§ 3.1 Tangents and the Derivative at a Point

Formal Definition

DefinitionThe derivative of a function at a point x0, denoted f ′(x0), is

f ′(x0) = limh→0

f (x0 + h)− f (x0)

h

provided the limit exists.

This is nothing new - we are just formally naming the derivative nowinstead of talking about average rates of change over infinitely smallintervals.

Formal Definition

DefinitionThe derivative of a function at a point x0, denoted f ′(x0), is

f ′(x0) = limh→0

f (x0 + h)− f (x0)

h

provided the limit exists.

This is nothing new - we are just formally naming the derivative nowinstead of talking about average rates of change over infinitely smallintervals.

What This Definition Means

As we have stated throughout the semester, this difference quotientrepresents different but equivalent statements:

1 the slope of the graph of f (x) at the point x = x0.2 the slope of the tangent line to f (x) at the point x = x0.3 The instantaneous rate of change of the function f (x) at the point

x = x0.4 f ′(x0)

What This Definition Means

As we have stated throughout the semester, this difference quotientrepresents different but equivalent statements:

1 the slope of the graph of f (x) at the point x = x0.

2 the slope of the tangent line to f (x) at the point x = x0.3 The instantaneous rate of change of the function f (x) at the point

x = x0.4 f ′(x0)

What This Definition Means

As we have stated throughout the semester, this difference quotientrepresents different but equivalent statements:

1 the slope of the graph of f (x) at the point x = x0.2 the slope of the tangent line to f (x) at the point x = x0.

3 The instantaneous rate of change of the function f (x) at the pointx = x0.

4 f ′(x0)

What This Definition Means

As we have stated throughout the semester, this difference quotientrepresents different but equivalent statements:

1 the slope of the graph of f (x) at the point x = x0.2 the slope of the tangent line to f (x) at the point x = x0.3 The instantaneous rate of change of the function f (x) at the point

x = x0.

4 f ′(x0)

What This Definition Means

As we have stated throughout the semester, this difference quotientrepresents different but equivalent statements:

1 the slope of the graph of f (x) at the point x = x0.2 the slope of the tangent line to f (x) at the point x = x0.3 The instantaneous rate of change of the function f (x) at the point

x = x0.4 f ′(x0)

Finding Tangent Lines

In order to find the equation of a linear function, what values do weneed to find?

Finding Tangent Lines

In order to find the equation of a linear function, what values do weneed to find?

Finding Tangent Lines

In order to find the equation of a linear function, what values do weneed to find?

Finding Tangent Lines

In order to find the equation of a linear function, what values do weneed to find?

Finding Tangent Lines

Example

Find the equation of the tangent line to f (x) = x2 + 4 at the pointwhere x = 3.

f ′(x) = limh→0

f (x + h)− f (x)h

= limh→0

(x + h)2 + 4− (x2 + 4)h

= limh→0

x2 + 2xh + h2 + 4− x2 − 4h

= limh→0

2xh + h2

h= lim

h→0(2x + h)

= 2x

So, at the point where x = 3, f ′(3) = 2(3) = 6.

Finding Tangent Lines

Example

Find the equation of the tangent line to f (x) = x2 + 4 at the pointwhere x = 3.

f ′(x) = limh→0

f (x + h)− f (x)h

= limh→0

(x + h)2 + 4− (x2 + 4)h

= limh→0

x2 + 2xh + h2 + 4− x2 − 4h

= limh→0

2xh + h2

h= lim

h→0(2x + h)

= 2x

So, at the point where x = 3, f ′(3) = 2(3) = 6.

Finding Tangent Lines

Example

Find the equation of the tangent line to f (x) = x2 + 4 at the pointwhere x = 3.

f ′(x) = limh→0

f (x + h)− f (x)h

= limh→0

(x + h)2 + 4− (x2 + 4)h

= limh→0

x2 + 2xh + h2 + 4− x2 − 4h

= limh→0

2xh + h2

h= lim

h→0(2x + h)

= 2x

So, at the point where x = 3, f ′(3) = 2(3) = 6.

