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INSTRUCTORSCarleen EatonGrant Fraser
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Lecture Comments (12)

2 answers

Last reply by: Cole Lovin
Tue Sep 20, 2016 9:08 PM

Post by Francisco Ramirez Cruz on July 12, 2014

i need help with  2/x + 6/x-2 =-5/2

1 answer

Last reply by: Dr Carleen Eaton
Mon Mar 19, 2012 6:54 PM

Post by Karen Shipp on March 18, 2012

How did you find the Y-intercept? (Karen is my mothers name. I am Gabriel)

1 answer

Last reply by: Dr Carleen Eaton
Mon Nov 14, 2011 11:17 PM

Post by Mary Moss on November 14, 2011

How do you find the x-int.?

4 answers

Last reply by: julius mogyorossy
Tue Sep 23, 2014 2:11 PM

Post by Jeff Mitchell on March 12, 2011

At about 26:40 into lecture in example 3, you said f(6) = -62 + 12(6)+8 = 44 because -6^2 = -36 but doesn't a negative number squared equal a +36? and therefore the answer would be +36+72+8 = 116 ?

Jeff

Graphing Quadratic Functions

  • The graph of a quadratic function is a parabola.
  • Use the axis of symmetry to help you graph a parabola. Graph the right or left half and then reflect the graph across the axis of symmetry.
  • The maximum or minimum value of the function occurs at the vertex. Use the formula for the vertex to find the maximum or minimum.

Graphing Quadratic Functions

Find the equation of the axis of symmetry, the coordinates of the vertex and graph f(x) = x2 − 4
  • Axis of Symmetry:
    x=−[b/2a]
    a = 1, b = 0,
    x = − [0/2] = 0
    Draw vertical, dashed line through x = 0
  • Vertex Coordinates:
    (− [b/2a],f( − [b/2a] )
    a = 1, b = 0,
    (0,f(0))
    f(0) = − 4
    (0, − 4)
  • Points to Graph: Choose 3 points after the axis of symmetry. Since axis is located at x = 0, choose three points after the x = 0 mark. You may also choose 3 points before

  • x
    1
    2
    3
    y = f(x) = x2 − 4
    − 3
    0
    5
  • You will notice that only half the parabola is shown with points x = 1, 2, 3. To Draw the other half
  • use the fact that Quadratics are Symmetric about their axis of Symmetry.
  • Point Symmetric with A will be 1 unit on the left of x = 0 at ( − 1, − 3)
  • Point Symmetric with B will b 2 units on the left of x = 0, at ( − 2,0)
  • Point Symmetric with C will be 3 units on the left of x = 0, at ( − 3,5)
  • Plot and draw a smooth curve.
Find the equation of the axis of symmetry, the coordinates of the vertex and graph f(x) = x2 − 6x + 4
  • Axis of Symmetry:
    x = − [b/2a]
    a = 1, b = − 6,
    x = − [( − 6)/2] = 3
    Draw vertical, dashed line through x = 3
  • Vertex Coordinates:
    ( − [b/2a],f( − [b/2a] )
    a = 1, b = − 6,
    (3,f(3))
    f(3) = 32 − 6(3) + 4
    = 9 − 18 + 4 = − 5
    (3, − 5)
  • Points to Graph: Choose 3 points after the axis of symmetry. Since axis is located at x = 3, choose three points after the x = 3 mark. You may also choose 3 points before

  • x
    4
    5
    6
    y = f(x) = x2 − 6x + 4
    − 4
    − 1
    4
  • You will notice that only half the parabola is shown with points x = 4, 5, 6. To Draw the other half
  • use the fact that Quadratics are Symmetric about their axis of Symmetry.
  • Point Symmetric with A will be 1 unit on the left of x = 3 at (2, − 4)
  • Point Symmetric with B will b 2 units on the left of x = 3, at (1, − 1)
  • Point Symmetric with C will be 3 units on the left of x = 3, at (0,4)
  • Plot and draw a smooth curve.
Find the equation of the axis of symmetry, the coordinates of the vertex and graph f(x) = 2x2 − 8x + 8
  • Axis of Symmetry:
    x = − [b/2a]
    a = 2, b = − 8,
    x = − [( − 8)/2(2)] = [8/4] = 2
    Draw vertical, dashed line through x = 2
  • Vertex Coordinates:
    ( − [b/2a],f( − [b/2a] )
    a = 2, b = − 8,
    (2,f(2))
    f(2) = 2(2)2 − 8(2) + 8
    = 8 − 16 + 8 = 0
    (2,0)
  • Points to Graph: Choose 3 points after the axis of symmetry. Since axis is located at x = 2, choose three points after the x = 2 mark. You may also choose 3 points before

