Class 10 Notes

I. The Quadratic Formula
   A. Completing the Square
   B. Quadratic Formula
   C. The Discriminant
II. Graphic Quadratics
   A. The Parabola
   B. Effects of Coefficients
      1. Varying a
      2. Varying c
      3. Varying b
   C. Zeros
   D. Vertex
   E. Axis of Symmetry


I. The Quadratic Formula

A. Completing the Square

Thus far we've been working with examples that factor nicely into
integers.

That's not an accident -- we've carfully chosen the things we're
factoring.

In the real world, though, things don't always factor so nicely into
binomials with integer coefficients.

We need a procedure for dealing with quadratic equations that aren't
so neat.

Let's consider an example:

(on board)
Solve x^2 - 4x - 6 = 0

Let's try to approach this by our standard method. Someone tell me
what to do. (let class walk through this)

(on board)
Possibilities for -6:
1,-6
-1,6
2,-3
-2,3

We can see immediately there's going to be problem. None of these
pairs of numbers add up to -4. Our usual factoring method won't work.

Instead, we use a technique called completing the square.

To complete the square, the first thing we do is take that -6 and move
it to the other side of the equal sign. It isn't giving us
possibilities we like, so we're going to get rid of it.

(on board)
x^2 - 4x - 6 = 0
          +6  +6
x^2 - 4x     = 6

That's step one.

Step two is to figure out what we want in place of that -6.

We need to put there a number whose factors add up to -4.

Well, suppose we take -2 and -2 as the factors: they add up to
-4. We'll see below why taking half of the coefficient of x as our
factors is useful.

When we multiply -2 times -2, we get 4, so let's try adding 4 to
both sides.

(on board)
x^2 - 4x     = 6
          +4  +4
x^2 - 4x + 4 = 10

Now let's factor the left side. How do we do that:

(on board)
Possibilities for 4:
-2,-2 -> (x - 2)(x - 2) = x^2 - 4x + 4

No surprise, this factors into (x - 2)(x - 2), since we chose the 4 by
saying we wanted to get factors of -2 and -2.

So we now have

(on board)
(x - 2)(x - 2) = 10
(x - 2)^2 = 10

Note that we get a square on the left side because we chose our
factors to be equal: -2 and -2. This turns out to be the best choice
to make in any situation.

We can solve this by taking the square root of both sides!

(on board)
x - 2 = +-sqrt(10)
   +2   +2
x = 2 +- sqrt(10)

There are two solutions, as usual, but they are not integers.

Based on our experience, we can write down the general procedure.

Let's do that while going through another example:

(on board)
Solve x^2 + 6x + 12 = 0

What's the first step? (let class answer)

(on board)
1. Move the constant term to the other side.

(on board)
x^2 + 6x - 12 = 0
          +12  +12
x^2 + 6x      = 12

What's next? (let class answer)

We want factors that are the same, and that add up to 6. Thus, we
choose half of 6, or 3.

(on board)
2. The factor you want is half of the coefficient of x. This factor
need not be an integer.

Now we need to add 3 . 3, or 3^2 = 9, to both sides.

(on board)
3. Add that factor squared to both sides.

(on board)
x^2 + 6x     = 12
          +9   +9
x^2 + 6x + 9 = 21

What's next? (let class answer)

(on board)
4. Factor into a perfect square.

(on board)
(x + 3)^2 = 21

Next? (let class answer)

(on board)
5. Solve by taking the square root.

(on board)
x + 3 = +- sqrt(21)
   -3   -3
x = -3 +- sqrt(21)


B. The Quadratic Formula

It is possible to go through this procedure in a completely general
manner for the equation ax^2 + bx + c = 0.

We'll go through it quickly in class, and you can review it in the
book in section 8.2.

The steps are as follows:

1. Starting from ax^2 + bx + c = 0, divide both sides by a:

ax^2 + bx + c = 0
x^2 + (b/a) x + c/a = 0

2. Move the (c/a) term to the other side:

x^2 + (b/a) x + c/a = 0
	      - c/a   -c/a
x^2 + (b/a) x = -c/a

3. Complete the square by adding [(1/2) (b/a)]^2 = b^2 / 4a to both
sides:

x^2 + (b/a) x = -c/a
     + b^2/4a   + b^2/4a
x^2 + (b/a) x + b^2 / 4a = b^2/4a - c/a
			 = (b^2 - 4ac) / (4a)

4. Factor the left side:

x^2 + (b/a) x + b^2 / 4a = (b^2 - 4ac) / (4a)
(x + b/2a)^2 = (b^2 - 4ac) / (4a)

5. Take the square root of both sides:

x + b/2a = +- sqrt(b^2 - 4ac) / 2a

6. Move the b/2a to the other side:

x + b/2a = +- sqrt(b^2 - 4ac) / 2a
  - b/2a   - b/2a

x = (-b +- sqrt(b^2 - 4ac)) / 2a

The result is something called the quadratic formula, which gives us
the solution to any quadratic equation.

