Download Finding the Equation of the Tangent Line to a Curve:

Remember: Derivative=Slope of the Tangent Line


What is the equation for the slope of the line tangent
to the curve at point A using points A and B?
y  f ( a)
slope of line 
x a
What is the equation for the slope of the line tangent
to the curve f(x)=x2+1 at point A using points A and
B?
y  f (a) y  2
slope of line 

x a
x 1
What is another way to find the slope of this line?
The DERIVATIVE!!!!
f ' (a )
What is another way to find the slope of this line?
f '( x )  2x
f ' (a)  f ' (1)  2(1)  2
Both ways give you the slope of the tangent to
the curve at point A.
set them equal to each other
That means you can _____________________________.
That means you can set them equal to each other:
y  2 y  f (a )
f ' (a)  2 

x 1
x a
That means you can set them equal to each other:
y 2
2
x 1
y  2  2( x  1)
Therefore,
y  2  2( x  1)
Is the slope of the tangent line
for f(x)=x2+1
y-f(a)=f’(a)(x-a)
Step 1: Find the point of contact
by plugging in the x-value in f(x).
This is f(a).
f (a)  f (3)  3(3)  4(3)  39
2
Step 2: Find f’(x). Plug in x-value
for f’(a)
f ' ( x )  6x  4
f ' (a)  f ' ( 3)  6( 3)  4  22
Step 3: Plug all known values into
formula
y-f(a)=f’(a)(x-a)
y  39  22( x  ( 3))
y  39  22( x  3)
 Find the equation of the tangent to y=x3+2x
at:

x=2

x=-1

x=-2
f’(x)=0
Step 1: Find the derivative, f’(x)
Step 2: Set derivative equal to zero and
solve, f’(x)=0
Step 3: Plug solutions into original
formula to find y-value, (solution, yvalue) is the coordinates.
Note: If it asks for the equation then you
will write y=y value found when you
plugged in the solutions for f’(x)=0
What do you notice about
the labeled minimum and
maximum?
They are the coordinates
where the tangent is
horizontal
Where is the graph increasing?
{x| x<-3, x>1}
What is the ‘sign’ of the
derivative for these intervals?
+
+
-3
1
This is called a
sign diagram
Where is the graph decreasing?
{x| -3<x<1}
What is the ‘sign’ of the
derivative for this interval?
+
+
–
-3
1
What can we hypothesize
about how the sign of the
derivative relates to the
graph?
f’(x)=+, then graph increases
f’(x)= – , then graph decreases
We can see this:
When the graph is increasing
then the gradient of the
tangent line is positive
(derivative is +)
When the graph is decreasing
then the gradient of the
tangent line is negative
(derivative is - )
So back to the
question…Why does the fact
that the relative max/min of
a graph have horizontal
tangents make sense?
A relative max or min is
where the graph goes from
increasing to decreasing
(max) or from decreasing to
increasing (min). This
means that your derivative
needs to change signs.
Okay…So what?
To go from being positive
to negative, the derivative
like any function must go
through zero. Where the
derivative is zero is where
the graph changes
direction, aka the relative
max/min
Take a look at f(x)=x3. What is the
coordinates of the point on the function
where the derivative is equal to 0? Find the
graph in your calculator, is this coordinate a
relative maximum or a relative minimum?
NO – the graph only flattened out then
continued in the same direction
This is called a HORIZONTAL INFLECTION
It is necessary to make a sign diagram to
determine whether the coordinate where
f’(x)=0 is a relative maximum, minimum, or
a horizontal inflection.
Anywhere that f’(x)=0 is called a stationary point; a
stationary point could be a relative minimum, a
relative maximum, or a horizontal inflection
 What do you know about the graph of f(x) when
f’(x) is
a) Positive
b) Negative
c) Zero
 What do you know about the slope of the tangent
line at a relative extrema? Why is this so?
 Sketch a graph of f(x) when the sign diagram of f’(x)
looks like
–
+
-5
–
1
 What are the types of stationary points? What do
they all have in common? What do the sign
?
diagrams for each type look like?
?
Stationary Point