In Exercises 57–62, use the vertex and the direction in which the...

Question

In Exercises 57–62, use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function?

y^2 + 6y - x + 5 = 0

Accepted Answer

hey everyone in this problem, we're given the equation of a parabola. Okay, we want to use its its vertex and it's the direction in which it opens to solve for the domain and range. And to tell if the given relation is a function or not. Okay, now the given equation is y squared plus four, Y minus x minus three is equal to zero. Okay. And so what you'll notice is that we have y squared term. And so what we want to write this in order to find the vertex, we're gonna have an equation that looks like X equals. Now let's recall the general equation for in vertex form when we have X equals, it's gonna be X equals a times y minus k squared plus H. Because we have X equals the vertex gets switched around a little bit. And so the vertex given this equation is going to be a church. Okay, Okay, Alright. So if we can get our equation into this form then we'll know the vertex A will figure out its domain and range and we can go from there. Alright, so let's start with our equation Y squared plus four, Y minus x minus three equals zero. And we know we want to write this as X equals something. Okay, so if we move the X over, we get x is equal to y squared plus four, Y minus three. Alright, so now we have our X equals in the equation that we're trying to match in vertex form, we need to complete the square. Now what this means is we want to have why? Plus or minus some terms squared. So in our case we have y squared plus four Y. Those are the terms with why we need to be concerned about. And so in order to have a term that when squared gives us y squared plus four, Y We're gonna need y plus two squared. Alright? So then when we expand this y plus two squared we're gonna get y squared plus four. Y plus some constant term. Okay, But we get the y terms back that we wanted. Alright now let's deal with that constant. So if we have y plus two squared we get again y squared plus four. Y. But we also get plus four. Okay. So in this equation we have plus four that we didn't have before. So to counteract that we need to subtract for And then we have our -3. When we do completing the square, we only deal with the white terms. And so we still have that -3 at the end. And so just simplifying this equation can be written as y plus two squared minus seven. Alright, so let's compare our equation then. So we have this K. Here as y minus K. In our case we have y plus two. So r K is going to be negative two then we have our H. Here. In our case this is -7 and these dictate what the vertex is. So in our case the vertex Given our H&K is gonna be negative seven negative two. Okay. All right. So we're done with the vertex we found the vertex we gotta look at what direction this opens because we have an X equals equation. We know that this parable is gonna open either to the right or to the left. Okay. When we have Y equals like the typical quadratic equations were used to seeing we know that those open either up or down when we have X equals and the Y. Term is squared. We're going to get opening sideways left or right now. Our A value here out in front of our equation is one. Okay, this is this a value out in front is one. So we have a is equal to one which is positive. So A is bigger than zero. And so this Parabola is going to open to the right. If a was negative, it would open to the left but it's positive. It's opening to the right. So let's just draw a brief little sketch. So we know that the vertex is at -7 -2. So negative 7 -2. Can say that that's about here and then it opens to the right. So it's gonna do something like this. Alright, so we have the vertex we have the direction we have an idea of what this looks like. Let's figure out the domain and range I'm gonna give ourselves some more room. Okay. Alright. So the domain. Well, for the domain we need to be concerned about what's going on with our X values. Okay. No, if you look at our Picture, we know that the vertex is at an x value of -7. And this Prabal opens right? So for our function we will have X values negative seven or anything bigger, but we don't have any X values to the left of negative seven. Okay. Or less than negative seven. So we're going to have X bigger than or equal to negative seven if we want to write this, not in set notation. What we have is X is going to go from negative seven and we have a square bracket because we know that the vertex is at negative seven. So we can be at negative seven and this is gonna go all the way up to infinity. Alright, now we look at our range, we're concerned with the wide eyes, all of the Y values. Now, let's look at the Y values and you can were asked to do it using the vertex and the way that it opens our vertex at negative two. This opens to the right, okay, so you can kind of imagine these branches going off to infinity. Ok. So we can have any Y value we want this is just gonna be all of the real numbers. And so this is gonna be negative infinity to positive infinity and you can see this from the equation to we can plug any Y value into this equation and we will get an X. Value out of it. Okay? All right. The last thing we want to do is figure out if this is a function. Okay, how do we do that while we use the vertical line test? So we draw a vertical line any vertical line and we need to make sure this passes for all vertical lines. And we see how many times that vertical line crosses our equation crosses here and it crosses here. So for this vertical line we're hitting the function twice. That means that this fails the vertical line test and so it's not a function. Okay. And I'm gonna put in brackets here. That that's by the vertical line test. And what that vertical line test is doing is essentially saying that for any X value we chose the X value. We chose here we have two different values for why that means that this isn't a function. Alright? So let's go back up and choose our answer. So the answer here, we found that the domain is from -7 to infinity including -7. The range is from negative infinity to infinity. And this is not a function. And so we have the answer is B Thanks everyone for watching. I hope this video helped see you in the next one

In Exercises 57–62, use the vertex and the direction in which the parabola opens to determine the relation's domain and range. Is the relation a function? y^2 + 6y - x + 5 = 0

Key Concepts

Parabola and its Vertex

Direction of Opening

Function Definition

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