ACCELERATED PRE-CALCULUS SUMMER STUDY GUIDE

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SECTION 15.1: EQUATIONS THAT PRODUCE LINES
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In an earlier section you learned how to evaluate algebraic expressions. We will look further into this concept by looking at algebraic expressions that generate ordered pairs that when graphed will fall along a straight line.

Let's consider the following equation y = 2x + 5. You can choose any value for x, substitute that value into the variable x of the algebraic expression 2x + 5, and then evaluate 2x + 5, which will produce a value that we will assign to y since y is equal to 2x + 5. From the resulting x-and y-value you can create an ordered pair (x, y). Consider figure 1 below.

figure 1

Notice in figure 1 we seem to have created a set of ordered pairs that name points that fall along a line.

Lets examine a few characteristics of the above line. First, look at the above graph, can you see that the y intercept is 5. Interestingly the equation y= 2x + 5 contains the y-intercept.

Let's consider another example: y = -2x + 1. We will construct an "ordered pair table" and then plot the ordered pairs associated points.

SO WHAT HAVE WE DISCOVERED?

It appears that we have found equations that will produce ordered pairs that will create lines! There also seems to be a pattern to these equations.In the next section will will uncover the pattern behind these equations that produce lines.

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SECTION 15.2: THE SLOPE - INTERCEPT FORMULA, y = mx + b
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In the equations above, we notice how both sets of ordered pairs produced a straight line. The above equations, y = 2x + 5 and y = -2x + 1, have the pattern which we will represnt by y = mx + b, where m and b are real numbers.

You can be guaranteed that any equation that has the pattern y = mx + b, where m
and b are real numbers, will produce ordered pairs that fall along a line.

Notice that in the pattern y = mx + b, m is the number that is the coefficient of x and b is the number "all by itself." A term that is a lone number in an equation is called the "numerical part." So b can be called the numerical part of the equation y = mx + b. You will find in time that the number m is the slope of the line produced by the ordered pairs and the number b is the y-intercept created by the line going through all the points.

So here are some more examples of y = mx + b patterns.

It is your job to verify the above graphs are accurate representations of
each of the equations of lines in the form of y = mx + b.

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SECTION 15.3: WHEN m = 0 IN y = mx + b
A HORIZONTAL LINE IS CREATED
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In figure 5 above in section 15.2 you created a graph of a "horizontal line." 

Let's consider the following equation y = 0x + 2. This equation fits the pattern of y = mx + b, the equation of a line. Notice in the algebraic expression 0x + 2 that m, the coefficient of x, in this instance is 0. In general, you would not see this equation written as y = 0x + 2: this equation would generally be written as y = 2, and it would be assumed that m was 0, and 0x which means 0 would be dropped from y = 0x + 2.

ex. 1) Let me convince you that y = 4 is a horizontal line.

step 1) First lets rewrite y = 4 as y = 0x + 4 so we have a placeholder for x.

step 2) Lets create a table of ordered pairs.

step 3) Plot the ordered pairs and see what you get.

A horizontal line!

ex. 2) Let me convince you that y = -5 is a horizontal line.

step 1) First lets rewrite y = -5 as y = 0x + -5 so we have a placeholder for x.

step 2) Lets create a table of ordered pairs.

step 3) Plot the ordered pairs and see what you get.

A horizontal line!

In y = 3x - 2 the coefficient of x, in this case 3, can be shown to be the average rate of change of y with respect to x, also called the slope, the rise over the run, and symbolized by the letter m.

Also, if you rewrite y = 3x - 2 as y = 3x + -2, the numerical part, -2, the number by itself in y = 3x + -2 is the y intercept or b.

My goal is to convince you of the above two mathematical facts by graphing y = 3x - 2.

First let's examine the relationship between the equation y = 3x -2 and the graph of y = 3x -2. Consider figure 1 below.

figure 1

Notice where the y-intercept is in the above graph? It's at y = -2 on the graph above.. The y-intercept is the b value in y = 3x - 2 or y = 3x + -2. SO BEFORE YOU GRAPH AN EQUATION OF A LINE YOU ARE ABLE TO DETERMINE WHERE THE GRAPH WILL CROSS THE Y-AXIS.

THE WORK TO DETERMINE THE SLOPE

Now I'll consider two of the above ordered pairs and find the average rate of change of y with respect to x, also called the slope or rise over the run.

Point 1: (1,1)  x₁ = 1 and y₁ = 1

Point 2: (0, -2)  x₂ = 0 and y₂ = -2

m = (y₂ - y₁ ) / (x₂ - x₁ ) =  (-2 -  1 ) / (0 - 1 ) =  -3/ -1 = 3

Well, as we can see the slope came out to 3 and is the coeficient of x in y = 3x - 2. SO BEFORE YOU GRAPH AN EQUATION OF A LINE YOU ARE ABLE TO DETERMINE THE SLOPE OF THE LINE YOU ARE GRAPHING.

TO SUM UP

Before I graph  y = 3x - 2 I was able to notice that the line would cross the y axis at -2, that the slope of the line would be 3 and since the slope is positive the top of the line will tilt to the right.