  • x
    3
    4
    5
    f(x) = 2x2 − 8x + 8
    2
    8
    18
  • You will notice that only half the parabola is shown with points x = 3, 4, 5. To Draw the other half
  • use the fact that Quadratics are Symmetric about their axis of Symmetry.
  • Point Symmetric with A will be 1 unit on the left of x = 2 at (1,2)
  • Point Symmetric with B will b 2 units on the left of x = 2, at (0,8)
  • Point Symmetric with C will be 3 units on the left of x = 2, at ( − 1,18)
  • Plot and draw a smooth curve.
Find the equation of the axis of symmetry, the coordinates of the vertex and graph f(x) = x2 + 6x + 8
  • Axis of Symmetry:
    x = − [b/2a]
    a = 1, b = 6,
    x = − [6/2] = − 3
    Draw vertical, dashed line through x=-3
  • Vertex Coordinates:
    ( − [b/2a],f( − [b/2a] )
    ( − 3,f( − 3))
    f( − 3) = ( − 3)2 + 6( − 3) + 8
    = 9 − 18 + 8 = − 1
    ( − 3, − 1)
  • Points to Graph: Choose 3 points after the axis of symmetry. Since axis is located at x = 3, choose three points after the x = 3 mark. You may also choose 3 points before

  • x
    − 2
    − 1
    0
    y = f(x) = x2 + 6x + 8
    0
    3
    8
  • You will notice that only half the parabola is shown with points x = − 2, − 1, 0. To Draw the other half
  • use the fact that Quadratics are Symmetric about their axis of Symmetry.
  • Point Symmetric with A will be 1 unit on the left of x = − 3 at ( − 4,0)
  • Point Symmetric with B will b 2 units on the left of x = − 3, at ( − 5,3)
  • Point Symmetric with C will be 3 units on the left of x = − 3, at ( − 6,8)
  • Plot and draw a smooth curve.
Determine whether the function has a maximum or a minimum.
Find the max or min, and state the domain and range.
f(x) = − x2 + 10x − 2
  • Max or Min:
    a = -1, b = 10, c = -2
    Because ä" is negative, this quadratice is a maximum
  • Find Max:
    a = -1, b = 10, c = -2
    x=−[b/2a]
    x = − [b/2a] = − [10/(2( − 1))] = 5
    To find the maximum, plug in x = 5 into the quadratic
    f(x) = − x2 + 10x − 2
    f(5) = − (5)2 + 10(5) − 2
    f(5) = − 25 + 50 − 2 = 23
  • Domain: All Real Numbers
  • Range: Given that this quadratic is a maxium, the values taken up by ÿ" cannont exceed the maximum value, therefore,
  • Range: All Real Numbers for y ≤ 23
The Quadratic is a Maximum;
Max = 23;
Domain = All Real Numbers;
Range = All Real Numbers for y ≤ 23
Determine whether the function has a maximum or a minimum.
Find the max or min, and state the domain and range.
f(x) = x2 − 2x − 1
  • Max or Min:
    a = 1, b = -2, c = -1
    Because ä" is positive, this quadratice is a minimum
  • Find Min:
    a = 1, b = -2, c = -1
    x=−[b/2a]
    x = − [b/2a] = − [(−2)/2] = 1
    To find the minimum, plug x=1 into the quadratic
    f(x) = − x2 + 10x − 2
    f(1) = (1)2 − 2(1) − 1
    f(1) = 1 − 2 − 1 = − 2
  • Domain: All Real Numbers
  • Range: Given that this qudratic is a minimum, the value taken up by ÿ" cannont be below the minimum value, therefore,
  • Range = All Real Numbers for y ≥ −2
This quadratice is a minimum;
Min = -2;
Domain = All Real Numbers;
Range = All Real Numbers for y ≥ −2
Determine whether the function has a maximum or a minimum.
Find the max or min, and state the domain and range.
f(x) = − 3x2 − 12x − 1
  • Max or Min:
    a = -3, b = -12, c = -1
    Because ä" is negative, this quadratice is a maximum
  • Find Max:
    a = -3, b = -12, c = -1
    x=−[b/2a]
    x = − [b/2a] = − [(−12)/(2(−3))] = −2
    To find the maximum, plug x=-2 into the quadratic
