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Welcome to Algebra I: Finding Values for Elements in the Domain of Linear Functions

12 minutes

Hey, guys. Welcome to Algebra 1. Today's lesson is going to focus on finding values for elements in a domain of linear functions. Your knowledge of how to evaluate expressions will come in handy during this lesson. Ready to get started? Let's go.

(Describer) She uses a stylus on a screen.

So to warm up to exactly what we'll be doing here, I want to throw back to finding the Y-intercept of liner functions. So, given this function, F of X equals 5X minus 10, if I were being asked to find the Y-intercept-- so I have in mind "Y-intercept." That means that-- write myself a note here... that X is definitely 0 and Y has some numerical value. So knowing what we do about slope intercept form, we kind of already know that the Y-intercept's going to be negative 10, but I want to show you what I mean about warming up for what we're doing. So if I actually went through this problem and replaced my X with 0, I would have F of 0 equals 5 times 0 minus 10, right? Because I'm substituting 0 for the value X. So 5 times 0 is 0, minus 10. And 0 minus 10 is just negative 10. So I know this function has a Y-intercept of negative 10, right? Because we substituted that value 0 for X. Now, you're able to substitute any value in the domain of a function for X in order to determine what the output value is that corresponds to that value. Let me show you want I mean. Look at this example. We still have our function F of X equals 5X minus 10, but now we need to find F of 3; that's how we read that-- that's what that interprets to mean. You see how the 3 took the place of the X, and we're just looking at that F of X part of our function? That indicates that you need to replace X with 3 in your function, and see what you get when you work the problem out. This is what I mean. My function is F of X equals 5X minus 10. So I'm going to find F of 3. I'm going to substitute 3 every time I see X. So I am going to have 5 times 3 minus 10. 5 times 3, that's 15, minus 10. 15 minus 10, that's 5. So that means I can say, okay, this function evaluated at 3, or F of 3, equals 5. And you're all done. You just evaluated this function at 3 in order to determine what its output value is, okay? So let's try another one. We'll stick with function F of X equals 5X minus 10, but this time we're going to find F of negative 2. So each time we see an X in our function, we're going to substitute negative 2. So F of negative 2 equals 5 times negative 2 minus 10. So 5 times negative 2, that's negative 10. And then minus 10. Negative 10 minus 10... is negative 20. So this function, evaluated at 2-- I'm sorry, evaluated at negative 2-- is negative 20. So F of negative 2 is negative 20. And you're all done. You see, it's just a matter of substituting whatever you're given for the value of X into your function, working everything out, and then just seeing what you get at the end, okay? It is time for you to try a couple. You have the function F of X equals 6X minus 4, and you need to find F of 1 and find F of negative 2. Press pause, take a few minutes, work through these problems, and when you're ready to compare your answers with me, press play.

(Describer) Titles: Given: F of x equals 6x minus 4. Letter a: Find f of 1 Letter b: Find f of negative 2.

(female narrator) Given: F of X equals 6X minus 4. Letter A: Find F of 1. Letter B: Find F of negative 2. All right? Let's see what you got. Let's move these little... hiders away here. Okay, so F of 1-- let's delete that. F of 1 equals 2. And F of negative 2-- let's delete that-- is negative 16. If you want to see how I got this, I'll show you my work. Okay, so to find F of 1-- scroll down a little bit-- I just replaced X with 1 in my function. So I have-- oops, get the pen... So I have F of 1 equals 6 times 1 minus 4. So 6 times 1 is 6 minus 4, and then 6 minus 4 is 2. So that's how I got the F of 1 equals 2, okay? It's a matter of what math users like to call "plug and chug." Just substitute that value for 1 in there for X and see what you get for your answer. Now, in B, what I did is I followed that same process. I substituted X with a negative 2. So F of negative 2 equals 6 times negative 2 minus 4. 6 times negative 2 is negative 12 minus 4. And negative 12 minus 4 is negative 16. So that's how I found that F of negative 2 equals negative 16. Just follow that same process. Substitute negative 2 in there for X, and then see what you get at the end, okay? Okay, let's look at some graphs of some linear functions, and I'll show you how to figure out the same thing, how you can be given a value for X and figure out what your output value is. So here you need to find F of 4. When you see something like that, they're asking you to figure out, when X is 4, what is Y? That's how you interpret that "find F of 4." We'll go to 4 on our X-axis. I'll just scroll a bit. So I'm going to go to 4, so one, two, three, four. Then look for your function. I see here's my line. In order to hit my line I have to drop down... So here X is 4 and my line's underneath me, so I'll drop down. And I need to find that Y value, right there, whenever X is 4. Basically I want to know what's this point, right here?

(Describer) ...where the lines cross.

Okay. And if I trace this to the Y-axis, then I'll know the Y value.

(Describer) She draws across.

It's negative 1. And that makes sense

because this is the point (4, negative 1). So when my input value is 4, my output value is negative 1, okay? I've evaluated this function at 4 and been able to tell what my output value is there, okay? So then I could come back to my problem and say, F of 4... is negative 1. Because when X is 4, Y is negative 1, okay? Let's try another one, make sure you got the hang of that. Okay, now you're being asked to find F of negative 1. I'll scroll down a bit. Okay. Let me get my highlighter out. So when X is negative 1-- that's my point here-- I see that my line, again, it's underneath me. So I'm going to drop down.

(Describer) She draws down to the line.

And I see that I hit that line, right there. I'm going to darken that point, okay? So when X is negative 1, I need to find this Y value here. I'm going to trace this over to the Y-axis and see where I'm at. And it looks like I'm at negative 3. I can say, all right,

well, this is the point, (negative 1, negative 3). So when X is negative 1, Y is negative 3. Then I could come back to my problem and say, okay, then F of negative 1 is negative 3. When negative 1's my input, negative 3 is my output, okay? And you were able to tell that from looking at the graph of that function. All right? Good job. Now you try one, see how you do. You're being asked to find F of negative 3. Press pause, take a few minutes, and analyze this graph. When you're ready to compare your answer against mine, press play.

(Describer) Title: find f of negative 3. The line on the graph crosses the x-axis at 4 and a half, and crosses the y-axis at 3.

(female narrator) Find F of negative 3. The line in the graph crosses the X-axis at 4-1/2, and crosses the Y-axis at 3. All right, let's see how you did. Let me switch to my highlighter. I'm being asked to find F of negative 3. I'll go to negative 3 on my X-axis-- negative 1, negative 2, negative 3-- this time my line is above me, so I'm going to go up and I end up right there.

(Describer) She draws up to the line.

I'm going to darken that point. And then I'll trace over to the Y-axis so I can see exactly what's going on there.

(Describer) She draws over.

I see when I'm at negative 3 for X, I'm at one, two, three, four, five for Y.

This is the point (negative 3, 5). So I can say, all right, then F of negative 3... is 5. Because when X is negative 3, Y has a value of 5, all right? Good job on that! Okay, guys, you've reached the end of our lesson on finding values for elements in the domain of our linear functions. I hope you saw how your knowledge of the coordinate plane, and of function notation, and how to evaluate expressions came in handy for you. See you back here soon for more Algebra 1. Bye!

(Describer) Accessibility provided by the US Department of Education.

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The graph of a linear function is a straight line and has the form y = f(x) = a + bx. These functions have one independent variable and one dependent variable. In this program, students learn how to find the values for elements in the domain of linear functions. Part of the "Welcome to Algebra I" series.

Media Details

Runtime: 12 minutes

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