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EXAMPLES
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EXAMPLE 1) Given y = -2x + 4 determine

a) By examining y = -2x + 4 what is the slope and y intercept

b) Graph y = -2x + 4 with the following x values: x = 0, x = 2 and x = -1

c) Using two of the ordered pairs in part b) and the slope formula verify the slope of y = -2x + 4 is -2.

d) Verify the y intercept, 4, with the graph

ANSWERS)

a) The slope m = -2 and the y-intercept b = 4 since y = -2x + 4 fits the pattern y = mx + b.

b)

c) I'll use ordered pairs (0, 4) and (2, 0)

Point 1: (0, 4) therefore x₁ = 0 and y₁ = 4.

Point 2: (2, 0) therefore x₂ = 2 and y₂ = 0.

m = (y₂ - y₁ ) / (x₂ - x₁ ) = (0 - 4) / (2 - 0 ) = -4 / 2 = -2  Q.E.D.

d) Notice the line in the graph below cuts the y axis at y = 4 which is the y intercept. Q.E.D.

So we have learned that the pattern y = mx + b produces lines. There is a name for this pattern and you must know this name.

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The pattern y = mx + b is called the

SLOPE INTERCEPT FORM OF AN EQUATION OF A LINE

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SECTION 15.5: STANDARD FORM OF AN EQUATION
OF A LINE, ax + by + c = 0
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I am going to assume that you have aquired basic symbol manipulation skills from your Algebra 1 and Algebra 2 courses.

Let's play with the equation y = 4x + 2.
y = 4x+2
→ y - 4x = 4x - 4x + 2 
→ y - 4x = 2
→ y - 4x - 2 = 2 - 2
→ y  -4x - 2 = 0
→ - 4x + y - 2 = 0
→ -1(-4x + y - 2) = -1(0)
→ 4x - y + 2 = 0
→ 4x + -1y + 2 = 0

IN THE ABOVE PROCESS I DID THE FOLLOWING:

1. I gathered all terms to the left hand side of the equation
2. I ordered the terms so that the x-term came first, the y-term second and the numerical part last.
3. I made the x-term positive by multiplying both sides of the equation by 1.
4. I made sure no coefficients were fractions
5. I set the equation equal to 0.
6. I made every operation addition

The resulting equation, 4x + -1y + 2 = 0, was constructed from the equation y = 4x + 2 of a line in the slope intercept form. This resulting equation 4x + -1y + 2 = 0 is called the "STANDARD FORM OF AN EQUATION OF A LINE" and symbolized by ax + by + c = 0 where a, b and c are integers.

Given 4x + -1y + 2 = 0, a = 4, b = -1 and c = 2.

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STANDARD FORM OF AN EQUATION
OF A LINE(SFEL) IS WRITTEN AS

ax + by + c = 0

where a, b and c are integers and the operations
between the terms are addition
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HERE'S ANOTHER EXAMPLE

Given y = (2/3)x - 5, place this is the standard form of an equation of a line.

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ANSWER
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y = (2/3)x - 5
→ 3(y) = 3((2/3)x - 5)     I multiplied both sides by 3 to remove the denominator 3 in 2/3
→ 3y = 3(2/3)x - 3(5)     I distributed the 3
→ 3y = 2x - 15
→ 3y - 2x = 2x -2x - 15
→ 3y - 2x = -15
→ 3y -2x + 15 = -15 + 15
→ 3y - 2x + 15 = 0
→ -2x + 3y + 15 = 0     I reordered my terms carefully carrying the appropriate signs and operations
→-1(-2x + 3y + 15) = -1(0)  
→ 2x - 3y - 15 = 0
→ 2x + -3y + -15 = 0

Hence, y = (2/3)x - 5 is an equation of a line in slope intercept form and was rewritten as 2x + -3y + -15 = 0, an equation of a line in Standard Form, and a = 2; b = -3; c = -15

IMPORTANT IDEA ON SIGNS OF THE STANDARD
FORM OF AN EQUATION OF A LINE
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IN ORDER TO CORRECTLY GET a, b AND c, FROM THE STANDARD
FORM OF THE EQUATION OF A LINE ALL
SUBTRACTION MUST BE CHANGED TO ADDITION.
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HERE'S ANOTHER EXAMPLE

Given 3x + 2y + -7 = 0, an equation of a line in Standard Form,

a) Rewrite in the Slope Intercept Form Of A Line.

b) Name the slope and y intercept

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ANSWER
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a) 3x + 2y + -7 = 0

→  3x + 2y + -7+ 7 = 0 + 7
→  3x + 2y = 7
→  3x + 2y - 3x = 7 - 3x
→  2y = 7 - 3x     or     2y = - 3x + 7
→ (1/2) 2y= (1/2)(- 3x + 7)
→ y= (- 3/2)x + 7/2

b) So, 3x + 2y + -7 = 0 was transformed into  y= (- 3/2)x + 7/2. Hence m = -3/2 and b = 7/2

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SECTION 15.6: THE POINT-SLOPE FORMULA OF
AN EQUATION OF A LINE
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If a number of points fall on a straight line then there's a way to find the equation that produces the ordered pairs that represent those points. Consider the graph below with the points represented by the ordered pairs (1, 0), (0, -3) and (-1, -6). These points fall on a straight line.