    f(x) = − 3x2 − 12x − 1
    f( − 2) = − 3( − 2)2 − 12( − 2) − 1
    f( − 2) = − 12 + 24 − 1 = 11
  • Domain: All Real Numbers
  • Range: Given that this quadratic is a maxium, the values taken up by ÿ" cannont exceed the maximum value, therefore,
  • Range: All Real Numbers for y ≤ 11
This quadratice is a maximum; Max = 11;
Domain = All Real Numbers;
Range = All Real Numbers for y ≤ 11
Determine whether the function has a maximum or a minimum.
Find the max or min, and state the domain and range.
f(x) = − 3x2 − 6x
  • Max or Min:
    a = -3, b = -6, c = 0
    Because ä" is negative, this quadratice is a maximum
  • Find Max:
    a = -3, b = -6, c = 0
    x=−[b/2a]
    x = − [b/2a] = − [(−6)/(2(−3))] = −1
    To find the maximum, plug x=-1 into the quadratic
    f(x) = − 3x2 − 6x
    f( − 1) = − 3( − 1)2 − 6( − 1)
    f( − 1) = − 3 + 6 = 3
  • Domain: All Real Numbers
  • Range: Given that this quadratic is a maxium, the values taken up by ÿ" cannont exceed the maximum value, therefore,
  • Range: All Real Numbers for y ≤ 3
This quadratice is a maximum; Max = 3;
Domain = All Real Numbers;
Range = All Real Numbers for y ≤ 3
Determine whether the function has a maximum or a minimum.
Find the max or min, and state the domain and range.
f(x) = 2x2 − 8x + 4
  • Max or Min:
    a = 2, b = -8, c = 4
    Because ä" is positive, this quadratice is a minimum
  • Find Min:
    a = 2, b = -8, c = 4
    x=−[b/2a]
    x = − [b/2a] = − [(−8)/2(2)] = 2
    To find the minimum, plug x=2 into the quadratic
    f(x) = 2x2 − 8x + 4
    f(2) = 2(2)2 − 8(2) + 4
    f(2) = 8 − 16 + 4 = − 4
  • Domain: All Real Numbers
  • Range: Given that this qudratic is a minimum, the value taken up by ÿ" cannont be below the minimum value, therefore,
  • Range = All Real Numbers for y ≥ −4
This quadratice is a minimum;
Min = -4;
Domain = All Real Numbers;
Range = All Real Numbers for y ≥ −4
Determine whether the function has a maximum or a minimum.
Find the max or min, and state the domain and range.
f(x) = 5x2 + 20x + 4
  • Max or Min:
    a = 5, b = 20, c = 4
    Because ä" is positive, this quadratice is a minimum
  • Find Min:
    a = 5, b = 20, c = 4
    x=−[b/2a]
    x = − [b/2a] = − [20/2(5)] = −2
    To find the minimum, plug x=-2 into the quadratic
    f(x) = 5x2 + 20x + 4
    f( − 2) = 5( − 2)2 + 20( − 2) + 4
    f( − 2) = 20 − 40 + 4 = − 16
  • Domain: All Real Numbers
  • Range: Given that this qudratic is a minimum, the value taken up by ÿ" cannont be below the minimum value, therefore,
  • Range = All Real Numbers for y ≥ −16
This quadratice is a minimum;
Min = -16;
Domain = All Real Numbers;
Range = All Real Numbers for y ≥ −16

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Graphing Quadratic Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

  • Intro 0:00
  • Quadratic Functions 0:12
    • A is Zero
    • Example: Parabola
  • Properties of Parabolas 2:08
    • Axis of Symmetry
    • Vertex
    • Example: Parabola
  • Minimum and Maximum Values 9:02
    • Positive or Negative
    • Upward or Downward
    • Example: Minimum
    • Example: Maximum
  • Example 1: Axis of Symmetry, Vertex, Graph 12:41
  • Example 2: Axis of Symmetry, Vertex, Graph 17:25
  • Example 3: Minimum or Maximum 21:47
  • Example 4: Minimum or Maximum 27:09

Transcription: Graphing Quadratic Functions

Welcome to Educator.com.0000

Today, we are going to start talking about quadratic equations and inequalities.0002

