Domain and range of a function given a formula | Algebra II | Khan Academy

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.

15 Replies to “Domain and range of a function given a formula | Algebra II | Khan Academy”

  1. 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

  2. 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

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