(on board)
Quadratic Formula: the solution to the equation ax^2 + bx + c = 0 is
x = (-b +- sqrt(b^2 - 4ac)) / (2a).

We can use this to read off the answer to any quadratic equation once
we have put it in the form ax^2 + bx + c.

Let's try this.

(on board)
Solve x^2 - 4x = -4.

What do we do with this? (let class answer)

The first step is to put it into ax^2 + bx + c form.

(on board)
x^2 - 4x = -4
      +4   +4
x^2 - 4x + 4 = 0

Now we just read off a, b, and c and put them into the formula. What
are a, b, and c here? (let class answer)

(on board)
a = 1, b = -4, c = 4
x = (-b +- sqrt(b^2 - 4ac)) / (2a)
  = (-(-4) +- sqrt((-4)^2 - 4(1)(4))) / (2(1))
  = (4 +- sqrt(16 - 16)) / 2
  = 2

In this case there is only one solution because the square root term
cancels out. The solution is x = 2.

This problem is factorable, so we can use factoring to check that this
works.

(on board)
x^2 - 4x + 4 = 0
(x - 2)(x - 2) = 0
x - 2 = 0 or x - 2 = 0
x = 2

Of course the advantage of the quadratic formula is that it works even
for things that aren't factorable over integers.

It is also possible to wind up with a negative number under the square
root, in which case the solutions will be complex numbers.

For example:

x^2 + 2x + 5 = 0

Here a = 1, b = 2, and c = 5.

Using the quadratic formula, we get

x = [-2 +- sqrt(2^2 - 4(1)(6))] / (2)
  = [-2 +- sqrt(4 - 20)] / 2
  = [-2 +- sqrt(-16)] / 2
  = [-2 +- 4 i] / 2
  = -1 +- 2i

Let's check that this works:

(x + 1 + 2i)(x + 1 - 2i)
   = x^2 + x - 2i x + x + 1 - 2i + 2i x + 2i - 4i^2
   = x^2 + 2x + 5

So this works.


C. The Discriminant

Notice that the type of solution we get is determined by what is under
the square root.

This quantity, b^2 - 4ac, is called the discriminant.

By looking at it, we can tell what type of solution there will be.

We've already seen two cases.

When the discriminant was positive, we took the square root and got a
real number, so the result was two real solutions.

When it was negative, we took the square root and got something
imaginary, so the result was two complex solutions.

There's one more possiblity, which is that it could be zero.

For example:

x^2 - 4x + 4 = 0

Here a = 1, b = -4, c = 4, so the solution is

x = (4 +- sqrt[(-4)^2 - 4(1)(4)]) / 2
  = [4 +- sqrt(16 - 16)] / 2
  = 2

There's only one solution, because the +- term disappeared. And this
makes sense. If we factor, we get

x^2 - 4x + 4 = (x - 2)^2 = 0,

so there really is only one solution.

To summarize, there are three possibilities:

b^2 - 4ac > 0 ==> Two real solutions
b^2 - 4ac = 0 ==> One real solution
b^2 - 4ac < 0 ==> Two complex solutions


II. Graphing Quadratics

Now let's talk for a bit about what graphs of quadratics look like.

We'll see that the graphs allow us to make sense of some of the
strange properties we've discovered for quadratics.


A. The Parabola

Let's start by graphing the simplest quadratic we can. Consider:

(on board)
y = x^2

We graph this just like we would graph a line. What should we do? (let
class answer)

First we make a table:

(on board)
x   x^2 = y
0   0
1   1
-1  1
2   4
-2  4

Then we draw the graph.

(draw graph on board)

That's pretty straightforward.

This curved shape is characteristic of all quadratics. It's called a
parabola.

One interesting thing to note about the parabola: it's symmetric.

In other words, if we flip it over, left to right, it remains
unchanged.

This reflects a symmetry in the equation: if we square a negative
number, we get the same thing as when we square a positive one.

For example, we got the same thing for y when we plugged in 2 and -2.

This invariance under changing x from positive to negative is
responsible for the symmetry of the graph.

We can also figure out the domain and range of this function.

The domain is easy: we can put in any x, so the domain is any real
number.

We'll get back to the range in a bit.


B. Effects of the Coefficients

Recall that in general a quadratic is of the form

(on board)
y = ax^2 + bx + c

The coefficients a, b, and c can be any numbers.