And the first thing we are going to review is graphing quadratic functions.0006

Recall from Algebra I that a quadratic function has the form f(x) = ax2 + bx + c.0012

So, that is the quadratic function in standard form, where a does not equal 0.0022

And if a were to equal 0, this section would drop out; and this is actually what makes it a quadratic function--the x2.0027

Otherwise, you would just end up with a linear function, which we have described earlier on in the course.0036

So, let's first talk about an example of a quadratic function.0045

It could be something such as f(x) = x2 + 4x - 3, or f(x) = -x2 + 1.0053

And here, in this second example, b is actually equal to 0; and that is allowed, so this drops out.0069

But as I just discussed, a cannot equal 0.0076

We are going to be working with the graphs of these functions today, and first just talking about the general shape of these curves.0080

The curve of a graph of a quadratic function is a shape called a parabola, and they may open upward;0087

a parabola can open upward; it could open downward; and you will see varying shapes, such as wider parabolas and narrower parabolas.0097

And as we get deeper into discussing this topic, you will be able to tell, just from looking at the quadratic function,0116

roughly what the shape is going to be and which direction it is going to curve in.0123

So, getting started with some properties of parabolas: a parabola has an axis of symmetry, and that axis is described by the equation x = -b/2a.0128

And what the axis of symmetry is: it is a vertical line that divides the parabola into two symmetric halves.0142

And the parabola intersects this axis of symmetry at what is called the vertex.0153

Recall that the x-coordinate of the vertex is given by x = -b/2a.0160

To illustrate this with an example: if I were looking at the function f(x) = -x2 - 2x + 1,0168

then first let's look for the axis of symmetry; and the axis of symmetry is up at -b/2a.0184

Well, here a is equal to -1, because recall that standard form is ax2 + bx + c, so a is -1, b is -2, and c is 1.0192

So, if I want -b/2a, that is going to give me -(-2)/2(-1).0211

So, looking at that, that will be a negative and a negative, to give me a positive, over -2, which equals -1.0220

Therefore, the axis of symmetry is going to be a vertical line right here at -1; so, this is the axis of symmetry.0230

Now, this point where x equals -1 is the x-coordinate of the vertex.0249

However, I need to find the y-coordinate; so in order to do that, I am going to go back and look for f(-1).0258

So, I am replacing my x terms with -1; that is going to give me -1 squared, is 1, so that is -1.0267

And then, that gives me -2 times -1, which will become a positive, so that is + 2, + 1.0284

So, this is 3 minus 1; so that equals 2.0291

What I did is found the x-coordinate for the vertex, substituted that in for my x-value, and found that f(-1) = 2.0296

So, the y-value for the vertex is going to be 2; the vertex is going to be at the point (-1,2).0304

So, that is (-1,2) right here.0312

OK, now, to further graph this, let's go ahead and find some points.0316

And it is best to pick some points around the vertex, because if I pick points way out here, my graph is not going to be as accurate.0323

I want to get a general shape of the curve in the area of the vertex.0329

OK, so my vertex is right here; pick some x-values, and then let's go ahead and determine what the y-values are, based on that.0333

When x is -2 (you can go ahead and work this out for yourself), if you worked the whole thing out, you would find that y is 1.0348

OK, when x is 0, you can see that these two are going to drop out, and y is going to be 1.0359

When x is 1, that is going to give me -1, minus 2, plus 1; that is going to come out to -2.0368

So, one more point, maybe: when x is 2, working this out, this is going to give me 2 squared; that is -4.0391

And then, 2 times 2, with a negative, is also -4; plus 1--that is going to equal -7; so here we get -7.0403

Go ahead and plot out the rest of these points.0415

When x is -2, y is 1; when x is 0, y is 1; when x is 1, y is -2.0417

And then, this one is way down here, so we will leave that off.0431

But looking at how the axis of symmetry can help you with graphing:0434

here I have one point on the left, and I am going...I have a couple points on the right.0438

Because I know that the axis of symmetry divides this into two symmetric halves, I can actually add another point right over here.0445

So, because this is symmetrical, I have a point here on one side of the vertex; I have another one here.0455

Looking at the axis of symmetry, if I have a point right here, which is an x-value that is 1, 2 away from the vertex,0462

so on this side it is 1, 2 away from the vertex at -3, and then the y is 1, 2, 3, 4 down;0474

here I would have it, again, compared to the vertex, 1, 2, 3, 4 down.0482

So, that is going to reflect it right across here.0488

So, using the axis of symmetry, I can find the mirror image point.0491

Even if I have just graphed this half, if I have the axis of symmetry, I can reflect across to the other half.0499

