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Hello, and welcome Bay College's online lectures for
College Algebra.
This is Section 2.1, which deals with distance, midpoint,
graphs, intercepts, and symmetry.
It seems like a lot, but we're just going to touch on them.
You should have some background in this at this
level of math.
This is college algebra.
The first thing we're going to look at is the Cartesian
coordinate system.
This is just a way that we can use to graph points that
represent either an equation, or the distance between two
points, or maybe we want to find what's the midpoint in
between them.
We're going to use the Cartesian coordinate system.
The Cartesian coordinate system is essentially made of
two number lines set at 90 degrees from each other.
One we call the x-axis, and that is our horizontal axis,
and the other one is the y-axis.
In quadrant I, all the points that we can graph in here have
an x value and a y value, where the x value is positive,
and the y value is positive.
That is the definition of quadrant I. Both our x and y
are positive.
In quadrant II, as we move counterclockwise here, our x
values are negative, and our y values are positive.
Anything to the left of the y-axis is a negative x value.
Moving down the quadrant III, here both the x and the y are
negative values because we're to the left of the y, but
we're below x.
So these are both negative, negative x
values, negative y values.
And then, finally, in the fourth quadrant, our x values
are positive because we're to the right of the axis, and our
y values are negative because we're below the x-axis.
So this is our quadrants of our
Cartesian coordinate system.
Next thing we're going to do is we're going to move over
here, and we're going to look at distance.
If we want to find the distance between two points,
we can use something called the distance formula.
And the distance formula isn't really anything new.
You should be somewhat familiar with it.
It is just a variation of Pythagorean theorem.
At some point in your math career, you should've seen or
come across Pythagorean theorem.
So we're going to take a look at these two points.
I have the point 1, 2, where x is the 1, and
the y value is 2.
We're in the first quadrant, so both are positive.
And then we're down here in the third quadrant.
Our second point is negative 3, negative 2.
Maybe I want to find the distance between them.
And I made reference to Pythagorean theorem because if
I want to find this distance between here and here,
essentially what I have to do is say, how far over in x do I
have to go and how far up in y?
If we think about it, I just made a 90-degree triangle, so
this can be described using Pythagorean theorem.
If we move over here, we can see c squared equals a squared
plus b squared.
That is Pythagorean theorem.
Essentially, our a is our horizontal change from this
point to this point.
And our horizontal change can be described as the change in
x, x2 minus x1, that quantity squared.
And our y values is the change in y.
How far up do I have to go to get from this
point to that point?
There is some change in y.
So y2 minus y1, that's my change in y,
that quantity squared.
It just replaces b.
Now, because this is squared--
and we call it a distance because
distances are always positive--
so we're not too concerned in that negative distance because
there's no such thing.
Really, it's just direction.
So to solve for this, we'd take the square
root of both sides.
So d equals the square root of the change in x squared plus
the change in y squared.
This is the same as that.
It's just another way of expressing it.
So let's find the distance between these two points.
Well, my change in x, if I call this point 1 and this
point 2, just moving left to right, I can find the distance
by saying, the square root of x2 minus x1--
1 minus a negative 3, well, that'd be 1 plus
3, which is 4 squared--
plus the change in y, y2 minus y1.
So that'd be 2 minus a negative 2.
Well, 2 minus a negative 2 is 2 plus 2, which is 4, that
quantity squared.
So 4 squared is 16 plus 4 squared is 16.
16 plus 16 is 32.
And hopefully, we remember when we have radicals, we
should simplify them.
This is 16 twice, as we've seen, and 16
is a perfect square.
The square root of 16 is 4 times the square root of 2,
that factor that I could not take the square root of.
So this would be our distance between those two points, the
distance from here to here.
The length of this line, or our hypotenuse of our
triangle, would be 4 square root of 2 units, whatever
value that is.
Now, what if we're interested in finding the midpoint of a
line segment?
What's the halfway point?
And maybe that's not it.
That's why we're going to find it algebraically.
But what's the middle of this line, the midpoint?
What's somewhere in between?
Well, here's the midpoint formula.
And honestly, I feel you do not have to
memorize any new formulas.
If you know Pythagorean, you can derive
your distance formula.
But this one here, this is essentially asking us, what is
the average x and average y?
If you know how to find an average, you just sum the
values together and divide it by the number of values.
That is the midpoint formula.
I have two x values, so I'm going to sum them together.
Negative 3 plus 1 is negative 2.
Negative 2 over two points is going to be negative 1.
And then to find the average y value, I
just sum them together.
Negative 2 and positive 2 is 0, and 0 divided by 2 is 0.