Finding Tangent Lines

Example

Find the equation of the tangent line to f (x) = x2 + 4 at the pointwhere x = 3.

f ′(x) = limh→0

f (x + h)− f (x)h

= limh→0

(x + h)2 + 4− (x2 + 4)h

= limh→0

x2 + 2xh + h2 + 4− x2 − 4h

= limh→0

2xh + h2

h= lim

h→0(2x + h)

= 2x

So, at the point where x = 3, f ′(3) = 2(3) = 6.

Finding Tangent Lines

Example

Find the equation of the tangent line to f (x) = x2 + 4 at the pointwhere x = 3.

f ′(x) = limh→0

f (x + h)− f (x)h

= limh→0

(x + h)2 + 4− (x2 + 4)h

= limh→0

x2 + 2xh + h2 + 4− x2 − 4h

= limh→0

2xh + h2

h

= limh→0

(2x + h)

= 2x

So, at the point where x = 3, f ′(3) = 2(3) = 6.

Finding Tangent Lines

Example

Find the equation of the tangent line to f (x) = x2 + 4 at the pointwhere x = 3.

f ′(x) = limh→0

f (x + h)− f (x)h

= limh→0

(x + h)2 + 4− (x2 + 4)h

= limh→0

x2 + 2xh + h2 + 4− x2 − 4h

= limh→0

2xh + h2

h= lim

h→0(2x + h)

= 2x

So, at the point where x = 3, f ′(3) = 2(3) = 6.

Finding Tangent Lines

Example

Find the equation of the tangent line to f (x) = x2 + 4 at the pointwhere x = 3.

f ′(x) = limh→0

f (x + h)− f (x)h

= limh→0

(x + h)2 + 4− (x2 + 4)h

= limh→0

x2 + 2xh + h2 + 4− x2 − 4h

= limh→0

2xh + h2

h= lim

h→0(2x + h)

= 2x

So, at the point where x = 3, f ′(3) = 2(3) = 6.

Finding Tangent Lines

Example

Find the equation of the tangent line to f (x) = x2 + 4 at the pointwhere x = 3.

f ′(x) = limh→0

f (x + h)− f (x)h

= limh→0

(x + h)2 + 4− (x2 + 4)h

= limh→0

x2 + 2xh + h2 + 4− x2 − 4h

= limh→0

2xh + h2

h= lim

h→0(2x + h)

= 2x

So, at the point where x = 3, f ′(3) = 2(3) = 6.

Finding Tangent Lines

Now that we have the slope, what else do we need?

f (x) = x2 + 4

f (3) = 32 + 4 = 13

So ...

y = mx + b

13 = 6(3) + b

13 = 18 + b

b = −5

Therefore, the equation of the tangent line we need is y = 6x− 5.

Finding Tangent Lines

Now that we have the slope, what else do we need?

f (x) = x2 + 4

f (3) = 32 + 4 = 13

So ...

y = mx + b

13 = 6(3) + b

13 = 18 + b

b = −5

Therefore, the equation of the tangent line we need is y = 6x− 5.

Finding Tangent Lines

Now that we have the slope, what else do we need?

f (x) = x2 + 4

f (3) = 32 + 4 = 13

So ...

y = mx + b

13 = 6(3) + b

13 = 18 + b

b = −5

Therefore, the equation of the tangent line we need is y = 6x− 5.

Finding Tangent Lines

Now that we have the slope, what else do we need?

f (x) = x2 + 4

f (3) = 32 + 4 = 13

So ...

y = mx + b

13 = 6(3) + b

13 = 18 + b

b = −5

Therefore, the equation of the tangent line we need is y = 6x− 5.