Just reviewing: an axis of symmetry for a parabola is given by the equation right here; and I found my axis of symmetry.0504

And that is a vertical line; the parabola actually intersects that axis at the vertex.0515

The x-value of the vertex is also given by this equation; and once I have found that (which was -1),0521

I can substitute that in and find the function's value for y at that point (which turned out to be 2).0530

The axis of symmetry can help you with graphing.0538

Now, we talked a little bit about the vertex; but the vertex can give you either a maximum or minimum value of the function,0544

depending on the shape of the curve--whether it opens upward or downward.0554

That tells you if you have a maximum value or a minimum value.0561

So, investigating this a little bit further: if the coefficient of the x2 term is positive, the graph will open upward,0564

whereas if the coefficient of the x2 term is negative, the graph opens downward.0577

So, recall that the standard form is ax2 + bx + c, so that is the standard form of the function f(x).0585

And if I look at a, if a is greater than 0 (that means that the x2 term is positive), the parabola opens upward.0599

If this is a negative value--if a is less than 0--the parabola is going to open downward.0613

Just by looking at the equation I am given, I will have a general sense of the shape--whether it opens upward or downward.0621

Now, if this is positive, and it opens upward, let's give an example.0631

If I have a parabola like this, and it opens upward, this tells me that a is greater than 0.0638

The vertex is going to be a minimum; and thinking about what that means--0645

that means the minimum value that y can have--the minimum value of the function.0654

If the coefficient of x2 is positive, the graph opens upward, and the function has a minimum value.0659

Right here at the vertex, that is as small a value as you will find for y.0666

y only gets bigger from there, no matter what x-value you put in.0672

Conversely, if I have a parabola that opens downward (here a is less than 0--I have a parabola that opens downward),0676

in this case, the vertex is going to give me a maximum value.0687

So, right here, y is approximately equal to 1; that is as large as y is going to get.0698

No matter what value I plug in for x, y is just going to be smaller than that.0705

So again, this gives us more information about the graph, just by looking at what this coefficient is right here-- the coefficient of x2.0711

So, in order to graph, as I said, once you find the vertex, you look at the equation and say, "OK, it is going to open [upward or downward]."0722

And then, look for some points around the vertex to get the shape of the graph.0730

Does it open very wide? Is it a very shallow graph, or is it a very steep graph?0735

And we will talk more about that later on--about the shape.0745

But for now, just focus on the vertex--whether the parabola opens upward or downward,0749

and then on finding some x and y values in order to help you make the graph.0755

OK, let's practice these with some examples.0762

Find the equation of the axis of symmetry, the coordinates of the vertex, and the graph of this function.0765

OK, first recall that the axis of symmetry is given by the vertical line at x = -b/2a.0773

And then, always just review what standard form is in order to find this: it is ax2 + bx + c.0784

So here, I have x =, and then my a value is 1; there is no bx term, therefore b must be 0.0793

There is no term with a coefficient and then just x; that implies that b is 0.0805

And then, here, c is -9; but I really don't need to worry about that right now.0811

So, I am going to go ahead and substitute these in; and as soon as I see that b is 0,0815

it honestly doesn't matter what a is, because if you take 0 and divide it by anything, you are going to get 0.0822

Now, this is the equation for the axis of symmetry; it is a vertical line at x = 0, right here.0828

The axis of symmetry is just going to follow this y-axis.0843

The coordinates of the vertex: well, I have my x-coordinate right here, which is 0.0848

That means that this axis of symmetry is going to intersect this parabola at x = 0, and then at some y-coordinate.0855

In order to find the y-coordinate, let's find f(0).0863

f(0) is -9; so, let's make this -2, -4, -6, -8, -10; this is the y-coordinate for the vertex.0876

The x-coordinate is x = 0, so the coordinates of the vertex are (0,-9); (0,-9) is right there.0898

Now, let me go ahead and write that in--that is what that is--that is the vertex, right there.0911

Now, in order to graph this, I need to find some other points; so let's find some values for x, and then figure out what f(x) is at those points.0918

When x is 1, that would be 1 squared minus 9; you are going to get -8 for y.0931

When x is 2, 2 squared is 4, minus 9--that is going to give me 5.0940

Let's see: when x is 3, 3 squared is 9, minus 9 is 0.0948

So, let's try these points; and that is going to give me: when x is 1, y is -8; when x is 2, y is going to be...actually, it should be -5 right here.0954

