In this video, I want to show some examples of graphing
quadratic functions.
But before I do that, I want to make one fairly minor
correction to the end of the last video where we did some
factoring with grouping.
When I showed you the proof of why that works, I multiplied
fx plus g times hx plus j.
When I multiplied them, I did the first term, I
said fx times hx.
When I did it in the last video, I just wrote fhx.
And we know that that's not true, we have f times h
times x times x.
It's fhx squared.
That times that is fhx squared and then the rest of it I had
gotten right, it's fx times j, so plus we get fjx and then we
had g times hx so we could write that as plus ghx.
And then finally g times j, which gives us gj.
So I apologize for that error, I had forgotten to write the
exponent there and I think I'd forgotten to write it in the
step after that, but it didn't change the argument for the
proof, I at least didn't mess up there.
So I just want to make that quick correction in case that
confused anyone.
And I also made a small annotation in that video if
you're watching it on YouTube.
With that out of the way, let's learn to graph some of
these quadratic functions.
So let's say I have the function y is equal to x
squared minus 2x minus 8.
And what we're going to do is we're going to rewrite this
quadratic expression on the right-hand side.
We're going to write it in intercept form.
That essentially means just writing it
in it factored form.
Then doing that, we're able to figure out the x-intercept,
where we're intersecting the x-access, and then using that
we'll be able to figure out the vertex.
And I'll show you you what the vertex is in a second.
So let's just factor this quadratic on
the right-hand side.
So we get y is equal to-- so what are two numbers where
their product is a negative 8, so they're going to have
different signs and then their difference is negative 2?
Let's see, if we have a negative 4, and plus 2.
When you take the product, you get negative 8, when you add
them, you get negative 2.
So we could rewrite y as being equal to x minus 4
times x plus 2.
So this is essentially writing this quadratic function in
intercept form.
And now let's figure out where this intersects the x-axis So
let me label my axis.
This is, of course, the x-axis, that is the y-axis.
So when will something intersect the x-axis?
The x-axis is when y is equal to 0, right?
So let's set this equal to 0.
0 is equal to x minus 4 times x plus 2.
We've seen this multiple times before, this means that either
x minus 4 or x plus 2 or both of them have to be equal to 0.
So x minus four is equal to 0 or x plus two is equal to 0.
So that means x could be equal to 4-- we're just adding 4 to
both sides of that equation-- or subtracting 2 from both
sides of this equation, x could be equal to negative 2.
So these points where we intersect, when x is 4, this
term is 0 so the whole thing is 0.
That's how we solved for this.
So the points 4, 0, and the point 2, 0 are going to both
be on this parabola.
So let's graph them.
So 4, 0-- let me do this in a darker color-- 1, 2 3, 4,
that's 4, 0.
And then the other point is negative 2, 0.
1, 2 negative 2, 0 is right there.
So that's the two points where we intersect the x-axis.
And now we're going to determine something called the
vertex of our parabola.
And the vertex, just to give you a sneak preview, parabolas
are the graphs of quadratic functions and so the graphic
is going to look either like an upward u or a downward u,
and it's going to be shifted around on this.
But the vertex is either this minimum point or
this maximum point.
And the word parabola just describes this function shape.
It's the shape of the graph of a quadratic.
Let me write that word down, these are parabolas.
So to figure out the vertex, you actually just take the
x-value that's halfway in between the two intercepts.
No matter what, the intercepts are going to be equidistant
from the vertex.
So if that is the graph, those would be our intercepts and
then our vertex is going to be an x-value exactly halfway
between them.
So to figure out the x-value of our vertex, we just average
these two x-values of our x-intercepts.
So let's do that.
We could say the x for the vertex, let's call that x for
the vertex-- that's why I wrote a v there-- is going to
be equal to the average of these.
So 4 plus negative 2 over 2.
Which is equal to this 4 minus 2, which is 2 over
2 is equal to 1.
