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[SOUND] Let's look at graphing a parabola.

[SOUND] For example, let's sketch the graph of this quadratic function, f(x) =

-3x^2 + 18x - 22. And then we'll find the maximum or

minimum value of f, as well as its domain and range and the intervals where it's

increasing or decreasing. Now to sketch the graph of a quadratic

function or a parabola, we plot the vertex and a few points on either side of

it. Our quadratic function here is written in

standard form, and what we need to do is convert it to vertex form, that is f(x) =

a(x - h)^2 + k, where (h, k) here is the vertex.

And the way we do that is by completing the square.

So we have f(x) = -3^2 + 18x - 22. And now the first step in completing the

square is to make sure that the coefficient of the square term is one.

And right now it's -3. So what we'll do first is factor in -3

out of these first 2 terms. That is, this is equal to, -3(x^2 - 6x).

And let's leave a little bit of room here.

And then we still have the -22. And now the next step in completing the

square is we take 1 half, of the coefficient of x, which in this case is

negative -6. Which gives us -3.

And then we square this. So we have -3^2 which is equal to 9, and

then we add and subtract that within the parentheses here.

That is, we have +9 -9. And now these first 3 terms will form a

perfect square. But we'll need to get this -9 out of the

parentheses. But remember we have to multiply by this

-3. That is f(x) = -3(x - 3)^2.

And then, -3 * -9 is 27. And then we still have -22.

Or, f(x) = -3(x - 3)^2 + 5. And now comparing this, to our vertex

form, over here on the left, we see that a = -3.

h = to 3 and k = 5. And since a<0 our parabola will be

opening downward. And our vertex is at (3, 5).

So let's plot the vertex. Let's say that this is the y axis, and

this is the x axis. So we have 1, 2, 3.

And then, 1, 2, 3, 4, 5. So the vertex is here, at (3, 5).

And when finding other points, we're going to use the fact, that the parabolla

will be symmetric, about this axis of symmetry.

Namely, x = 3. Namely, if we choose, x values, to the

left of 3 like 1 and 2, for example, that's going to give us corresponding

values, to the right of 3. So let's compute, what y is equal to,

when x = 1. Which we can do by plugging in x = 1 down

here in our vertex form of our function. So we'll do that, we plug 1 in here, we

have 1 - 3 which is -2. -2^2 is 4.

4 * -3 is -12, and -12 + 5 = -7 and when we plug in x=2, we have 2 - 3 = (-1),

(-1)^2 = 1, 1(-3) = (-3) + 5 = 2. So lets plot 1, -7.

So this is -1, -2, -3, -4, -5, -6, -7. So 1, -7 will be here.

And, 2, 2, will be up here. And now, because of symmetry, we know

that (4, 2) also has to lie on our graph, as well as (5, -7).

So our parabola will look like this. So it's very helpful to use this axis of

symmetry to get the other points. That is, by knowing this point 2, 2 lies

on our graph, we can conclude that 4, 2 does as well.

And that we knew this point 1, -7 lay on our graph, then we can conclude that 5,

-7 does as well. And we'll just darken in the vertex here,

too. So here's the graph of our quadratic

function or our parabola, but we're still asked to find the maximum or minimum

values of f, its domain and range, and intervals of increase or decrease.

Now, because this parabola is opening downward, it will have a maximum, not a

minimum, and the maximum will occur here at the vertex.

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That is the maximum value of our function is the y coordinate of that vertex or 5.

It occurs that x = 3 but the actual maximum value of the function is the y

coordinate or 5. And the domain is all real numbers.

And the range is from that vertex down or the interval from megative infinity up to

5. And we see that to the left of that

vertex the graph is rising and to the right it's falling.

Which means, that the function is increasing, on the interval from negative

infinity, up to 3. And it's decreasing, from 3 to infinity.

And remember, that, increasing and decreasing, are defined, on an interval.

Not pointwise, but in terms of intervals. So we could have included 3, on either of

these intervals, and we would have been correct.

Alright. And this is how we work with the graph of

a parabola. Thank you.

And we'll see you next time.

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