When x is 3, y is 0; 3 is about here.0974

OK, now recall that the axis of symmetry can help me to graph--it can save me work,0980

because here, I have some points on one side of the axis of symmetry, and that means that on the other side...0986

I can just reflect across to get points that are mirror-image--approximately here, and then here at...this is here,0994

and then here is at (3,0), so I am going to go over here to (-3,0) and put a point.1009

Therefore, I was able to cut down my work by knowing that the axis of symmetry divides this into two symmetrical halves.1017

So, checking, I did find the equation of the axis of symmetry, x = 0.1026

I found the coordinates of the vertex, and that was (0,-9).1032

And I graphed by finding some points, and then reflecting across, using the axis of symmetry.1037

Again, find the equation of the axis of symmetry, the coordinates of the vertex, and the graph.1046

The equation of the axis of symmetry is given by x = -b/2a.1054

And here, I have a = 1, b = -4, because f(x) in standard form is ax2 + bx + c.1066

So, I am going to go ahead and find this; the axis of symmetry is x = -b (that is negative -4), over 2 times a (which is 1).1077

Therefore, x equals...that negative and negative gives me...positive 4 over 2, which equals 2.1090

So, this is the axis of symmetry; I'll go ahead and form an x-y axis.1099

The axis of symmetry is going to be right here at x = 2, and there is going to be a vertical line passing through that point.1132

My next thing that I need to do is find the coordinates of the vertex.1142

And I have the x-coordinate (that is x = 2), but I need to find the y-coordinate.1146

And the y-coordinate is going to be given for f(2), which equals 2 squared, minus 4, times 2, plus 4.1150

So, that gives me 4 minus 8, plus 4, so that is 4 + 4 is 8, minus 8 is 0.1164

So, this equals 0; so the y-coordinate f(2) for the vertex is 0.1172

So, the vertex is at (2,0); that is where the axis of symmetry is going to intersect with the graph.1180

Then, to finish graphing this, I just need to find some points--some x-values and some corresponding y-values.1189

So, let's start out with some points around the vertex.1200

When x is 1, that is going to give me 1 squared (is 1), minus 4, plus 4; so these two are going to cancel, and I am just going to get 1 right here.1205

OK, I already have 2; that is my vertex; how about 3?1219

When you go ahead and plug this in and figure it out, you are going to get 3 times 3 (that is 9), minus 3 times 4 (is 12), plus 4.1225

So, that is 9 minus 12 (is going to give me -3), plus 4: that is going to give me 1.1234

OK, when x is 4, if you work this out, you will find that y is 4; and then another easy point--when x is 0, these drop out; you get 4 here also.1241

When x is 1, y is 1; when x is 3, y is 1; when x is 4, y is 4; when x is 0, y is 4.1252

Now again, I could have used the axis of symmetry; I could have found even fewer points,1266

and then just used symmetry to find the points on the other side of that axis.1269

So, I completed the tasks that I was asked to do, which are to find the axis of symmetry (and that is right here at x = 2);1279

to find the vertex (the vertex is here at (2,0)), and to graph this (which is a parabola opening upward,1288

with a minimum value--the vertex here is a minimum, because this opens upward--at (2,0)).1299

In this next one, we are asked to determine whether the function has a maximum or a minimum,1310

Find the maximum or minimum, and state the domain and range.1315

So, we don't even have to graph this one; but we first need to look at it and determine if it has a maximum or a minimum.1319

And the way we figure that out is by seeing what this coefficient of x2 is--what a is.1326

Here, a equals -1; recall that, when a is less than 0--when a is negative--the parabola opens downward.1337

Now, just sketching this out to think about it: if the parabola opens down--it faces downward--1354

the vertex is going to give me a maximum value; that is as large as y is going to get; y only gets smaller.1365

So here, the vertex is a maximum value--it doesn't get any larger.1380

If this were to open upward, then the vertex would be a minimum value.1393

So, here the parabola opens downward; my vertex is a maximum.1404

Now, I am asked to find that maximum or minimum--in other words, to find the vertex.1409

And recall that the vertex is given by -b/2a.1413

So, I have that my a is -1; here, b is 12; so let's find the vertex.1423

It is going to be a vertical line at x = -b, over 2 times -1.1431

So, that is going to give me x = -12/-2, or x = 12/2, or x = 6.1443

That gives me the equation; there is going to be a line at x = 6--that is the axis of symmetry.1454