Finding Tangent Lines

Now that we have the slope, what else do we need?

f (x) = x2 + 4

f (3) = 32 + 4 = 13

So ...

y = mx + b

13 = 6(3) + b

13 = 18 + b

b = −5

Therefore, the equation of the tangent line we need is y = 6x− 5.

Finding Tangent Lines

Now that we have the slope, what else do we need?

f (x) = x2 + 4

f (3) = 32 + 4 = 13

So ...

y = mx + b

13 = 6(3) + b

13 = 18 + b

b = −5

Therefore, the equation of the tangent line we need is y = 6x− 5.

Another Example

Example

Find the equation of the tangent line to f (x) = x+1x+2 at the point where

x = 2.

f ′(x) = limh→0

x+h+1x+h+2 −

x+1x+2

h

= limh→0

x+h+1x+h+2 ·

x+2x+2 −

x+1x+2 ·

x+h+2x+h+2

h

= limh→0

x2+3x+xh+2h+2(x+2)(x+h+2) −

x2+3x+xh+h+2(x+2)(x+h+2)

h

= limh→0

h(x+2)(x+h+2)

h

Another Example

Example

Find the equation of the tangent line to f (x) = x+1x+2 at the point where

x = 2.

f ′(x) = limh→0

x+h+1x+h+2 −

x+1x+2

h

= limh→0

x+h+1x+h+2 ·

x+2x+2 −

x+1x+2 ·

x+h+2x+h+2

h

= limh→0

x2+3x+xh+2h+2(x+2)(x+h+2) −

x2+3x+xh+h+2(x+2)(x+h+2)

h

= limh→0

h(x+2)(x+h+2)

h

Another Example

Example

Find the equation of the tangent line to f (x) = x+1x+2 at the point where

x = 2.

f ′(x) = limh→0

x+h+1x+h+2 −

x+1x+2

h

= limh→0

x+h+1x+h+2 ·

x+2x+2 −

x+1x+2 ·

x+h+2x+h+2

h

= limh→0

x2+3x+xh+2h+2(x+2)(x+h+2) −

x2+3x+xh+h+2(x+2)(x+h+2)

h

= limh→0

h(x+2)(x+h+2)

h

Another Example

Example

Find the equation of the tangent line to f (x) = x+1x+2 at the point where

x = 2.

f ′(x) = limh→0

x+h+1x+h+2 −

x+1x+2

h

= limh→0

x+h+1x+h+2 ·

x+2x+2 −

x+1x+2 ·

x+h+2x+h+2

h

= limh→0

x2+3x+xh+2h+2(x+2)(x+h+2) −

x2+3x+xh+h+2(x+2)(x+h+2)

h

= limh→0

h(x+2)(x+h+2)

h

Another Example

Example

Find the equation of the tangent line to f (x) = x+1x+2 at the point where

x = 2.

f ′(x) = limh→0

x+h+1x+h+2 −

x+1x+2

h

= limh→0

x+h+1x+h+2 ·

x+2x+2 −

x+1x+2 ·

x+h+2x+h+2

h

= limh→0

x2+3x+xh+2h+2(x+2)(x+h+2) −

x2+3x+xh+h+2(x+2)(x+h+2)

h

= limh→0

h(x+2)(x+h+2)

h

Another Example

= limh→0

h(x+2)(x+h+2)

h

= limh→0

h(x + 2)(x + h + 2)

· 1h

= limh→0

1(x + 2)(x + h + 2)

=1

(x + 2)2

So, when x = 2, we have f ′(2) = 116 .

Another Example

= limh→0

h(x+2)(x+h+2)

h

= limh→0

h(x + 2)(x + h + 2)

· 1h

= limh→0

1(x + 2)(x + h + 2)

=1

(x + 2)2

So, when x = 2, we have f ′(2) = 116 .

Another Example

= limh→0

h(x+2)(x+h+2)

h

= limh→0

h(x + 2)(x + h + 2)

· 1h

= limh→0

1(x + 2)(x + h + 2)

=1

(x + 2)2

So, when x = 2, we have f ′(2) = 116 .

