WEBVTT mathematics/algebra-2/eaton
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Welcome to Educator.com.
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Today, we will be talking about square root functions and inequalities.
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First, defining what a square root function is: a **square root function** contains a square root involving a variable.
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So, let's look at some examples before we go on to talk more about these.
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A square root function could be something like f(x) = √(2x + 1), or g(x) = √(x² + 4) - 2x.
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Notice that it says "a square root involving a variable"; so there is a variable under the square root sign.
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If I had a function--say h(x) =...let's say, instead of x² + 4, I said √3, minus 2x.
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This is not a square root function; and it is not a square root function because there is no variable in the radicand.
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So, recall that whatever is under the square root sign here is the radicand.
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And recall that terminology, because here it says that the radicand must be non-negative.
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So, the domain is restricted to values that make the radicand non-negative.
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Let's look at this first one, f(x) = √(2x + 1): if this expression were to become negative,
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then what I would end up with here is...let's say I ended up with something like...this was a value like -4:
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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.
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The problem is that that is not a real number; and although we have talked about complex numbers
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and imaginary numbers, we are going to restrict our discussion of square root functions to only functions that result in real numbers.
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We are not going to allow values such as this.
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Instead, when we look at a square root function, the first thing we are going to do is determine what the domain will be.
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And we can do that by saying that this domain, this radicand, needs to be greater than or equal to 0.
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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.
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If I have f(x) = √(2x + 1), I am saying that I want 2x + 1 to be greater than or equal to 0.
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Therefore, subtracting 1 from both sides gives me that 2x must be greater than or equal to -1.
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Dividing both sides by 2 gives me x ≥ -1/2.
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So, the domain is restricted to values of x that are greater than or equal to -1/2, for this function.
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This function is not defined; we are not defining the function for values of x that are less than -1/2.
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OK, in graphing a square root function, we are going to exclude values of x that make the radicand negative.
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So, I just discussed how to find what the domain is; and values outside the domain are restricted.
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And we won't even include those values on the graph.
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For example, looking at the function f(x) = √(x + 3) - 2: what I want to do is make sure that x + 3 ≥ 0,
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so that I don't end up with a negative radicand, and then an imaginary or complex number.
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I am going to subtract 3 from both sides; and this is telling me that I must restrict my domain to x ≥ -3.
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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.
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Values of x that are less than -3 are excluded from the domain.
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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.
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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.
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The square root of that is 0; minus 2 is going to give me -2 for y.
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OK, when x is -2, the radicand will be 1; the square root of that is 1; minus 2 gives me -1.
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Let's let x be -1: when x is -1, what I am going to end up with is the square root of 2.
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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.
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OK, when x is 1, this becomes 4; the square root of that is 2; minus 2 is 0.
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Now, it is good to pick values that are easy to work with, so I am going to think about perfect squares.
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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)?
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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.
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Another perfect square is 16: I just think, "OK, 16 - 3 is 13; 13 would be a good value to work with."
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13 + 3 is 16; the square root of that is 4; subtract 2 from that, and I have 2.
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Now, you will notice, I picked a large value here; and it was a convenient value;
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but also, if you look at the coordinate axis I am using here, it is short but long,
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because, to get a good graph, you are going to need to pick some big x-values,
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because often, these graphs are not very tall, but they are wide--the slope is not very great.
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For each change in x, I don't get a big change in y.
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And then, if I don't pick enough values, I am not going to be able to find the shape of my graph.
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Here, when x is -1, 2...-3 is right here...y is -2; that is (-3,-2), right here.
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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.
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When x is 6, y is 1--not much change in the graph there--not much of a slope.
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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.
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So, just keep that in mind.
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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.
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Beyond this is not defined, so I can't graph the function past x = -3, for values of x less than -3.
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This is graphing square root functions: now, talking about graphing square root inequalities:
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when we are graphing an inequality involving square roots, we are going to use
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the same techniques that we used to graph linear and quadratic inequalities.
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And recall that, when we handled these, what we did is...let's say quadratic inequalities:
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the first thing we would do is graph the corresponding quadratic equation.