So x is going to be equal to 1 and what is y when
x is equal to 1?
So the y for the vertex is going to be equal to 1
squared-- I'll just write that as 1-- minus 2 times 1, so
minus 2, minus 8.
So what is this?
This is 1 minus 10, which is equal to negative 9.
So the vertex is going to be at the point 1, negative 9.
So we go one and then we go down nine.
So x is 1 and then y is negative 9.
1, 2, 3, 4, 5, 6, 7, 8, 9.
So it's right there.
And so if we were to graph this parabola, the line of
symmetry is on the x-value of the vertex.
So, I want to be doing it in a darker color than that.
So the graph should be symmetric around x is equal to
1, and the graph will look something like this.
I'll try my best to draw it.
Bam!
It'll intersect the x-axis there, it'll be symmetric,
it'll look just like that on the other side.
And it'll go bam, just like that.
I think you get the general idea.
And then the other point that you may or may not be
interested in is the actual y-intercept.
If x is equal to 0, you immediately see that y is
equal to negative 8.
So the point 0, negative 8 should also be on this graph.
Let's see if I drew it right.
0, negative eight right there should also be on that graph.
Let's do another one of these.
I'm kind of taking up some of the space for the next
problem, so let me clear this.
Let's do another one.
All right.
So let's say we have y is equal to 2x squared
plus 6x plus 4.
So here, the thing that immediately jumps out at me is
that I can factor a 2 out of everything.
So I could rewrite this as y is equal to 2 times x squared
plus 3x plus 2.
That saves me the pain of having to do the factoring by
grouping that we talked about in the last video.
And this we can factor in a pretty straightforward way.
This is 2 times 1 is 2, 2 plus 1 is 3, so this is x plus 2
times x plus 1, right?
2 plus 1 is 3, 2 times 1 is 2.
So that's what y is equal to.
So if we wanted to know the x-intercepts, we figure out
where this expression is equal to 0, so we said 0 is equal to
2 times x plus 2 times x plus 1.
Remember, we want to figure out where we intercept the
x-axis, that's the x-axis there.
Y is 0 on the x-axis.
So let's get back to this problem.
So either of these can be equal to 0.
Obviously the 2 can't be equal to 0, so we get the situation
where x minus x plus 2 is equal to 0 or x plus 1 is
equal to 0.
I didn't have to write the parentheses there.
Subtract 2 from both sides of this equation, we get x is
equal to negative 2.
Subtract 1 from both sides of this question or x could be
equal to negative 1.
So the point negative 2, 0.
You put negative 2 here, this is obviously going to make y
equal to 0 and the point negative 1, 0, same argument.
Those are our x-intercepts.
Let me graph it.
We have negative 2, 0 right there.
Actually let me do this graph a little bit bigger.
Let's say that this is right here is-- let me label it-- in
double steps.
So let's say this is negative 1, let's say this is negative
2, that is negative 3, 1, 2, this is our 3.
It's a 1, 2, 3 just like that, negative 1, negative 2,
negative 3.
So, I'm using two blocks to represent one.
So the point negative 2, 0 is on our graph, with our
x-intercept, so negative 2, 0 is right there.
And then we have negative 1, 0, which is right there.
And then we want to figure out the vertex.
Well, the vertex is going to be the average of those two,
so the x for the vertex is going to be negative 2 plus
negative 1 over 2, that's negative 3/2 which is the same
thing is negative 1.5.
Now what is y equal to when x is negative 1.5?
This is going to be a little bit involved.
I'll use actual 3/2.
So so we get y is equal to 2 times negative 3/2 squared
plus 6 times negative 3/2 plus 4.
And let us figure this out.
So this is equal to 2 times 9 over 4 minus-- you have that
negative sign there-- and then we could make this 6 divided
by 2 is just the same thing is 3/1.
So this is minus 9, 3 times 3 plus 4.
This 2/4 the same thing as 1/2, so this becomes 9/2
minus 9 plus 4.