Another Example

= limh→0

h(x+2)(x+h+2)

h

= limh→0

h(x + 2)(x + h + 2)

· 1h

= limh→0

1(x + 2)(x + h + 2)

=1

(x + 2)2

So, when x = 2, we have f ′(2) = 116 .

Another Example

= limh→0

h(x+2)(x+h+2)

h

= limh→0

h(x + 2)(x + h + 2)

· 1h

= limh→0

1(x + 2)(x + h + 2)

=1

(x + 2)2

So, when x = 2, we have f ′(2) = 116 .

Another Example

Now, we need what?

f (2) =2 + 12 + 2

=34

Now, we need b.

y = mx + b34=

116

(2) + b

34=

18+ b

58= b

So, the tangent line we seek is y = 116 x + 5

8 .

Another Example

Now, we need what?

f (2) =2 + 12 + 2

=34

Now, we need b.

y = mx + b34=

116

(2) + b

34=

18+ b

58= b

So, the tangent line we seek is y = 116 x + 5

8 .

Another Example

Now, we need what?

f (2) =2 + 12 + 2

=34

Now, we need b.

y = mx + b34=

116

(2) + b

34=

18+ b

58= b

So, the tangent line we seek is y = 116 x + 5

8 .

Another Example

Now, we need what?

f (2) =2 + 12 + 2

=34

Now, we need b.

y = mx + b

34=

116

(2) + b

34=

18+ b

58= b

So, the tangent line we seek is y = 116 x + 5

8 .

Another Example

Now, we need what?

f (2) =2 + 12 + 2

=34

Now, we need b.

y = mx + b34=

116

(2) + b

34=

18+ b

58= b

So, the tangent line we seek is y = 116 x + 5

8 .

Another Example

Now, we need what?

f (2) =2 + 12 + 2

=34

Now, we need b.

y = mx + b34=

116

(2) + b

34=

18+ b

58= b

So, the tangent line we seek is y = 116 x + 5

8 .

Another Example

Now, we need what?

f (2) =2 + 12 + 2

=34

Now, we need b.

y = mx + b34=

116

(2) + b

34=

18+ b

58= b

So, the tangent line we seek is y = 116 x + 5

8 .

Another Example

Now, we need what?

f (2) =2 + 12 + 2

=34

Now, we need b.

y = mx + b34=

116

(2) + b

34=

18+ b

58= b

So, the tangent line we seek is y = 116 x + 5

8 .

Horizontal Tangent Lines

Example

Find when f (x) = x2 + x has a horizontal tangent.

How does this problem differ from the last two? What are we lookingfor here?

We need to know when f (x) = 0.

Horizontal Tangent Lines

Example

Find when f (x) = x2 + x has a horizontal tangent.

How does this problem differ from the last two? What are we lookingfor here?

We need to know when f (x) = 0.

Horizontal Tangent Lines

Example

Find when f (x) = x2 + x has a horizontal tangent.

How does this problem differ from the last two? What are we lookingfor here?

We need to know when f (x) = 0.

Horizontal Tangent Lines

f ′(x) = limh→0

(x + h)2 + (x + h)− (x2 + x)h

= limh→0

x2 + 2xh + h2 + x + h− x2 − xh

= limh→0

2xh + h2 + hh

= limh→0

2x + h + 1

= 2x + 1

So, we have that 2x + 1 = 0 when x = −12 and that is where this

function has a horizontal tangent line.

Horizontal Tangent Lines

f ′(x) = limh→0

(x + h)2 + (x + h)− (x2 + x)h

= limh→0

x2 + 2xh + h2 + x + h− x2 − xh

= limh→0

2xh + h2 + hh

= limh→0

2x + h + 1

= 2x + 1

So, we have that 2x + 1 = 0 when x = −12 and that is where this

function has a horizontal tangent line.