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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.
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And we are going to use those same techniques here, except now we are working with square root inequalities.
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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.
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I am going to need to graph this corresponding equation, which is going to be y = √2x.
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But I have to make sure that I don't end up with some excluded value.
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So, I know that, when I look at the radicand, I need 2x (the radicand) to be greater than or equal to 0.
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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.
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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.
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First, graph the corresponding equation; and in this case, the corresponding equation here is y = √2x.
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And I can plot points, now that I have figured out what my domain is.
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And the smallest value for my domain is 0; so when x is 0, y is 0.
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When x is 1, that gives me the square root of 2, which is about 1.4.
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When x is 2, that gives me a perfect square, 4; the square root of that is 2.
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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.
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And the square root of 9 is 3, so that makes it easy on that end of things.
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And I will pick one more value; and again, I want to look for perfect squares, if I can, to make my life easy.
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So, we used 4; we ended up with 4 when we made x 2.
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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;
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that is a perfect square, and the square root of that is 4.
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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.
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When x is 4.5, y is going to be 3; and then, when x is 8, y will be 4, right here.
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Now, this is where the graph starts--or ends, depending on how you want to look at it.
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I cannot define values of this function--evaluate the function--for any value smaller than x; the domain is restricted.
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So, this is what the graph is going to look like.
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Now, something else to point out: we are working with an inequality, and I did make this a solid line.
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As previously, when we are working with inequalities, if it is ≥ or ≤, that means that the boundary line is part of the solution set.
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If this had been a strict inequality, if it was greater than or just less than, I would have made this a dashed line
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to indicate that the boundary line is not part of the solution set.
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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.
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Now, often, we use (0,0), the origin, as the test point.
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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.
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So, I am going to choose (right here) (3,0) as a test point.
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And recall that, to use a test point, we go back to the inequality, y ≥ √2x, and we insert these values.
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We substitute these values: so, I am going to let y be 0, and I am going to let x be 3.
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That is going to give me 2 times 3; and this is going to end up being 0 ≥ √6.
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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.
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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.
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So, I know that this is not true: so this is not valid; therefore, the test point is not part of the solution set.
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When I inserted these values, I came up with something not valid; so it is not part of the solution set.
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So, the solution set is actually up here, not down here.
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And I am going to shade in this region, and the boundary line is actually going to be included in my solution set.
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But I need to keep in mind that this inequality is not going to be defined for any values of x smaller than 0.
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So, I am not going to shade over here: the graph ends here; my shading ends there.
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OK, so that was inequalities with square roots.
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And with square root inequalities, you graph the corresponding equation, find the boundary line,
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look at the inequality to determine if you should use a dashed line or a solid line,
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and then use a test point to find if the solution set is above or below the boundary line.
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The first example is asking me to graph y = √4x.
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First, we are going to define the domain; and the domain has to be such that 4x is greater than or equal to 0.
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I am dividing both sides by 4, so x has to be greater than or equal to 0.
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All values for x that are less than 0 are excluded.
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Now, I can plot points, starting with 0: 0 times 4 is 0; the square root of that is 0.
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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,
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and figure out that it is approximately 2.8; or you can just estimate by knowing that
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it is going to be greater than a square root you know of, such as the square root of 4,
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but less than another square root that you are familiar with.
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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...
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that is 24, and the square root of 24 is approximately equal to 5--a little bit less than 5, actually...4.9.
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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.
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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.
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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.
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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.
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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.
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So, we are beginning the graphing by finding the domain, x ≥ 0, and then plotting some points.
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OK, here y is less than -√6x; so I am working with a strict inequality.
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And I first need to think about excluded values, because I want to graph the corresponding equation, y = -√6x.
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And I know that the radicand, 6x, must be greater than or equal to 0.
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Dividing both sides by 6 tells me that x has to be greater than or equal to 0; so this is the domain.
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So, as I plot points, I am only going to choose values for x that are at least 0.