So, the axis of symmetry is at x = 6.1463

And that is a maximum value; so the axis of symmetry is x = 6.1468

The next thing I am asked to find is the domain and the range.1475

Well, looking at the domain, I could say that x is really any real number.1480

I could say it is one, a thousand, negative two hundred...so the domain is going to be all real numbers.1489

It may be tempting to say, "Oh, the range is also all real numbers."1498

But if you think about it, it is actually not: now, let's go a little bit deeper and find this maximum point--let's find the vertex.1502

I found that x equals 6, so I know that my x-coordinate is 6; but let's find the y-coordinate, which is going to be at f(6).1512

So, that f(6) equals -6 squared, plus 12, times 6, plus 8, which is going to equal...-6 times -6 is 36; 12 times 6 is 72.1521

6 times 6 is 36, and that is negative, so it is -36, plus 72, plus 8.1541

6 squared; take a negative; plus 12 times 6; plus 8; this is going to give me -36 + 72 + 8, which actually comes out to 44.1551

So, my vertex is actually going to be at the point (6,44); and what that is telling me is that this is the largest y is ever going to get.1562

Now, y can be an infinite number of smaller values; but this is the maximum it is going to get.1574

Therefore, the range is actually all real numbers, where y is less than or equal to 44.1582

So, it might be tempting to just look at this and say "all real numbers" for y, but that, in fact, is not true,1594

because we have a maximum here that y cannot be greater than.1599

Therefore, answering all of this, I determined that I have a vertex that is a maximum.1603

I found what the maximum is, and it was x is 6 and y is 44; so the maximum is y = 44.1609

And the domain and the range--the domain is all real numbers; the range is all real numbers such that y is less than or equal to 44.1621

OK, Example 4: Determine whether the function has a maximum or a minimum.1630

Find the maximum or minimum, and state the domain and range.1636

Recall standard form, and that I need to look at the coefficient of x--I need to look at a; and here, a equals 3.1642

And since that is greater than 0--since a is positive--this tells me that this parabola opens upward.1653

And since it opens upward, just to help visualize it, the vertex right here is going to be a minimum.1660

It is going to be the smallest value that y can achieve for any x.1674

They want me to find that minimum value; so I am going to recall that the x-coordinate of the minimum value is defined by -b/2a.1679

I know what a is; b is -12; so, just figuring that out, that is negative, times -12, over 2 times a (which is 3).1690

That is going to give me...a negative and a negative is a positive 12, divided by 2 times 3, which is 6; so this is going to give me 12/6; that is 2.1708

OK, so the x-coordinate for the minimum is 2; to figure out the y-coordinate, I am going to figure out f(2),1723

which is going to be 3 times 2 squared, minus 12 times 2, plus 7.1737

This is going to give me 3 times 4, minus 12 times 2 (is 24), plus 7; so this is 12 minus 24, plus 7.1746

So, that is going to give me -12 + 7, which is -5.1759

Therefore, the minimum value is going to be y = -5; so this is my minimum.1765

Now, looking at this, in reality, actually, this is going to be at x = 2, y = -5; it is actually going to be more like this.1777

We weren't asked to graph it, but just to help visualize it: -5, let's say, is down here; this is actually going to be down here, in this quadrant.1787

OK, now, I found that I have a minimum; I have found that the minimum occurs at y = -5.1797

What is the domain? Looking, I can make x anything I want--any real number--so the domain is going to be all real numbers.1804

What is the range? Well, there is a limit on what that can be, because I just said that the minimum value for y is -5, right here.1818

That is the minimum value; therefore, y can be greater, but it can never be less than that.1834

So, actually, the range is all real numbers, where y is greater than or equal to -5.1841

The domain is all real numbers; the range is more restricted--it is all real numbers greater than or equal to -5.1853

OK, so we found that we had a minimum, and we know that because a is positive.1866

We found that using the formula -b/2a to get the x-coordinate, which is 2.1872

Then, we found f(x) using 2, so f(2), to tell me that y equals -5.1879

Since y is -5, then there is a limit here on the range.1887

The domain is all real numbers, whereas the range is only those real numbers greater than or equal to -5.1894

That concludes this session of Educator.com, where we covered graphing quadratic functions.1901

I will see you next lesson!1907