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# Domain and range of quadratic functions

Determine the domain and range of the function f of x is equal to 3x squared plus 6x minus 2. So, the domain of the function is: what is a set of all of the valid inputs, or all of the valid x values for this function? And, I can take any real number, square it, multiply it by 3, then add 6 times that real number and then subtract 2 from it. So essentially any number if we re talking about reals when we talk about any number. So, the domain, the set of valid inputs, the set of inputs over which this function is defined, is all real numbers. So, the domain here is all real numbers. And, for those of you who might say, well, you know, aren t all numbers real? You may or may not know that there is a class of numbers, that are a little bit bizarre when you first learn them, called imaginary numbers and complex numbers. But, I won t go into that right now. But, most of the traditional numbers that you know of, they are part of the set of real numbers. It s pretty much everything but complex numbers. So, you take any real number and you put it here, you can square it, multiply it by 3, then add 6 times it and subtract 2. Now, the range, at least the way we ve been thinking about it in this series of videos– The range is set of possible, outputs of this function. Or if we said y equals f of x on a graph, it s a set of all the possible y values. And, to get a flavor for this, I m going to try to graph this function right over here. And, if you re familiar with quadratics– and that s what this function is right over here, it is a quadratic– you might already know that it has a parabolic shape. And, so its shape might look something like this. And, actually this one will look like this, it s upward opening. But other parabolas have shapes like that. And, you see when a parabola has a shape like this, it won t take on any values below its vertex when it s upward opening, and it won t take on any values above its vertex when it is downward opening. So, let s see if we can graph this and maybe get a sense of its vertex. There are ways to calculate the vertex exactly, but let s see how we can think about this problem. So, I m gonna try some x and y values. There s other ways to directly compute the vertex. Negative b over 2a is the formula for it. It comes straight out of the quadratic formula, which you get from completing the square. Lets try some x values and lets see what f of x is equal to. So, let s try, well this the values we ve been trying the last two videos. What happens when x is equal to negative two? Then f of x is 3 times negative 2 squared, which is 4, plus 6 times negative 2, which is 6 times negative 2, so it s minus 12 minus 2. So, this is 12 minus 12 minus 2. So, it s equal to negative 2. Now, what happens when x is equal to negative 1? So, this is going to be 3 times negative 1 squared, which is just 1, minus, or I should say plus 6 times negative 1 which is minus 6 and then minus 2, and then minus 2. So, this is 3 minus 6 is negative 3 minus 2 is equal negative 5, and that actually is the vertex. And, you know the formula for the vertex, once again, is negative b over 2 a. So, negative b. That s the coefficient on this term right over here. It s negative 6 over 2 times this one right over here, 2 times 3. 2 times 3, this is equal to negative 1. So, that is the vertex, but let s just keep on going right over here. So, what happens when x is equal to 0? These first two terms are 0, you re just left with a negative 2. When x is equal to positive 1. And, this is where you can see that this is the vertex, and you start seeing the symmetry. If you go one above the vertex, f of x is equal to negative 2. If you go one x value below the vertex, or below the x value of the vertex, f of x is equal to negative 2 again. But, let s just keep going. We could try, let s do one more point over here. So, we have, we could try, x is equal to 1. When x is equal to 1, you have 3 times one squared which is 1. So, 3 times 1 plus 6 times 1, which is just 6, minus 2. So, this is 9 minus 2 it s equal to 7. And, that I think is enough points to give us a scaffold of what this graph will look like. What the graph of the function would look like. So, it would look something like this. I do my best to draw it. So, this is a x equals negative 2. We draw the whole axis. This is x is equal to negative 1, this is x is equal to, this is x is equal to 0 and then this is x is equal to 1 right over there and then when x is equal to, we go from negative 2 all the way to positive. Or, we should go from negative 5 all the way to positive 7. So, let s say this is negative 1,2,3,4,5. That s negative five over there on the y axis, y axis and then it will go to positive 7. One, two, three, four, five, six, seven. I could keep going, this is in the y, and we re going to set y equal to whatever our output of the function is. Y is equal to f of x. And this is one right here. So, lets plot the points. You have the point negative 2, negative 2. When x is negative 2, this is the x axis. When x is negative 2, y is negative 2. Y is negative 2 so that is that right over 3. So, that is the point, that is the point negative 2, negative 2. Fair enough? Then, we have this point that we have this pink or purplish color. Negative, when x is negative 1, f of x is negative 5. When x is negative 1, f of x is negative 5. And, we already said that this is the vertex. And, you ll see the symmetry around it in a second. So, this is the point negative 1, negative 5. And then, with the point 0, negative 2. 0, negative when x is a 0, y is negative 2, for f of x is negative 2 or f of 0 is negative 2, so this is the point 0, negative 2, and then finally when x is equal to 1 and f of 1 is 7, f of 1 is 7. So, that s right there it s a point 1, 7 and it gives us a scaffold for what this parabola, what this curve will look like. So, I ll try my best to draw it respectably. So, it would look something, something like that, and keep on going in that direction. Keep on going in that direction. But, I think you see the symmetry around the vertex. That if you were to. If you were to put a line right over here, the two sides are kind of the mirror images of each other. There, you can flip them over, and that s how we know it s the vertex. And, that s how we also know, because this is an upward opening parabola, I mean, there is formulas for vertex, and there are multiple ways of calculating it. But, since it s an upward opening parabola, where the vertex is going to be, the minimum point. This is the minimum value that the parabola will take on. So, going back to the original question, this is all for trying to figure out the range, the set of y values, the set of outputs that this function can generate. You see that the function, it can get as low as negative 5. It got all the way down to negative 5 right at the vertex. But, as you go to the right, as x values increase to the right or decrease to the left, then the parabola goes upwards. So, the parabola can never give you values– f of x is never going to be less than negative 5. So, our domain, but it can take on all the vaues. It can keep on increasing forever as x gets larger, x gets smaller farther away from the vertex. So, our range, so we already said our domain is all real numbers. Our range, the possible y values is all real numbers greater than or equal to negative 5. It can take on the value of any real number greater than or equal to negative 5. Nothing less than negative 5.

Range of quadratic functions