Let me introduce you to the following helpful formula.

ex. 1) Given two ordered pairs (1,0) and (0,-3), find the equation of the line going through the points named by these ordered pairs.

Step 1) Find the slope between these two ordered pairs.

and

Step 2) We know

Step 3) Place m = 3 and into y - y₁ = m( x - x₁ ). Doing so results in y - 0 = 3( x - 1).

Step 4) Solve for y.

which is an equation that will the ordered pairs (1,0) and (0,-3), and will have a slope of 3.

ex. 2) Given one ordered pair and a slope of a line ,m = -2 and (-6, 8), find the equation of the line.

Step 1) and m = -2.

Step 3) Place m = -2 and into y - y= m( x - x). Doing so results in .

Step 4) Solve for y.

will produce a line that will have a slope of -2 and will create the ordered pair (-6, 8).

The following chart and graph below show the distance from Deering High of a person moving away from Deering High over a period of time. The position is in feet and the time in seconds. Notice that on the graph I am using a d-axis instead of a y-axis and a t-axis instead of a x-axis. So, at t = 0 seconds the person is 7 feet away from Deering High school; at t=1 second the person is 10 feet away from Deering High school. I think you get the gist!

Notice that the above ordered pairs model points that follow a linear path, i.e. a path of a line. This is great because we can create a formula that will create these ordered pairs because of what we learned in the last section. So let's do it!

I'll use the ordered pairs (0, 7) and (1,10), find the equation of the line going through these ordered pairs. (See graph above)

Step 1) Find the slope between these two points.

and

Notice that I reconstructed the slope formula since I am using d instead of y and t instead of x.

Step 2) Choose one of the given ordered pairs: I'll use (0, 7) for (t₁, d₁): i.e. t₁ = 0 and d₁= 7 and we know m = 3.

Remember our graph uses different variables. Instead of y and x we are using d and t.

Step 3) Place m = 3 and (0,7) into . Doing so results in .

Notice how I reconstructed the point slope forula from to . The graph uses different variables. Instead of y and x, we are using d and t.

Step 4) Solve for d.

will produce the original table of ordered pairs! Very Cool! We can now predict a real world event with mathematics!

ex. 1) Predict the position of the person at t = 10 seconds.

feet

So, at 10 seconds the person is 37 feet from Deering High school.

Comment: when you choose a t value OUTSIDE the given t values you are "EXTRAPOLATING" information.

ex. 2) Predict the position of the person at t = 2.5 seconds.

feet

So, at 2.5 seconds the person is 14.5 feet from Deering High school.

Comment: When you choose a t value BETWEEN the given t values you are "INTERPOLATING" information.

SECTION 15.8: VERTICAL LINES

Look at the vertical line below going through the three points (2, 7), (2, 2) and (2, -1). Notice all the ordered pairs have the same x values.
Let's use (2, 7) and (2, 2) to find the slope m.

As you know division by 0 is "undefined." Any pair of points falling on a vertical line will produce undefined slopes. Whenever you have points that fall on a vertical line, we will use a special equation to define that vertical line.

The vertical line graphed above would have an equation written as x = 2.

SECTION 15.9: PARALLEL LINES

Compare the following two equations that model lines: y = 4x + 7 and y = 4 x - 6. Notice that both these equations have an m-value, i.e. slope, of 4. That m-value, as you know, represents the slope or rate of change of y with respect to x.

Let's see if this is true. We'll graph y = 4x + 7 and y = 4 x - 6 on the same graph.

y = 4x + 7
x = 0: y = 4(0) + 7 = 7; (0, 7)
x = 1: y = 4(1) + 7 = 11; (1, 11)

y = 4 x - 6
x = 0: y = 4(0) - 6 = -6; (0, -6)
x = 1: y = 4(1) - 6 = -2; (1, -2)

Here are the graphs of the ordered pairs. Notice how the lines are parallel because they have the same slopes.

Before we continue our next topic, perpendicular lines, I need to talk about the concept of "negative reciprocals."

The following pairs of fractions are negative reciprocals:

ex 1) 2/3 and -3/2 are negative reciprocals

ex 2) 3/7 and -7/3 are negative recciprocals

ex 3) 5 and -1/5 are negative reciprocals

ex. 4) -8 and 1/8 are negative reciprocals

SECTION 15.11: PERPENDICULAR LINES

Compare the two equations that model lines: and  y= 2x -2 . Notice that both these equations have m values, ie. slopes, that are negative reciprocals.

Let's see if this is true. We'll graph and y = 2x -2 on the same graph.

line 1: (red line in graph below)

x = 0: y = -1/2(0) + 1 = 1: (0, 1)
x = 2: y = -1/2(2) +1 = -1 + 1 = 0: (2, 0)

line 2: (blue line in graph below)

x = 0: y = 2(0) - 2 = -2: (0, -2)
x = 2: y = 2(2) - 2 = 4 - 2 = 2: (2, 2)

Notice in the graph below the lines intersect each other at an angle of 90⁰ which means the lines are perpendicular.