So negative 1, 0, actually this right here, would be my
midpoint, negative 1 in the x, 0 in the y.
I'm right on the x-axis.
I don't go up any.
I don't go down any.
0 in the y direction.
So this is my midpoint, and I can label that
point negative 1, 0.
Make sure when you're asked to label points in college
algebra, you actually write it as an ordered pair because
that's how we describe ordered pairs on a Cartesian
coordinate system.
We have to label them with an x value and a y value.
All right.
Let's move on and look at how graphs of these equations work
on a coordinate system.
The first thing we're going to do is let's just take, for
example, y equals x squared minus 1.
Here we have a parabola.
Hopefully, we recognize that as a parabola.
If not, you will throughout the semester.
We have a question here, is x equal to 2 and y
equal to 3 a solution?
Well, we can check it simply by putting it in.
If y is 3, I can say 3 equals this value here, which says x
squared minus 1.
Well, x is 2, so I replace the x term with a 2 minus 1.
Now, let's just see.
Is this a true statement?
If it is, that makes this a solution to this equation.
Well, 2 squared is 4.
4 minus 1 is, in fact, 3.
3 equals 3 is a true statement.
So what I'm going to do is I'm going to
build a table of values.
2, when it goes into this equation for the x value, we
get a y value of 3.
Those were the values given-- x is 2, y is 3.
Well, what if I chose more points?
What if I said, what if x is 1?
Well, 1 squared is 1, minus 1 is 0, y would be 0.
If I say x is 0, I can plug in that value.
0 squared is 0, 0 minus 1 is still negative 1, y equals
negative 1.
And I can continue on in the negative direction as well.
These values were positive, and now we're
going into the negative.
If I put in negative 1, I still get 0.
If I put in negative 2, well, negative 2 squared is still
positive 4, minus 1 is still 3.
So here's a question.
How many solutions does this equation have?
Well, if we think about this table, if I continue in the
positive direction, I'm going to find y values.
If I continue in the negative value, I'm still
going to find y values.
Essentially, this has infinite solutions.
So how can we describe an equation that
has infinite solutions?
Well, one way is by graphing it on a
Cartesian coordinate system.
If we go over to this graph here, let's start plotting
some points.
2, 3--
so that means positive 2 in the x, up 3 in the y.
And then I have 1, 0.
So when y is 1--
or excuse me, when x is 1, y is 0.
When x is 0, which is right on the y-axis, y is negative 1.
So we notice it's right on the y-axis because
the x value is 0.
And we have negative 1, 0.
When x is negative 1, y is 0.
And finally, negative 2 for x, positive 3 for y, negative 2,
positive 3.
And if we continue to do that, we'd see a pattern.
And hopefully, we already see a pattern.
If we're familiar with parabolas, we know that they
have this shape.
This is the parabolic shape, and I put these arrows on here
to indicate that that would continue on forever.
As x gets larger, y also gets larger.
As x goes into the negative direction, y also gets larger.
So let's define a few points on this graph
by going over here.
If we want to know where it crosses or touches the x-axis,
essentially what we're asking for is, what is the value of
the y-axis, or excuse me, the x-axis?
What is the y value at the x-axis?
Well, the y value is always 0 at the x-axis because we're
not going up any or down any.
We're at 0 in the y.
Now, if we ask that question, essentially what we're looking
for is the x-intercept.
To find the x-intercept, we essentially set y equal to 0.
And if we go back to our equation and
set y equal to 0--
x squared minus 1 equals 0--
we find that x can equal plus or minus 1 when y equals 0.
So let's look at this here.
This value here, we have two x-intercepts.
Where the y value is 0, it crosses at 1 and negative 1
when y is 0, so negative 1 and 1 when y is 0.
These are called the x-intercepts.
Where does it cross or touch the y-axis?
Well, if we go back to the graph, we can see the value
right here.
Well, what is the x value?
Well, the x value is 0 because we are right on the y-axis.
And if we plug 0 in for x--
0 squared is 0, minus 1 is negative 1--
we'd have a y value of negative 1.
And if we look at the graph, sure enough, it crosses the
y-axis where y is negative 1.
This is called the y-intercept.
Now, to find any of the intercepts, you always set the
other variable to 0.
If I want to find the x-intercept, I set y to 0.
If I want to find the y-intercept, I set x to 0.
All right.
So here's your quiz.
Using the example y equals x squared minus 1, I want you to
find the intercepts algebraically.
Actually plug in 0 and prove that this is true.
And I want you to show your work.
All right?
So that's your quiz.
Do that for yourself.
Make sure you show your work.
All right.
Moving over here, we're going to talk about symmetry.