So let's simplify this.
Minus 9 plus 4 is the same thing as minus 5, right?
Minus 9 plus 4.
Minus 5 we can rewrite is equal to minus 10/2.
9 minus 10, all of that over 2, is minus 1/2.
So, our vertex of this parabola is the point minus
1.5 or negative 1.5, negative 1/2.
This is our vertex, so we can graph it right here.
Negative 1.5, negative 1/2, our vertex
is right over there.
And then we can a graph this parabola.
If we want to do the y-intercept, we can.
The y-intercept is going to be here when x is 0,
y is equal to 4.
These terms cancel out.
X is 0, y is 1, 2, 3, 4 right over there.
And then we can graph this parabola.
It will look like that, that's my best, it should be
symmetric around x-coordinate of the vertex just like that.
Let's do one more of these.
And, once again, I've used all my real estate up, let me
clear this out of the way.
Let's do one more.
Let's say that I have y is equal to negative x squared
plus 10x minus 21.
Once again, let's write this in intercept form, we just
factor the right-hand side.
We get y is equal to-- well the first thing I want to do
is it is factor this negative 1 out-- negative 1 times x
squared minus 10x plus 21.
Now what two numbers, when I multiply them, they're
positive 21?
So they have to be the same side, and when I add them I
get negative 10.
At least in my brain, negative 7 and negative 3 jump out.
I mean, there's not that many factors of 21, so 7 and 3 are
probably two that usually jump out because they add to 10.
So you have y is equal to negative 1 times x minus 3
times x minus 7.
Negative 3 plus negative 7 is negative 10, negative 3 times
negative 7 is positive 21.
So if you want to find the x-intecepts, the x-intercepts
are going to be where-- so we could set this whole equation
equal to 0, negative 1 times x minus 3 times x minus 7 are
x-intercepts or the x-values that make y equal to 0 are
either the ones that make x minus 3 equal to 0 or the
x-values that make x minus 7 equal to 0.
Add 3 to both sides of this equation, you got x is equal
to 3, add 7 to both sides of this equation, you get x is
equal to 7.
So so we get our two x-intercepts, the point 3, 0--
1, 2, 3, 0-- and the point 7, 0-- 1, 2, 3, 4, 5, 6, 7-- 0.
Now, we know that our vertex is going to be right in
between them.
So the x for the vertex is going to be 3 plus 7 over 2,
which is equal to 10/2, which is equal to 5.
And to the y-value, let's figure it out, y for our
vertex going to be negative 5 squared, so negative 25 plus
10 times 5 plus 50 minus 21.
So what is this?
This is equal to-- we have a negative 25 and negative 21,
you add those together you get a negative 46.
Add that to 50, you get y is equal to 4.
So the vertex is at 5, 4.
Let me graph that, the vertex is 5, 4.
So 1, 2, 3, 4, 5, right in between.
And then up four, 1, 2, 3, 4.
So notice something, in this graph over here, the last two
graphs, actually I think I just cleared them, you saw
that the vertex was below the x-axis.
So, when we plotted the x-intercept and then the
vertex, the upward-forming u-shape was the only option.
But here, the only option is a downward u-shape.
This being either the minimum or maximum point and going for
both of these points and having a u-shape, our u-shape
is going to have to look something like this.
Now,-- let me draw that little bit better-- the u is going to
have to look something like that.
That's a better shot at it.
Now notice what was the difference between this
equation and the previous ones?
The difference was is that this one had a negative in
front of the x squared.
The coefficient in front of the x squared
was a negative number.
And that's why we have kind of a downward opening parabola.
If this was a positive number, we would have an
upward-opening parabola.
So that's why this one is kind of a downward shape and its
vertex represents a maximum point.
It's the highest point on the parabola.
In the previous videos, we had an upward opening, or in the
previous problems, with an upward-opening parabola and
our vertex was the minimum point.