this is the graph of

Y is equal to F of X.

Where F of X is equal to

X to the third

minus six X squared

plus X minus five.

What I want to do in this video

is think about what is the equation

of the tangent line

when X is equal to one?

So we can visualize that.

So, this is X equaling one right over here.

This is the value of the function.

When X is equal to one.

Right over there.

And then the tangent line

looks something like

will look something like.

I know I can do a better job than that.

It's going to look something like that.

And what we want to do is find the equation

the equation of that line.

And if you are inspired I encourage you to be,

pause the video and try to work it out.

Well the way that we can do this is

if we find the derivative at X equals one

the derivative is the slope of the tangent line.

And so we'll know the slope of the tangent line.

And we know that it contains that point

and then we can use that to find the equation

of the tangent line.

So let's actually just, let's just.

So we want the equation of the tangent line

when X is equal to one.

So let's just first of all evaluate F of one.

So F of one

is equal to one to the third power

which is one.

Minus six times ones squared,

so it's just minus six.

And then

plus one

plus one

minus five.

So, this is equal to what?

Two minus 11?

Which is equal to

negative nine.

And that looks about right.

That looks like about negative nine right over there.

The scales are different on the Y and the X axis.

So that is F of one.

It is negative nine.

Did I do that right?

This is negative five.

Yep, negative nine.

And now let's evaluate what the derivative is

at one.

So,

what is F prime of X?

F prime of X.

Well here it's just a polynomial.

You take the derivative of X to the third

while we apply the power rule.

We bring the three out front.

So you get three X

to the.

And then we go one less than three

to get the second power.

Then you have

minus six X squared.

So you bring the two times the six

to get 12.

So minus 12 X

to the

well two minus one is one power

so that's the same thing as 12 X.

And then plus the derivative of X

is just one.

That's just going to be one.

And if you view this as X to the first power

we're just bringing the one out front

and decrementing the one.

So it's one times X to the zero power

which is just one.

And then the derivative of a constant here

is just going to be zero.

So this is our derivative of F

and if we want to evaluate it at one

F prime of one

is going to be three times one squared

which is just three

minus 12 times one

which is just minus 12.

And then we have plus one.

So this is

three minus 12 is negative nine

plus one is equal to negative eight.

So we know the slope right over here

is the slope of negative eight.

We know a point on that line

it contains the point one negative nine

so we can use that information to find the

equation of the line.

The line, just to remind ourselves,

has the four.

Y is equal to M X plus B.

Where M is the slope.

So we know that Y is going to be

equal to negative eight X

plus B.

And now we can substitute

the X and Y value that we know

sits on that line to solve for B.

So we know that Y is equal to negative nine.

Let me just write this here.

Y is equal to negative nine

when X is equal to

when X is equal to one.

And so we get

we get

negative nine

is equal to negative eight times one.

So negative eight

plus B.

Well, let's see.

We could add

we could eight to both sides

and we get negative one

is equal to B.

So we're done.

The equation of the line

the equation of this line

that we have in magenta right over there

that is

that is

Y is equal to the slope

is negative eight X.

And then the Y-intercept

minus one.