For this equation, y equals x squared minus 1.
It does have some symmetry.
And what is symmetry?
Well, we're actually looking relative to
some axis or the origin.
Are there points on one side that are mirror images of the
points on the other side relative to an axis?
Well, y symmetry means that what's on the right side of y
is also on the left side of y.
Talking in terms of ordered pairs, the y value is the
same, but the x values, one's positive and one's negative of
equal distance from the y-axis.
And we can find these values, which we'll see in an
example coming up.
If I have a function, if I put in a negative value of x and
get out the same value of y had I put in a positive value
of x, we can see that this is symmetric with the y.
In this example, let's actually do that.
If I put in a negative value of x, what happens when I
square a negative?
It becomes positive.
So essentially, y doesn't change.
Its value is the same as the original function.
So it doesn't matter if x is positive or x is negative, we
will get the same result.
That means, algebraically, this is symmetric with respect
to the y-axis.
Well, we can also have symmetry with
respect to the x-axis.
If I replace y with a negative y value, I get the
same x value out.
Now, this wouldn't be symmetric with that because
whenever I square this negative, y is actually not
going to change.
So the signs won't change.
OK?
So this would not have y symmetry, or x
symmetry, excuse me.
And finally, origin, symmetry through the origin.
If we review the Cartesian coordinate system for a
moment, symmetry through the origin means if there's a
point up here that's positive and positive, because it's in
the first quadrant, we have that exact same point mirrored
where it's negative, negative.
OK?
So this point is mirrored, but it's mirrored through the
origin, our value 0, 0.
And hopefully, we remember that's the origin.
How can we test if something's symmetric through the origin?
Well, if I have some function with x, y values, if I put in
a negative x value, I get out a negative y value.
Both signs change if I put in a negative, and that tells me
that it's symmetric through the origin.
So we're going to go back to the other end of the board
here, and we're going to look at testing for symmetry.
How do we do this?
Well, if we look at this function, to test for x
symmetry, I'm going to put in a negative y value.
So to test for symmetry, I put in a negative of the opposite.
So I have x equals negative y squared
instead of just y squared.
Now, what happens when I square this negative?
I get x equals y squared.
This is the original function.
If that happens, it means we are symmetric with respect to
the x-axis.
So we have symmetry with respect to x.
If I wanted to test for y symmetry, well, I replace x
with a negative x.
Is this the same function?
There's no simplifying I can do here.
Is this the same function that I started with?
No, this is not.
So there is no symmetry with respect to y.
Finally, to test for symmetry through the origin, I replace
negative x and negative y, and then we do some simplifying.
Well, if I square a negative, I get negative x equals
positive y squared.
Is that the same as the original function?
No.
So it is not symmetric through the origin.
Now, we didn't actually have to test all three of these.
If it's symmetric with one, it will not be
symmetric with another.
It's one or another, so not both.
So we see, yep, it was symmetric with the x because
when we replaced the y, we got the original
function back out.
Let's do it again, but let's look at this example.
We have y equals x cubed.
Let's test for x symmetry.
If I put in negative y, because I'm testing for x
symmetry I replace the y value, is there any
simplifying I can do there?
Not really.
So does this result in the original equation?
No, it is not symmetric with x.
Well, let's test and see if it's symmetric with y.
If I put in a negative x value and then do a little
simplifying, what happens when I cube a negative?
Well, I get y equals negative x cubed.
When you cube a negative, it stays negative.
Is that the same as the original function?
No.
So that tells me it is not symmetric with y.
Finally, let's test symmetry through the origin.
If I put in a negative y value and a negative x value, and
then I do some simplifying, well, negative x cubed is
negative x cubed.
And there's yet some more simplifying I could do.
If both sides are negative, I can factor out a negative 1,
or divide through by negative 1, or multiply through by a
negative 1, and I get y equals x cubed.
This is the same function I started
with, y equals x cubed.
We can see, yes, there is symmetry with
respect to the origin.
Now, let's just go back for a moment, and I'm going to graph
this real quick.
This is the graph of x equals y squared.
And we can see what's above the x-axis is mirrored below
the x-axis.
So this is symmetric with respect to the x-axis.
If we look at this function--
I'll just graph it real quick--
it looks like this.
Which means for every value up here in the first quadrant,
plus 1, plus 1, or plus, plus in our first quadrant-- both
values are positive--
we have a value down here mirrored through the origin
that's negative, negative.
So if I put in a negative, I get out a negative.
If I put in its equivalent value but positive, I get the
equivalent value of x that's positive.
So this would be symmetric through the origin.
So that has been Section 2.1.
Thank you for watching.