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

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

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

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

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

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

Â it. Our quadratic function here is written in

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

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

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

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

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

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

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

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

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

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

Â negative -6. Which gives us -3.

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

Â then we add and subtract that within the parentheses here.

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

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

Â parentheses. But remember we have to multiply by this

Â -3. That is f(x) = -3(x - 3)^2.

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

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

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

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

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

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

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

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

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

Â will be symmetric, about this axis of symmetry.

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

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

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

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

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

Â have 1 - 3 which is -2. -2^2 is 4.

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

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

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

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

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

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

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

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

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

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

Â too. So here's the graph of our quadratic

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

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

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

Â 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.

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

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

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

Â 5. And we see that to the left of that

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

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

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

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

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

Â these intervals, and we would have been correct.

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

Â a parabola. Thank you.

Â And we'll see you next time.

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