1.1 : (the slope of a straight line) - m5zn.com · y mx b. y 3x 4 1.1 : (the slope of a straight...

Dr. Suha AlShaikh M129(Chapter 10)

1

Equations of Nonvertical Lines:

A nonvertical line L has an equation of the form

The number m is called the slope of L

and the point (0, b) is called the y-intercept.

The equation above is called the slope-intercept equation of L.

Example:

For this line, the slope m = 3 and the y-intercept b = -4.

Example

The following are graphs of equations of lines that have positive slopes.

Example

The following are graphs of equations of lines that have negative slopes.

.bmxy

43 xy

1.1 : (The Slope of a straight Line)


2

Properties of the Slope of a Nonvertical Line

We write: L1 // L2

In another word if the slope of the first line is (m) , then the slope of the second line is (-

).


3

We write: L1 L2

Example: the lines

Y = 2x +1 , Y = 2x , Y = 2x – 1 , are parallel lines.

Example: the line y = 3x+1 is perpendicular to the line y = -

x + 1

intercept of the line -yFind the slope and Example

Solution:

Since the number being multiplied by x is 1/3, 1/3 is the slope of the line. Since the other

1/3 is the number being added to the term containing x, 1/3, or (0, 1/3), is the y-intercept.

Example:

Sketch the graph of the line passing through (-1, 1) with slope ½.

Solution:

We use Slope Property 1. We begin at the given point (-1, 1) and from there, move up one

unit and to the right two units to find another point on the line.

Now we connect the two points that have already been determined, since two points

determine a straight line.

.3

1

xy


4

Example:

Find an equation of the line that passes through the points (-1/2, 0) and (1, 2).

Solution:

Using the two points we will determine the slope by using Slope Property 2.

We now use Slope Property 3 to find an equation of the line

Example:

Find an equation of the line that passes through the point (2, 0) and is perpendicular to the

line y = 2x.

Solution:

We can now find the slope of the line that we desire. Let the slope of the new line be m.

Now we can find the equation of the desired line using Property 3. As we have a point (2,0)

and the slope m = - 0.5.

3

4

3

22

23

2

211

02

m


5

Slope as a Rate of Change

=

=

= m

Example:

Compute the rate of change of the function over the given intervals.

Solution :

Since this is clearly a linear function by definition, it also has a constant rate of change, -2.

Therefore, no matter what interval is considered for this function, the rate of change will be

-2. Therefore the answer, for both intervals, is -2.

Tangent Lines

Slope of a Curve & Tangent Lines

5.0 ,0, 1 ,0272 xy

1.2 : (Slope of a curve at a point)


6

Example:

Find the slope of the tangent line to the graph of y = x2 at the point (-0.4, 0.16) and then

write the corresponding equation of the tangent line.

Solution:


7

Derivative: The slope formula for a function y = f (x), denoted:

Differentiation: The process of computing a derivative.

Differentiation Examples

Example:

Find the derivative of

Solution:

What we’ve done so far has been done for the sole purpose of rewriting the function in the

form of f (x) = xr.

Example:

Find the slope of the curve y = x5 at x = -2.

Solution:

We must first find the derivative of the given function

→

1.3 : (The Derivative)

. / xfy

.1

7 xxf

787771171

7

1

7

1

7

1 xxxxf


8

Example:

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

Solution:

Leibniz Notation for Derivatives

Equation of the Tangent Line to the Graph of y = f (x) at the point (a, f (a))


9

Definition of the Derivative:

And, (a) = →

(VII) Limit of a polynomial Function : Let P(x) be a polynomial function, a any number.

Then

→

(VIII) Limit of a Rational Function: Let r(x)=

be a rational function, where p(x) and q(x)

are polynomials. Let a be a number such that q(a)≠0 . Then

→

Examples:

(1) →

( 5 x3 – 15) = 5(2)3 – 15= 25

(2) →

=

→

=

→ (x+3) = (3+3)= 6

(3) →

,

Since the denominator approaches zero when taking the limit, we need to simplify

the expression.

=

∙

=

=

=

)DerivativesLimit and the )1.4 :


11

Thus,

→

= →

=

=

(4)

Both10x+ 100 and x2 – 30 increase without bound as x does. To determine the limit of their

quotient, we employ an algebraic trick. Divide both numerator and denominator by x2 (since

the highest power of x in either the numerator or the denominator is 2) to obtain

As x increases without bound, 10/x approaches 0, 100/x2 approaches 0, and 30/x2

approaches 0. Therefore, as x increases without bound, 10/x + 100/x2 approaches 0 + 0 = 0

and 1 - 30/x2 approaches 1 – 0 = 1. Therefore,

30

10010lim

2

x

xx

.30

1

10010

lim30

10010lim

2

2

2

x

xxx

xxx

.01

0

01

00

301

10010

lim30

10010lim

2

2

2

x

xxx

xxx


11

:Example

Use limits to compute the derivative f/(3) for the function

Solution:

.52

1

xxf


12

Example:

:Solution

Thus, f/ (-1) = -2(-1)= 2


13

:Example

:Solution

Example


14


15

Example:

Solution:

1.6 : ( Some Rules for Differentiation)


16

Example:

Solution:

Example:

(1)

(2)

(3)


17

(4)

The Derivative as a Rate of Change

Example:

Let S(x) represent the total sales (in thousands of dollars) for month x in the year 2005 at a

certain department store. Represent each statement below by an equation involving S or

(a) The sales at the end of January reached $120,560 and were rising at the rate of $1500

per month.

(b) At the end of March, the sales for this month dropped to $80,000 and were falling by

about $200 a day (Use 1 month = 30 days).

Solution:

(a) Since the sales at the end of January (the first month, so x = 1) reached $120,560 and

S(x) represents the amount of sales for a given month, we have: S(1) = 120,560.

Further, since the rate of change of sales (rate of change means we will use the

derivative of S(x)) for the month of January is a positive $1500 per month, we have:

(b) At the end of March (the third month, so x = 3), the sales dropped to $80,000.

Therefore, sales for the month of March was $80,000. That is: S(3) = 80,000.

Additionally, since sales were dropping by $200 per day during March, this means that

the rate of change of the function S(x) was (30 days) x (-200 dollars) = -6000 dollars per

month. Therefore, we have:

S

.15001 S

.60003 S


18

The second derivative , is defined to be the derivative of the derivative.

We write:

f// (x) =

or we write ,

f(x) to denote the second derivative of f(x)

Example:

Solution:

Example:

Solution:

1.7: More About Derivatives (Second Derivatives)

1.1 : (the slope of a straight line) - m5zn.com · y mx b. y 3x 4 1.1 : (the slope of a straight...

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