A Slide Show Demonstrating the Tangent Line Problem

The AcroTEX Web Site, 2000The AcroTEX Web Site, 2000

A Slide ShowA Slide ShowDemonstrating theDemonstrating the

Tangent Line ProblemTangent Line Problem

D. P. StoryD. P. StoryThe Department of MathematicsThe Department of Mathematics

and Computer Scienceand Computer ScienceThe University of Akron, Akron, OHThe University of Akron, Akron, OH

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate . . .

|

a

P ( a, f(a) )

� �

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x

msec

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

|

a

P ( a, f(a) )

� �

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x

msec

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

Q( x, f(x) )

|

x|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3msec

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

Q( x, f(x) )

|

x|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3msec −2.5

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

Q( x, f(x) )

|

x|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

• Repeat.

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3msec −2.5

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5msec −2.5

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5msec −2.5 −2

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

• Repeat.

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5msec −2.5 −2

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25msec −2.5 −2

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25msec −2.5 −2 −1.75

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

• Repeat.

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25msec −2.5 −2 −1.75

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2msec −2.5 −2 −1.75

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2msec −2.5 −2 −1.75 −1.5

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

• Repeat.

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2msec −2.5 −2 −1.75 −1.5

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75msec −2.5 −2 −1.75 −1.5

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75msec −2.5 −2 −1.75 −1.5 −1.25

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

|

x

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

• Repeat.

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75msec −2.5 −2 −1.75 −1.5 −1.25

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

b

|

x

b Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75 1.6msec −2.5 −2 −1.75 −1.5 −1.25

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

b

|

x

b Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75 1.6msec −2.5 −2 −1.75 −1.5 −1.25 −1.1

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

b

|

x

b Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

• Repeat.

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75 1.6msec −2.5 −2 −1.75 −1.5 −1.25 −1.1

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

b

b

|

x

b

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75 1.6 1.55msec −2.5 −2 −1.75 −1.5 −1.25 −1.1

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

b

b

|

x

b

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75 1.6 1.55msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

b

b

|

x

b

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

• Repeat.

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75 1.6 1.55msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

b

bb

|

x

b

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75 1.6 1.55 1.501msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

b

bb

|

x

b

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75 1.6 1.55 1.501msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05 −1.001

Tangent Line ProblemProblem: Given a point P ( a, f(a) ), we want to define andcalculate the slope of the line tangent the graph at P .

b

b

b

b

b

bb

|

x

b

Q( x, f(x) )

|

a

P ( a, f(a) )

� �

• Choose a point x near a andplot Q(x, f(x) ).

• Draw the secant line through Pand Q. The slope of this secantline is

msec =f(x)− f(a)

x − a

• Repeat.• Continue . . .

Example: f(x) = 5− (x − 1)2 and a = 1.5.

x 3 2.5 2.25 2 1.75 1.6 1.55 1.501msec −2.5 −2 −1.75 −1.5 −1.25 −1.1 −1.05 −1.001

DiscussionExample: f(x) = 5 − (x − 1)2 and a = 1.5. As we choose values ofx getting closer and closer to a = 1.5, the corresponding secant linesrotate around the point P and become more and more “tangent-like”.Therefore, it is not too surprising that the slopes of these secant linesare approaching a value we would want to call “the slope of the linetangent to the graph at P”.

There are more calculations for those who want to see more.

x < 1.5x msec

1 −0.51.4 −0.91.45 −0.951.49 −0.991.499 −0.9991.4999 −0.99991.49999 −0.99999

x > 1.5x msec

2 −1.51.6 −1.11.55 −1.051.51 −1.011.501 −1.0011.5001 −1.00011.50001 −1.00001

The values of msecappear to be gettingclose and closer to−1. In this case, wewrite:

limx→1.5

msec = −1