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.

our Domain and range only has {x|x} or {y|y}, no graphs or table of values, how about that???

I don't really know if my brain hates hard math or what. This is really hard bruh i don't like this. 😭

Thank you so much since I just started in the K to 12 curriculum, I have prelims to answer this for I'm not a good mathematician but a great listener. Thx

2019 squad where you at

Thxx khan

any shortcuts for range? i don't have time for a graph when i have to do like 15 questions in 30 minutes

I think that only my teacher knows the trick that she showed me it takes fewer times to do but more difficult to understand

You shoudn't put the arrow on the curve

good

Am I the only one who didn’t understand it lol

I’m watching this through my tears lolThanks

His writing is so tiny, neat, and cute.

I could help someone with algebra problems

i did understand a little bit on this now. but this isnt what im looking for,. sadly

thank you for this video its very helpful