Let's study how the shape of the parabola varies as we vary a, b, and
c.

1. Varying a

Let's start by varying a.

In the graph we just drew, what value of a did we use? (let class answer)

We used a = 1.

Now let's try a = 1/2:

(on board)
y = 1/2 x^2

Someone walk me through graphing this. (let class walk through it)

(on board)
x   2x^2 = y
0   0
1   1/2
-1  1/2
2   2
-2  2

(draw graph)

How does this compare to the graph of y = x^2? (let class answer)

The difference is that this parabola is wider.

That's one effect of changing a: values of a closer to zero lead to
wider parabolas, while values of a further from zero lead to narrower
ones.

How about if we make a negative?

(on board)
y = -x^2

Here we're using a = -1.

Let's graph this:

(on board)
x   -x^2 = y
0   0
1   -1
-1  -1
2   -4
-2  -4

(draw graph)

We can immediately see the effect of the sign: this parabola opens
downward.

Thus, there are two effects of a:

(on board)
1. Sign of a:
   a>0 => parabola opens upward
   a<0 => parabola opens downward
2. |a|: larger values lead to narrower parabolas


2. Effects of c

I'm going to skip b for a moment and come back to it.

First, I want to talk about c.

Let's take another example:

(on board)
y = x^2 - 2

What is c here? (let class answer)

Let's graph this:

(on board)
x   x^2 - 2
0   -2
1   -1
-1  -1
2   2
-2  2

(draw graph)

What did c do? (let class answer)

The effect of c is to shift the graph up or down without changing the
shape.

A positive value of c shifts the graph upward, a negative value shifts
it downward.

This makes sense: adding a constant just makes y larger of smaller,
regardless of x.


3. Effects of b

Finally, let's consider the effects of b.

(on board)
y = x^2 - 2 x

What is b here? (let class answer)

Let's graph this:

(on board)
x   x^2 - 2 x
0   0
1   -1
2   0
-1  3
3   3

(draw graph)

We can notice two things here:

First, the shape of the parabola is the same.

Second, the parabola has been shifted down and to the right.

To gain more insight, let's graph

(on board)
y = -x^2 + 2 x

(on board)
x   -x^2 + 2 x
-1  -3
0   0
1   1
2   0
3   -3

(draw graph)

This graph has been shifted up, and b has gone left instead of right.

Thus, we can summarize the effects of b:

(on board)
A more positive b shifts the graph to the left.
A more negative b shifts the graph to the right.

It turns out that whether the graph is shifted up or down depends not
just on b, but on a as well.


C. Zeros

One thing we're always interested in with graphs is their intercepts.

In particular, we'd like to find the x intercepts of this graph, the
points where y = 0.

(draw picture showing the zeros)

So, if we have, for example

(on board)
y = x^2 - 2x - 3

we want to find where y = 0.

Fortunately, we don't have to graph to find the zeros.

We want to find the value of x that gives

(on board)
y = x^2 - 2x - 3 = 0

But this is just a quadratic equation, and we know how to solve
these. Someone solve this for me. (let class walk through it)

(on board)
x^2 - 2x - 3 = 0
(x - 3)(x + 1) = 0
x = 3 or -1

Let's graph this just to make sure.

(on board)
x   x^2 - 2x - 3
0   -3
1   -4
2   -3
3   0
-1  0

(draw graph)

As we can see, this graph does indeed cross the x axis at x = 3 and
x = -1.

Thus, the two solutions of a quadratic correspond to the two places
where the parabola crosses the axis.

Let's consider a slightly different equation now:

(on board)
y = x^2 - 2x + 1

Someone solve for the zeros of this. (let class walk through it)

(on board)
x^2 - 2x + 1 = 0
(x - 1)(x - 1) = 0
x = 1

Here these is only one solution. What do we get if we graph this?

(on board)
x   x^2 - 2x - 3	x^2 - 2x + 1
0   -3			1
1   -4			0
2   -3			1
3   0			4
-1  0			4

(draw graph)

As we can see, this corresponds to the case of a parabola that just
touches the x axis at a single point.

Finally, let's consider

(on board)
y = x^2 - 2x + 2

Let's check this one with the quadratic formula. What are a, b, and c?
(let class answer)

(on board)
a = 1, b = -2, c = 2
x = (-b +- sqrt(b^2 - 4 a c)) / (2a)
  = (2 +- sqrt(4 - 8)) / (2)
  = 1 +- 1/2 sqrt(-4)

As we can see, this has no real solution.

And if we graph:

(on board)
x   x^2 - 2x - 3	x^2 - 2x + 1		x^2 - 2x + 2
0   -3			1			2
1   -4			0			1
2   -3			1			2
3   0			4			5
-1  0			4			5

(draw graph)

Thus, the case of no solutions corresponds to a parabola that is
always above the x axis.


