; when x is 2, y is a little bit below 3...2.8...around there.1024

When x is...let's move this over just a bit; it is going to be right about there...and when x is 4, y is 4.1036

And then, one more point here: we are going to have x be 6, and y is going to be a little bit below 5, just to give me one more point to work with.1051

Now, again, the graph begins right here; that is just going to be a point, and then this is going to be an arrow, continuing on out.1067

So, we are beginning the graphing by finding the domain, x ≥ 0, and then plotting some points.1074

OK, here y is less than -√6x; so I am working with a strict inequality.1084

And I first need to think about excluded values, because I want to graph the corresponding equation, y = -√6x.1093

And I know that the radicand, 6x, must be greater than or equal to 0.1106

Dividing both sides by 6 tells me that x has to be greater than or equal to 0; so this is the domain.1112

So, as I plot points, I am only going to choose values for x that are at least 0.1120

Starting out with 0: 6 times 0 is 0; the square root of that is 0; -0 is still 0.1129

So, when x is 1, this is going to give me the square root of 6, and that is going to be about 2.5.1138

The square root of 6 is about 2.5, and I need to make that negative--about -2.5.1149

2 times 6 is 12; the square root of 12 is around 3.5; I am making that negative: -3.5.1156

4 times 6 is 24; and the square root of that, we said, is a little bit less than 5; and making that negative, it is about -4.9.1166

And then, 5 times 6 is 30; the square root of that is about 5.5, so it is going to give me -5.5.1178

So, this is enough to plot; and I am going to use a dashed line, because it is a strict inequality,1186

meaning that the boundary line is not part of the solution set.1192

So, when x is 0, y is 0; when x is 1, y is going to be -2.5; when x is 2, y is going to be -3.5...about right there.1196

When x is 4 (that is 1, 2, 3, 4, 5...) y is going to be a little bit more than -5...right about there.1225

And then, I am going to put a -6 down here; and when x is 5, this is going to give me -5.5.1241

So, this is curving like this.1251

Now, I am going to use a dashed line for this, since it is a strict inequality; and the graph begins right there.1256

I graphed the boundary line; now I need to find a test point; and again, I am not going to use (0,0) as a test point, because it is on the boundary line.1270

Instead, I am going to go ahead and use (3,0), so my test point is (3,0).1278

I am going to look at this: y < -√6x; let y be 0; let x be 3.1290

Is 0 less than -√18? Well, the square root of 18 is about 4.2, and so this is not true.1302

Even if you didn't know what the square root of 18 is, you know that it is going to be greater than √16, which is 4.1318

So, it is going to be some number greater than 4; the square root of 18 will be a little bit more than 4.1324

And when we make that negative, it is going to be negative 4 point something, and you know that 0 is not less than a negative number.1330

This is not true; so the test point is not part of the solution set, which tells me that the solution set is down here.1337

And again, I am not going to shade in past x = 0, because this graph--this function--is not defined for values of x less than 0.1344

Those are excluded values.1356

OK, we are supposed to graph this square root function, y = √(x + 2) - 3.1362

Let's find the excluded values: x + 2 must be greater than or equal to 0, so x must be greater than or equal to -2.1371

This is my domain, so I am not going to attempt to plot points where x is less than -2.1380

Let's start out with -2: when x is -2, the radicand is 0; the square root of 0 is 0, minus 3 gives me -3.1390

When x is -1, -1 + 2 is 1; the square root of 1 is 1; minus 3 is -2.1403

When x is 2, 2 + 2 is 4; the square root of that is 2; minus 3 is -1.1412

When x is 3, this gives me 5; the square root of 5 is about 2.2; subtracting 3 from that...2.2 - 3 is -0.8.1420

Again, I like to look for perfect squares in the radicand to make things easy.1437

And 9 is a perfect square, so I am going to let x be 7; this will become 9; the square root of 9 is 3; minus 3 is 0.1440

And I am going to also let x be 14, because that will give me 16; 14 + 2 is 16; the square root of that is 4; minus 3 is 1.1449

OK, so when x is -2, y is -1, 2, 3, so right here; when x is -1, y is -2; when x is 2, y is -1.1462

When x is 3, y is -0.8, right about there; when x is 7, y is 0; and when x is 14, y is 1.1481

OK, so I am going to go ahead and draw a line through these points, and alter that just a little bit so it goes through there...1498

And this continues on: again, we have one of these short, very wide graphs, where right here, we don't have a lot of slope.1508

So, I needed to find some large values of x for x.1517

And this finishes out the graph; the graph starts right there at x = -2 (I can't graph anything past there) and continues on.1522

Here I have an inequality; and I am going to start out...what I want to do is graph the corresponding equation, y = √(x - 4) + 2.1538

But I know that I need to find excluded values; and I know that x - 4 must be greater than or equal to 0,1549

since I don't want to have a negative radicand; I need to have values in the radicand that are 0 or greater.1556

If I add 4 to both sides, I find that the domain is that x must be greater than or equal to 4.1563

So, I am going to plot points accordingly; and I know that x can be 4--that is the smallest value I can give it.1572

4 - 4 is 0; the square root of that is 0; plus 2 gives me 2.1581

OK, I am looking for perfect squares: a perfect square...if I have 1, I can easily get the square root of that.1589

So, if I let x be 5, and I subtract 4 from that, I am going to get 1; the square root is 1; plus 2 gives me 3.1596

Let's see, another perfect square is 4; so I am going to let x equal 8; 8 - 4 is 4; the square root of that is 2; plus 2 is 4.1610

Another perfect square would be 9, so I am going to let x equal 13, because 13 - 4 is 9; the square root of that is 3; add 2 to that--that gives me 5.1620

So, I have some values that I can plot out.1631

And I am going to use a solid line, because this inequality is greater than or equal to; it is not a strict inequality.1636

So, when x is 4, y is 2; and that is the start of my graph--I can't define this function for values less than 4.1643

OK, when x is 5, y is 3; when x is 8, then y is 4; and when x is 13, y is 5...right about there.1655

OK, and this is a solid line that I am going to use; and the graph begins here.1678

Now, this is an inequality, and I need to figure out which side of this boundary line the solution set is.1686

And I know that I am not going to worry about values over here--just ones where x is greater than or equal to 4.1696

So, I am going to go ahead and...I don't want to pick a point on the boundary line, and (0,0) is not even part of this graph at all;1703

so, I am going to go ahead and pick (5,0) as a test point.1709

Test point (5,0) for y ≥ √(x - 4) + 2: when y is 0, let's see if this holds true.1716

5 - 4 + 2: is 0 greater than the square root of 1, plus 2? Is 0 greater than or equal to 1 + 2?1730

Is 0 greater than or equal to 3? No, it is not, so the test point is not part of the solution set.1740

This is not part of the solution set; so what I need to do is shade in this region of the graph1754

above the boundary line, being careful not to go over here, because that is not part of the graph; those are not defined areas.1766

To find the graph of this inequality, I graphed the corresponding equation, first finding excluded values1777

(that values that were less than 4 are excluded), then using values that are allowed to plot points,1784

and then using a test point and finding that my test point is not part of the solution set down here.1791