Horizontal Tangent Lines

f ′(x) = limh→0

(x + h)2 + (x + h)− (x2 + x)h

= limh→0

x2 + 2xh + h2 + x + h− x2 − xh

= limh→0

2xh + h2 + hh

= limh→0

2x + h + 1

= 2x + 1

So, we have that 2x + 1 = 0 when x = −12 and that is where this

function has a horizontal tangent line.

Horizontal Tangent Lines

f ′(x) = limh→0

(x + h)2 + (x + h)− (x2 + x)h

= limh→0

x2 + 2xh + h2 + x + h− x2 − xh

= limh→0

2xh + h2 + hh

= limh→0

2x + h + 1

= 2x + 1

So, we have that 2x + 1 = 0 when x = −12 and that is where this

function has a horizontal tangent line.

Horizontal Tangent Lines

f ′(x) = limh→0

(x + h)2 + (x + h)− (x2 + x)h

= limh→0

x2 + 2xh + h2 + x + h− x2 − xh

= limh→0

2xh + h2 + hh

= limh→0

2x + h + 1

= 2x + 1

So, we have that 2x + 1 = 0 when x = −12 and that is where this

function has a horizontal tangent line.

Horizontal Tangent Lines

f ′(x) = limh→0

(x + h)2 + (x + h)− (x2 + x)h

= limh→0

x2 + 2xh + h2 + x + h− x2 − xh

= limh→0

2xh + h2 + hh

= limh→0

2x + h + 1

= 2x + 1

So, we have that 2x + 1 = 0 when

x = −12 and that is where this

function has a horizontal tangent line.

Horizontal Tangent Lines

f ′(x) = limh→0

(x + h)2 + (x + h)− (x2 + x)h

= limh→0

x2 + 2xh + h2 + x + h− x2 − xh

= limh→0

2xh + h2 + hh

= limh→0

2x + h + 1

= 2x + 1

So, we have that 2x + 1 = 0 when x = −12 and that is where this

function has a horizontal tangent line.

Tangent Lines Given Slopes

ExampleFind the equation of all lines with a slope of −4 tangent tof (x) = 16

x+1 .

Where do we begin?

We need to find the x values where the function has a slope of −4.

Tangent Lines Given Slopes

ExampleFind the equation of all lines with a slope of −4 tangent tof (x) = 16

x+1 .

Where do we begin?

We need to find the x values where the function has a slope of −4.

Tangent Lines Given Slopes

ExampleFind the equation of all lines with a slope of −4 tangent tof (x) = 16

x+1 .

Where do we begin?

We need to find the x values where the function has a slope of −4.

Tangent Lines Given Slopes

f ′(x) = limh→0

16x+h+1 −

16x+1

h

= limh→0

16x+h+1 ·

x+1x+1 −

16x+1 ·

x+h+1x+h+1

h

= limh→0

16(x+1)(x+h+1)(x+1) −

16(x+h+1)(x+h+1)(x+1)

h

= limh→0

−16h(x+h+1)(x+1)

h

= limh→0

−16(x + h + 1)(x + 1)

=−16

(x + 1)2

Tangent Lines Given Slopes

f ′(x) = limh→0

16x+h+1 −

16x+1

h

= limh→0

16x+h+1 ·

x+1x+1 −

16x+1 ·

x+h+1x+h+1

h

= limh→0

16(x+1)(x+h+1)(x+1) −

16(x+h+1)(x+h+1)(x+1)

h

= limh→0

−16h(x+h+1)(x+1)

h

= limh→0

−16(x + h + 1)(x + 1)

=−16

(x + 1)2

Tangent Lines Given Slopes

f ′(x) = limh→0

16x+h+1 −

16x+1

h

= limh→0

16x+h+1 ·

x+1x+1 −

16x+1 ·

x+h+1x+h+1

h

= limh→0

16(x+1)(x+h+1)(x+1) −

16(x+h+1)(x+h+1)(x+1)

h

= limh→0

−16h(x+h+1)(x+1)

h

= limh→0

−16(x + h + 1)(x + 1)

