00:01
So let's not wait any longer
and let's go straight
into an example.
00:04
Here we have a question where
you are given a curve C.
00:08
And curve c has the equation
2x squared - 6x + 4.
00:12
So it's going to be a quadratic,
a curve that we've been
dealing with.
00:17
And it's asking us to find
the gradient of the tangent
to see at point x equals to 3.
00:22
Now before we start, let's
just see if we can model this.
00:26
So we can kind of figure out
what's actually happening.
00:29
So we're not quite answering
the question yet.
00:32
We're just going to see if we
can think about,
what this question is actually
asking.
00:36
So I have here,
y = 2x2 - 6x + 4.
00:42
Just like we've mentioned
already is a quadratic.
00:46
We can solve this.
00:48
So if we say y equals to 0.
00:50
We can try to factorize it.
00:51
And see if we can
find the two factors.
00:54
So those points would be
the point of intersection
on the x axis.
00:57
If I try that here x and x.
01:00
And if we try our
two factors, 2 and 2.
01:03
And if you double check this
that gives me 2x squared
minus 4, minus 2 to give
me minus 6x.
01:12
And then I've got minus 2 times
minus 2 to give me plus 4.
01:15
So I can factorize it.
01:17
It's fairly nice equation.
01:19
From here you can say
you got two solutions.
01:21
So you can say that
either 2x - 2 = 0.
01:24
Or x - 2 = 0.
01:27
When you solve this, you get x
equals to 1 you get x = 2 here.
01:33
So remember I haven't quite
started answering
the question yet.
01:36
I'm just trying to imagine
what this actually looks like.
01:39
And for mathematicians it's
really important for us
to be able to visualize to
have graphical and algebraic
interpretations of the questions
we're looking at.
01:48
So here is what I think
this looks like.
01:54
So I've got x = 1 and x = 2.
01:57
So those are my two
solutions from there.
02:00
I know it's a quadratic
so it has a shape like so.
02:05
And now coming back
to the question.
02:08
It says find the gradient
of the tangent to see
at point x to 3.
02:12
So let's just say 3 is
somewhere around here.
02:14
So we are looking for
the gradient at that point.
02:18
And as a result, we are also
looking for the tangent here.
02:25
So this here is our
tangent line.
02:30
So we're actually looking
for the equation of this line
which you know takes the form,
y = mx + c.
02:36
Or the alternate form that I
mentioned earlier.
02:39
And we could see if we could
work with both.
02:41
So now that we understand
the problem,
let's try and answer
the question.
02:45
The question says find
the equation of the tangent.
02:49
So we're looking for
the equation of this blue line
here on this side.
02:53
Right, we've already mentioned
that the equation of a tangent
takes this form y = mx + c.
02:59
Or we're using y - y1 = m(x - x1).
03:07
Both methods again are valid.
03:09
The first thing that we need to
do is to actually
calculate the gradient.
03:14
Whether you're using this
equation or whether you're using
this equation we need to
calculate the gradient
of the tangent.
03:20
And here's what we start to use
all the knowledge that we have
just discussed.
03:23
The gradient of the tangent,
this tangent, I can calculate
by finding the gradient
of the curve at this point
x = 3.
03:32
And it will be exactly the same
gradient of the tangent
as it is of the curve at x = 3.
03:38
And we know how to do that.
03:40
So to calculate the gradient
of a curve,
I want to take my curve.
03:44
I'm going to dy by dx it.
03:47
So that's how I calculate
my gradient.
03:49
That gives me 4x - 6.
03:52
This is the general gradient.
03:55
So everywhere around this curve
I can use that gradient.
03:58
However, if I need the gradient
at the point that the question
asks, so we want to
know dy by dx at x = 3.
04:07
You simply substitute
3 into here.
04:12
So we will do 4 times 3 minus 6
which gives you 12 minus 6
to give you a positive
gradient of 6.
04:20
Which is good news because from
our picture it looks like this
graph has a positive gradient
anyway.
04:26
So it's a nice little
way of checking.
04:28
And that you can
see it graphically.
04:31
You expect a positive gradient.
04:32
And so you've got a positive
gradient.
04:34
Now we can put
into our equation.
04:38
So you could put it into
y = mx + c or we can substitute
this into y - y1 = m(x - x1).
04:49
Both of them are fine.
04:52
But you might find this a little
bit easier to use.
04:55
Because you don't have
to do it into two parts.
04:58
So you don't have to find your
m first and then find your c
at the end.
05:03
Okay, let's just talk about
what we have and what we need.
05:08
So we found the gradient.
05:09
We know what m is.
05:10
We know an x point here.
05:12
So we know that,
that point here is 3.
05:14
However, we don't know this yet.
05:16
So we're missing a y value.
05:19
Now how could you find
a y value.
05:22
You go back to your original
equation here.
05:24
Which says that y
equals to something.
05:27
So all we have to do here
is replace our x with 3.
05:31
So using your original function
we got y = 2x squared
minus 6x plus 4.
05:37
We want to know what
this is at x = 3.
05:40
So we're going to substitute
that in,
3 squared minus 6 times 3
plus 4.
05:47
That gives you two times 9 to
give you 18 minus 6 times 3
which also gives you 18 + 4
to give you an answer of 4.
05:57
Therefore, your y value,
y equals to 4, when x equals to 3.
06:06
Now we have everything.
06:08
So we've also calculated
at this y value.
06:11
You can put it all
into the equation.
06:13
So we have the y minus 4 equals
to the gradient which was
6(x minus 3).
06:20
The equation has been found.
06:23
But we can tidy up and make
it look more like y = mx + c.
06:28
So we make it look more
into that from there.
06:30
We've got y minus 4
equals to 6x minus 18.
06:35
Take the 4 over to the other
side, you get y equals to 6x
minus 18 + 4 to
give you minus 14.
06:43
And this equation now follows
the form, y = mx + c.
06:50
So the plus c is going
to be minus 14.
06:52
So it meets this y
axis here at -14.
06:57
So it gives you everything
you need to know
about that equation.