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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 the 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 all 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 if 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 3 then add 6 times and subtract 2 now the range at least the way we've been thinking about it in this series of videos the range is the set of possible the set of possible outputs of this function or if we said y equals F y equal f of X on a graph is 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 a it won't take on any values below its vertex when it's upward opening and it won't take any values take on any values above its vertex when it is downward opening so let's look at 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 how we can think about this problem so I'm going to try some x and y values there's other ways to direct we 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 but let's try some X values and then let's see what f of X is equal to so let's try well this the values we've been trying in the last few videos what happens when X is equal to negative 2 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 to negative 5 and that actually is the vertex and you know the formula for the vertex once again is negative B over 2a 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 2 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 cemetry if you go 1 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 well 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 1 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 let me do my best to draw it so this is x equals negative 2 let me 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 5 over there on the y axis y axis and then let me go to positive 7 1 2 3 4 5 6 7 I can 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 1 right here so let's 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 there so that is the point that is the point negative 2 comma negative 2 fair enough then we have at this point that we have in 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 we have the point 0 negative 2 0 negative 2 when x is 0 Y is negative 2 or 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 f of 1 is 7 f of 1 is 7 so that's right there is the point 1 comma 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 is there multiple ways of calculating it but since it's an upward-opening parabola 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 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 votes you can keep on increasing forever as X gets larger X gets smaller away from the vertex so our range here so we already said the domain is all real numbers our range the possible Y values is all real numbers 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