=−16

(x + 1)2

Tangent Lines Given Slopes

f ′(x) = limh→0

16x+h+1 −

16x+1

h

= limh→0

16x+h+1 ·

x+1x+1 −

16x+1 ·

x+h+1x+h+1

h

= limh→0

16(x+1)(x+h+1)(x+1) −

16(x+h+1)(x+h+1)(x+1)

h

= limh→0

−16h(x+h+1)(x+1)

h

= limh→0

−16(x + h + 1)(x + 1)

=−16

(x + 1)2

Tangent Lines Given Slopes

f ′(x) = limh→0

16x+h+1 −

16x+1

h

= limh→0

16x+h+1 ·

x+1x+1 −

16x+1 ·

x+h+1x+h+1

h

= limh→0

16(x+1)(x+h+1)(x+1) −

16(x+h+1)(x+h+1)(x+1)

h

= limh→0

−16h(x+h+1)(x+1)

h

= limh→0

−16(x + h + 1)(x + 1)

=−16

(x + 1)2

Tangent Lines Given Slopes

f ′(x) = limh→0

16x+h+1 −

16x+1

h

= limh→0

16x+h+1 ·

x+1x+1 −

16x+1 ·

x+h+1x+h+1

h

= limh→0

16(x+1)(x+h+1)(x+1) −

16(x+h+1)(x+h+1)(x+1)

h

= limh→0

−16h(x+h+1)(x+1)

h

= limh→0

−16(x + h + 1)(x + 1)

=−16

(x + 1)2

Tangent Lines Given Slopes

What good did this do us?

−16(x + 1)2 = −4

1(x + 1)2 =

14

(x + 1)2 = 4

x + 1 = ±2

x = −1± 2

So, x = 1 and x = −3 are the values we need.

Are we done?

Tangent Lines Given Slopes

What good did this do us?

−16(x + 1)2 = −4

1(x + 1)2 =

14

(x + 1)2 = 4

x + 1 = ±2

x = −1± 2

So, x = 1 and x = −3 are the values we need.

Are we done?

Tangent Lines Given Slopes

What good did this do us?

−16(x + 1)2 = −4

1(x + 1)2 =

14

(x + 1)2 = 4

x + 1 = ±2

x = −1± 2

So, x = 1 and x = −3 are the values we need.

Are we done?

Tangent Lines Given Slopes

What good did this do us?

−16(x + 1)2 = −4

1(x + 1)2 =

14

(x + 1)2 = 4

x + 1 = ±2

x = −1± 2

So, x = 1 and x = −3 are the values we need.

Are we done?

Tangent Lines Given Slopes

What good did this do us?

−16(x + 1)2 = −4

1(x + 1)2 =

14

(x + 1)2 = 4

x + 1 = ±2

x = −1± 2

So, x = 1 and x = −3 are the values we need.

Are we done?

Tangent Lines Given Slopes

What good did this do us?

−16(x + 1)2 = −4

1(x + 1)2 =

14

(x + 1)2 = 4

x + 1 = ±2

x = −1± 2

So, x = 1 and x = −3 are the values we need.

Are we done?

Tangent Lines Given Slopes

What good did this do us?

−16(x + 1)2 = −4

1(x + 1)2 =

14

(x + 1)2 = 4

x + 1 = ±2

x = −1± 2

So, x = 1 and x = −3 are the values we need.

Are we done?

Tangent Lines Given Slopes

If x = 1, y =

8.

y = mx + b

8 = −4(1) + b

12 = b

This gives the equation of the one the tangent lines as y = −4x + 12.

Tangent Lines Given Slopes

If x = 1, y = 8.

y = mx + b

8 = −4(1) + b

12 = b

This gives the equation of the one the tangent lines as y = −4x + 12.

