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

12 minutes

Hi, guys. Welcome to Algebra 1. Today's lesson focuses on finding values for elements in the domain of quadratic functions. Your knowledge of the coordinate plane, of evaluating expressions will come in handy during this lesson. You ready to get started? Let's go. Okay. To warm you up to what we're doing here, I want to throw back a bit to find y-intercepts of quadratic functions. Here we have the function: f of x equals x squared, plus x minus 20. Because of what we know about quadratic functions, we know the y-intercept is typically our constant term right here, on the right side. Here I could say, "All right, it's negative 20." What we want to do, to warm you up to what we're doing, is I'll show why it's always that constant term, that negative 20, okay? When we have a y-intercept, we know that x must be 0 and y is some numerical value, right? In order to show the algebra behind this, we'll substitute 0 in for x and see why it is that we get negative 20. Okay? Let's say f of 0-- because we're replacing that x with 0-- equals 0 squared plus 0 minus 20, right? So 0 squared--that's 0-- plus 0 minus 20. 0 plus 0 minus 20 is just negative 20. We'd say, "Okay, our y-intercept is negative 20." What we did to verify that our y-intercept is really going to be negative 20, like we thought, we did that algebra, and substituted 0 for x. You can substitute any value in the domain of a function, evaluate it at that value, and see what you get for the output. That's what we'll do. I wanted to warm you up by showing what I meant with y-intercepts in doing the same thing. Look at this example. Here you have: f of x equals x squared plus 5x plus 6, and you're asked to find f of 3. What I'm going to do is substitute 3 for my x value and see what I get for my output. I'll find "f of 3"-- that's how it's read...

(female narrator) It's written as f with a 3 in parenthesis. ...equals 3 squared plus 5 times 3 plus 6. Okay? So, 3 squared-- that's 9-- plus 5 times 3-- that's 15-- plus 6. 9 plus 15--that's 24-- and 24 plus 6 is 30. I'd say, "Okay, f of 3 equals 30." And you'd be all done with that one. You'd evaluate this function at 3 and you saw its output was 30. Then you'd read that final answer as "f of 3 equals 30." Okay? Let's try another one together. Here's the same function, f of x equals x squared plus 5x plus 6, but now you're asked to find f of negative 1. Our process is the same; it's just our number that's changed. Just how we started the last one. We're going to substitute negative 1 for x. We'll have f of negative 1 equals negative 1 squared, plus 5 times negative 1, plus 6. Now we just want to evaluate this right side. So negative 1 squared-- that's positive 1. 5 times negative 1 is negative 5. And then we have plus 6. So, 1 minus 5, that's negative 4, and negative 4 plus 6 is positive 2. So we'd say, "Okay, f of negative 1 is positive 2." If negative 1's my input for this function, then 2 is my output. That's exactly what this means. Good job. It's time to try a couple on your own. Go ahead and press pause. Take a few minutes and work through these problems. To compare your answers with me, press play.

(female narrator) Given: f of x equals x squared minus 3x plus 2. Letter A: Find f of negative 2. Letter B: Find f of 4. Let's see what you got. Let's switch to our pointer tool to move these out of our way. For f of negative 2, you should have got 12-- f of negative 2 equals 12. And for B, f of 4 equals 6. To see how I got these, stick with me, and I'll show you my work. Let's switch to my pen here. For f of negative 2, what we'll do is substitute negative 2 for x in our function. Okay? So f of negative 2 equals negative 2 squared minus 3 times negative 2 plus 2. Right? Okay. So, negative 2 squared, that's positive 4. Negative 3 times negative 2 is positive 6. And then we have our plus 2. Then just left to right: 4 plus 6 is 10, 10 plus 2 is 12. That's how I got that f of negative 2 was positive 12. When negative 2 is my input, 12 is my output. All right? Now let me show you how I got B. Here I was asked to find f of 4. So f of 4 equals 4 squared minus 3 times 4 plus 2. Okay? So 4 squared, that's 16. Negative 3 times 4 is negative 12. Then we've got our plus 2. So, 16 minus 12, that's 4, and then 4 plus 2 is 6. That's how I got that f of 4 equals 6. All right? When 4 was my input, 6 was my output. Good job on that. Let's look at some graphs. Given the graph of a quadratic function, you should be comfortable with being given an input and being able to find the output. Here you're asked to find f of negative 1. I'm going to switch to my highlighter here. Okay. Now, f of negative 1-- so that's an input value. I need to find negative 1 on my x-axis-- here--on my graph it's above me at this point. So I'll trace it up. There--I land right there at this point. I need to find out the y value right there. I'm going to trace that over to the y-axis to figure out that this is the point

(negative 1, positive 1). I know that when negative 1 is my input, 1 is my output. Okay? I'd return to my problem and I'd write that as "f of negative 1 equals positive 1." When negative 1's my input, 1 is my output for this function. Let's try another one together, then I'll let you chart one on your own. Now we've got f of 2. Let me switch to my highlighter. I'm going to 2 on my x-axis, because 2 is an input value here. At 2, my parabola is above me, so I'm going to trace up, and I land right there. I'm going to put a point. Now I'll trace that over to the y-axis to figure out the output. And it looks like I am at 1, 2, 3, 4. When 2 is my input, 4 is my output because this is the point (2, 4). To rewrite it in function notation, going back to the form of our problem, I'd say that f of 2 equals 4 because when 2 is my input for this function, 4 is my output. Okay? All right? Time for you to try one. Press pause, take a few minutes and work through this problem. You're asked to find f of 1. To check your answer against mine, press play.

(female narrator) On the graph, the parabola curves up from the point of origin. On one side, the curve goes to minus 1, plus 1, and then minus 2, plus 4. On the other side of the parabola, the curve goes to plus 1, plus 1, and then plus 2, plus 4. Find f of 1. Let's take a look at how you did. I'll switch to my highlighter here. So f of 1-- 1's an input value, so I'll go to the x-axis. At 1, my parabola's above me, so I'll go up. I landed right there; I'm going to put a point. And I'll go back to the highlighter tool and trace it over to the y-axis to see where I'm at. It looks like I'm at 1, right there. So when my input value's 1, my output value is also 1 for this function. I'd rewrite the answer, coming back to my problem: f of 1 equals 1. When 1 is my input, 1 is also my output for this function. All right? Good job on that! Great job, guys, evaluating functions at values given from their domain. I hope your knowledge of the coordinate plane and how to evaluate expressions really came in handy. See you soon for more Algebra 1. Bye!

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In this program, students learn how to find the values for elements in the domain of quadratic functions. Part of the "Welcome to Algebra I" series.

## Media Details

Runtime: 12 minutes