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Chapter 1 Functions

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**Chapter Outline The Slope of a Straight Line**

The Slope of a Curve at a Point The Derivative Limits and the Derivative Differentiability and Continuity Some Rules for Differentiation More About Derivatives The Derivative as a Rate of Change

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§ 1.1 The Slope of a Straight Line

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**Section Outline Nonvertical Lines**

Positive and Negative Slopes of Lines Interpretation of a Graph Properties of the Slope of a Nonvertical Line Finding the Slope and y-Intercept of a Line Sketching the Graph of a Line Making Equations of Lines Slope as a Rate of Change

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**Nonvertical Lines Definition Example**

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. For this line, m = 3 and b = -4.

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**Lines – Positive Slope EXAMPLE**

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

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**Lines – Negative Slope EXAMPLE**

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

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**Interpretation of a Graph**

EXAMPLE A salesperson’s weekly pay depends on the volume of sales. If she sells x units of goods, then her pay is y = 5x + 60 dollars. Give an interpretation of the slope and the y-intercept of this straight line. SOLUTION First, let’s graph the line to help us understand the exercise. (80, 460)

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**Interpretation of a Graph**

CONTINUED The slope is 5, or 5/1. Since the numerator of this fraction represents the amount of change in her pay relative to the amount of change in her sales, the denominator, for every 1 sale that she makes, her pay increases by 5 dollars. The y-intercept is 60 and occurs on the graph at the point (0, 60). This point suggests that when she has executed 0 sales, her pay is 60 dollars. This $60 could be referred to as her base pay.

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**Properties of the Slope of a Nonvertical Line**

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**Properties of the Slope of a Line**

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**Finding Slope and y-intercept of a Line**

EXAMPLE Find the slope and y-intercept of the line SOLUTION First, we write the equation in slope-intercept form. This is the given equation. Divide both terms of the numerator of the right side by 3. Rewrite 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. Incidentally, it was a complete coincidence that the slope and y-intercept were the same number. This does not normally occur.

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**Sketching Graphs of Lines**

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. (-1, 1) (-1, 1)

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**Sketching Graphs of Lines**

CONTINUED Now we connect the two points that have already been determined, since two points determine a straight line.

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**Making Equations of Lines**

EXAMPLE Find an equation of the line that passes through the points (-1/2, 0) and (1, 2). SOLUTION To find an equation of the line that passes through those two points, we need a point (we already have two) and a slope. We do not yet have the slope so we must find it. 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. To use this property we need the slope and a point. We can use either of the two points that were initially provided. We’ll use the second (the first would work just as well).

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**Making Equations of Lines**

CONTINUED This is the equation from Property 3. (x1, y1) = (1, 2) and m = 4/3. Distribute. Add 2 to both sides of the equation. NOTE: Technically, we could have stopped when the equation looked like since it is an equation that represents the line and is equivalent to our final equation.

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**Making Equations of Lines**

EXAMPLE Find an equation of the line that passes through the point (2, 0) and is perpendicular to the line y = 2x. SOLUTION To find an equation of the line, we need a point (we already have one) and a slope. We do not yet have the slope so we must find it. We know that the line we desire is perpendicular to the line y = 2x. Using Slope Property 5, we know that the product of the slope of the line desired and the slope of the line y = 2x is -1. We recognize that the line y = 2x is in slope-intercept form and therefore the slope of the line is 2. We can now find the slope of the line that we desire. Let the slope of the new line be m. (slope of a line)(slope of a new line) = -1 This is Property 5. 2m = -1 The slope of one line is 2 and the slope of the desired line is denoted by m. m = -0.5 Divide. Now we can find the equation of the desired line using Property 3.

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**Making Equations of Lines**

CONTINUED This is the equation from Property 3. (x1, y1) = (2, 0) and m = -0.5. Distribute.

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**Slope as a Rate of Change**

EXAMPLE Compute the rate of change of the function over the given intervals. SOLUTION We first get y by itself in the given equation. This is the given equation. Subtract 2x from both sides. Since this is clearly a linear function (since it’s now in slope-intercept form) it has constant slope, namely -2. Therefore, 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.

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Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 § 1.8 The Derivative as a Rate of Change.

Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 11e – Slide 1 of 115 § 1.8 The Derivative as a Rate of Change.

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