This video is going to be about finding
the quadratic equation when you're given
the roots of the equation and a point.
So here's what this is about.
Let's say I've got this graph.
This graph has a downward opening parabola.
It goes through the horizontal axis, the
x-axis,
at the point where x equals 2
and x equals 5.
So what we know is that there 
are two roots
to the equation,
and the roots are 2 and 5.
It also goes through the vertical axis, the y-axis,
at the place where y equals negative 2.
So I can take those two roots...
I can write x = 2
and 5,
and as I've done in the previous video,
I can
rewrite them as
x = 2
and x = 5.
And then I can set both of those 
equations equal to zero.
I could have
x - 2
equals zero,
and x - 5
equals zero.
and then since both of the equations
are equal to zero, 
I'm just going to multiply them together.
So I've got 
x - 2
times
x - 5
equals zero.
And now I'm just gonna 
change its form to y =
x - 2
time x - 5.
Now I could multiply these out,
the x - 2 and the x - 5, but I'm not
ready for that yet.
Here's the next thing I've got to think about.
I've got a third point here,
(0, -2),
and I haven't accounted for that.
Here's what I'm getting at.
Let's look at 
this pair of axes.
So I've got an x- and y-axis,
and on the x-axis I'm going to mark 
the point where x equals 2
and where x equals 5.
And I could draw a parabola that goes through those two points. I could draw a parabola like this,
and that parabola would have the solutions
x equals 2 and 5.
But I could have drawn 
a totally different parabola.
I could have drawn an upward opening parabola that also went through those two points.
And that would have the same solutions, 
x equals 2 and 5.
I could have drawn a third parabola, 
a very steep downloading opening parabola
that goes through 2 and 5 on the x-axis,
and that would have the same solutions.
But I want an equation for a specific parabola.
I want the equation for the parabola 
that I was given.
and this parabola goes through the point
(0, -2).
So here's what I've got to do.
I've got to realize that that point, (0, -2)
is a coordinate pair, an ordered pair.
So I could think of the
first part as the part as the x and 
the second part as the y.
And now let's look at the equation I have so far.
Well, let's think about it. 
When I multiply this out, 
I'm going to get... If I multiply it just like this, 
I'm going to get an x-squared.
But it's a downward opening parabola,
So if it's a downward opening parabola,
I want the coefficient of the x-squared
to be negative.
So it seems like I would have to
multiply this whole equation
by something...
or rather, multiply the right side
by something
to make the x-squared negative.
Well, here's how we find out what the something is. Let's say that the something,
I'm going to call that 'a'.
Some books are gonna call it 'a', some books are gonna call it 'k'. I doesn't really matter what
you call it you can call it. You can't call it x or y, because we've used those two letters already.
I want to find out what I'm multiplying 
(x - 2) times (x - 5)
by
so that I'm going to get a negative
coefficient.
And it's got to be avery specific 
negative coefficient.
Okay, I know that I've got this point 
at (0,-2),
so I could think... at that point 
x equals zero and y equals -2.
So, any time I've got a coordinate pair, 
an ordered pair,
all I have to do
is take that pair and plug them in 
where I've got x's and y's.
So I'm gonna take the 
y equals -2,
I'm going to replace the y with -2.
I'll still have this 'a'.
And instead of x, I'm going to have zero.
So I'll have
zero minus 2
times zero minus 5.
And now all I've got to do is multiply this all out and find out what
'a' equals. So -2
equals
'a' times... zero minus 2 is -2, and 
zero minus 5 is -5.
So all I'm doing is multiplying 
-2 times -5,
and that's going to be a positive 10.
So  -2 equals 'a' times 10, or 10a.
And now I'll divide both sides by 10.
So I'm going to have
-2/10
-2/10
equals 'a',
and I'll reduce that 
-2/10. So it's going to be -1/5.
-1/5
-1/5
is 'a'.
and then I can go back 
where I've got this 'a'
and instead of the 'a'
I'm going to have
-1/5
-1/5.
Okay? So
this is going to be
the specific equation...
y = -1/5 times 
(x - 2) times (x - 5),  
the specific equation that gives
me
that specific
graph.
Let me go through this again.
We were given this graph.
From that graph, I find out what the
solutions are. I can find that just by
looking at the graph.
I take those solutions
and 
I set x equal to each one of the
solutions.
Then I take my equations,
and set the equations equal to zero.
And then I'm going to write out
y equals...
in this case, something like 
(x - 2) times (x - 5).
but I've got to find out
what's multiplying the right side of the
equation
to give me the specific graph.
Okay? So
I'm going to call the thing that's multiplying it 'a' or 'k', or any letter I want, just about.
Then I'm going to go back
and find a third point. In this case
I've got a point which is actually right
on the graph,
right on the y-axis.
It could have been any third point.
I take the x- and y-values from that third point,
and plug them in
where my x- and y-value were
in the solution I've gotten to so far.
And then,
I solve that
for 'a'.
Once I know what 'a' is,
I take the 'a', and I plug it back in
where I had the 'a'.
So I've got y equals whatever the 'a' was
times
the two binomials, (x - 2) and (x - 5).
And then if I want to, I could of course 
multiply (x - 2) times (x - 5)
and get something with an x-squared. And if I wanted to go to further I could multiply
everything by negative 1/5.
That's just a question of finding out
what form
you're supposed to put the final answer into.
Okay. So that's about it.
Take care. I'll see you next time.