Tangent Lines Given Slopes

If x = 1, y = 8.

y = mx + b

8 = −4(1) + b

12 = b

This gives the equation of the one the tangent lines as y = −4x + 12.

Tangent Lines Given Slopes

If x = 1, y = 8.

y = mx + b

8 = −4(1) + b

12 = b

This gives the equation of the one the tangent lines as y = −4x + 12.

Tangent Lines Given Slopes

If x = 1, y = 8.

y = mx + b

8 = −4(1) + b

12 = b

This gives the equation of the one the tangent lines as y = −4x + 12.

Tangent Lines Given Slopes

If x = 1, y = 8.

y = mx + b

8 = −4(1) + b

12 = b

This gives the equation of the one the tangent lines as y = −4x + 12.

Tangent Lines Given Slopes

If x = −3, y =

−8.

y = mx + b

− 8 = −4(−3) + b

− 8 = 12 + b

b = −20

This gives the equation of the one the tangent lines as y = −4x− 20.

Tangent Lines Given Slopes

If x = −3, y = −8.

y = mx + b

− 8 = −4(−3) + b

− 8 = 12 + b

b = −20

This gives the equation of the one the tangent lines as y = −4x− 20.

Tangent Lines Given Slopes

If x = −3, y = −8.

y = mx + b

− 8 = −4(−3) + b

− 8 = 12 + b

b = −20

This gives the equation of the one the tangent lines as y = −4x− 20.

Tangent Lines Given Slopes

If x = −3, y = −8.

y = mx + b

− 8 = −4(−3) + b

− 8 = 12 + b

b = −20

This gives the equation of the one the tangent lines as y = −4x− 20.

Tangent Lines Given Slopes

If x = −3, y = −8.

y = mx + b

− 8 = −4(−3) + b

− 8 = 12 + b

b = −20

This gives the equation of the one the tangent lines as y = −4x− 20.

Tangent Lines Given Slopes

If x = −3, y = −8.

y = mx + b

− 8 = −4(−3) + b

− 8 = 12 + b

b = −20

This gives the equation of the one the tangent lines as y = −4x− 20.

Tangent Lines Given Slopes

If x = −3, y = −8.

y = mx + b

− 8 = −4(−3) + b

− 8 = 12 + b

b = −20

This gives the equation of the one the tangent lines as y = −4x− 20.

Tangent Lines Given Slopes

Do these equations make sense?

Tangent Lines Given Slopes

Do these equations make sense?

Slopes from Graphs

How could we find the derivative at a point of a curve given only agraph?

x

y

(4,5)

(-6,-6)

Slopes from Graphs

How could we find the derivative at a point of a curve given only agraph?

x

y

(4,5)

(-6,-6)

Slopes from Graphs

How could we find the derivative at a point of a curve given only agraph?

x

y

(4,5)

(-6,-6)

Slopes from Graphs

How could we find the derivative at a point of a curve given only agraph?

x

y

(4,5)

(-6,-6)

Slopes from Graphs

How could we find the derivative at a point of a curve given only agraph?

x

y

(4,5)

(-6,-6)

Slopes from Graphs

How could we find the derivative at a point of a curve given only agraph?

x

y

(4,5)

(-6,-6)

Slopes from Graphs

How could we find the derivative at a point of a curve given only agraph?

x

y

(4,5)

(-6,-6)

Slopes from Graphs

So, what is the slope?

m =5− (−6)4− (−6)

=1110

What does this mean in terms of what we have been talking about inthis section?

Slopes from Graphs

So, what is the slope?

m =5− (−6)4− (−6)

=1110

What does this mean in terms of what we have been talking about inthis section?

Slopes from Graphs

So, what is the slope?

m =5− (−6)4− (−6)

=1110

What does this mean in terms of what we have been talking about inthis section?

Slopes from Graphs

So, what is the slope?

m =5− (−6)4− (−6)

=1110

What does this mean in terms of what we have been talking about inthis section?