D. Vertex

Another useful thing to find on a parabola is the point where it
reaches its maximum or minimum.

This point is called the vertex.

(draw a picture showing vertex)

Consider an example:

(on board)
y = x^2 + 2x - 2

Since a is positive, we know this parabola opens upward, and we're
looking for a minimum.

We can find the vertex of a parabola using our procedure for
completing the square.

We want to complete the square. Does anyone remember how to do this?
(let class answer)

First, we move the 2 to the other side.

(on board)
y = x^2 + 2x - 2
+2	      +2
y + 2 = x^2 + 2x

Next we find b. What is b here? (let class answer)

b is 2.

We take that and halve it:

(on board)
b/2 = 2/2 = 1

Then we square it:

(on board)
(b/2)^2 = 1^2 = 1

Then we add that to both sides:

(on board)
y + 2 = x^2 + 2x
   +1		 +1
y + 3 = x^2 + 2x + 1

We then factor the perfect square.

(on board)
y + 3 = x^2 + 2x + 1
      = (x + 1)^2

Now let's take that 3 and put it back on the other side, so we can
have y by itself.

(on board)
y + 3 = (x + 1)^2
   -3	-3
y = (x + 1)^2 - 3

What was the point of all this?

Well, in this form we can make a simple observation. The observation
is that the term (x+1)^2 is always positive.

Since it's a square, it can't ever be negative. The smallest it can be
is zero.

If we want to find the minimum value of y, it is therefore going to
have to occur wherever (x+1)^2 is 0.

This lets us solve for the x coordinate at which y has its minimum,
the x coordinate of the vertex.

It must be the value where

(on board)
(x + 1)^2 = 0
x = -1

We can then plug in to get the y coordinate of the vertex.

(on board)
y = (x + 1)^2 - 3
  = (-1 + 1)^2 - 3
  = -3

So the coordinates of the vertex are

(on board)
(-1, -3)

Let's graph the parabola to make sure that looks right.

(on board)
x   x^2 + 2x - 2
0   -2
1   1
-1  -3
-2  -2
-3  1

(draw graph)

As we can see, it looks like (-1,-3) is indeed the minimum.

Fortunately, we don't have to go through this procedure every time.

Just as it is possible to go through the process of completing the
square once in general and derive a general formula for the zeros --
the quadratic formula -- we can do the same thing for the vertex.

If we do so, we find that the x coordinate at which the vertex occurs
is

(on board)
x = -b / (2a)

For any quadratic, we can just find a and b and this will give us the
x coordinate of the vertex.

Then we just plug in to get the y value.

Note that c doesn't appear in this formula.

That's exactly what we expect: since changing c only shifts the graph
up and down, it won't affect the x coordinate, the horizonal position,
of the max or min.

Note that we can also figure out the range now.

The vertex is the point where the graph goes to its highest or lowest,
depending on x.

Thus, the y coordinate of the vertex is going to be one end of the
range.

If the parabola goes up, because a > 0, then the range is from that
point to plus infinity. Otherwise it's from -infinity to that point.

For example, what's the range of y = 2x^2 + 4x - 5?

Well, the vertex is at x = -b / 2a = -4 / (2*2) = -1.

The corresponding y is

y = 2x^2 + 4x - 5
  = 2(-1)^2 + 4(-1) - 5
  = 2 - 4 - 5
  = -7

Since a > 0, this parabola goes upward, so the range is [-7, infinity).

E. Axis of Symmetry

Once we've got the vertex, we get a freebie with it.

Another thing that's useful to know is what's called the axis of
symmetry.

I noted earlier that a parabola is symmetric, meaning that if you flip
it over it stays the same.

When you flip it over, you're flipping it around some specific line.

This line is called the axis of symmetry.

(draw an example on the board)

From the graphs we've drawn, we can see that a parabola is always
symmetric about a vertical line through its vertex.

We just talked about how to find the coordinates of the vertex, so we
can find the x coordinate.

Recall that the equation of a vertical line is just x = some constant.

Thus, the equation of the axis of symmetry is

(on board)
x = -b / (2a)

We just have to fill in a and b.

So, for example, what are the coordinates of the vertex and the
equation of the axis of symmetry for this quadratic:

(on board)
y = x^2 - 6x + 5

(let class answer)
a = 1, b = -6
x coordinate of vertex: x = -b / (2a) = 6 / 2 = 3
y coorindate of vertex: y = 3^2 - 6(3) + 5 = -4
Vertex: (3, -4)
Axis of symmetry: x = 3