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Starting out with 0: 6 times 0 is 0; the square root of that is 0; -0 is still 0.
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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.
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The square root of 6 is about 2.5, and I need to make that negative--about -2.5.
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2 times 6 is 12; the square root of 12 is around 3.5; I am making that negative: -3.5.
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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.
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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.
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So, this is enough to plot; and I am going to use a dashed line, because it is a strict inequality,
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meaning that the boundary line is not part of the solution set.
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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.
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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.
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And then, I am going to put a -6 down here; and when x is 5, this is going to give me -5.5.
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So, this is curving like this.
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Now, I am going to use a dashed line for this, since it is a strict inequality; and the graph begins right there.
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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.
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Instead, I am going to go ahead and use (3,0), so my test point is (3,0).
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I am going to look at this: y < -√6x; let y be 0; let x be 3.
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Is 0 less than -√18? Well, the square root of 18 is about 4.2, and so this is not true.
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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.
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So, it is going to be some number greater than 4; the square root of 18 will be a little bit more than 4.
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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.
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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.
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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.
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Those are excluded values.
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OK, we are supposed to graph this square root function, y = √(x + 2) - 3.
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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.
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This is my domain, so I am not going to attempt to plot points where x is less than -2.
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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.
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When x is -1, -1 + 2 is 1; the square root of 1 is 1; minus 3 is -2.
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When x is 2, 2 + 2 is 4; the square root of that is 2; minus 3 is -1.
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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.
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Again, I like to look for perfect squares in the radicand to make things easy.
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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.
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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.
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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.
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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.
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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...
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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.
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So, I needed to find some large values of x for x.
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And this finishes out the graph; the graph starts right there at x = -2 (I can't graph anything past there) and continues on.
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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.
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But I know that I need to find excluded values; and I know that x - 4 must be greater than or equal to 0,
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since I don't want to have a negative radicand; I need to have values in the radicand that are 0 or greater.
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If I add 4 to both sides, I find that the domain is that x must be greater than or equal to 4.
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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.
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4 - 4 is 0; the square root of that is 0; plus 2 gives me 2.
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OK, I am looking for perfect squares: a perfect square...if I have 1, I can easily get the square root of that.
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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.
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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.
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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.
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So, I have some values that I can plot out.
00:27:16.300 --> 00:27:23.000
And I am going to use a solid line, because this inequality is greater than or equal to; it is not a strict inequality.
00:27:23.000 --> 00:27:35.400
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.
00:27:35.400 --> 00:27:57.900
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.
00:27:57.900 --> 00:28:06.200
OK, and this is a solid line that I am going to use; and the graph begins here.
00:28:06.200 --> 00:28:15.900
Now, this is an inequality, and I need to figure out which side of this boundary line the solution set is.
00:28:15.900 --> 00:28:23.400
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.
00:28:23.400 --> 00:28:29.500
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;
00:28:29.500 --> 00:28:36.100
so, I am going to go ahead and pick (5,0) as a test point.
00:28:36.100 --> 00:28:49.900
Test point (5,0) for y ≥ √(x - 4) + 2: when y is 0, let's see if this holds true.
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5 - 4 + 2: is 0 greater than the square root of 1, plus 2? Is 0 greater than or equal to 1 + 2?
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Is 0 greater than or equal to 3? No, it is not, so the test point is not part of the solution set.
00:29:14.100 --> 00:29:25.900
This is not part of the solution set; so what I need to do is shade in this region of the graph
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above the boundary line, being careful not to go over here, because that is not part of the graph; those are not defined areas.
00:29:37.600 --> 00:29:44.300
To find the graph of this inequality, I graphed the corresponding equation, first finding excluded values
00:29:44.300 --> 00:29:51.500
(that values that were less than 4 are excluded), then using values that are allowed to plot points,
00:29:51.500 --> 00:29:57.500
and then using a test point and finding that my test point is not part of the solution set down here.
00:29:57.500 --> 00:30:00.700
So, I went ahead and shaded in up there.
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That concludes this lecture for Educator.com; and thanks for visiting!