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Algebra 2 Writing Linear Functions

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Section 2: Linear Relations and Functions: Lecture 4 | 23:05 min

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Lecture Comments (6)

1 answer

Last reply by: Dr Carleen Eaton
Sat Jul 12, 2014 1:43 PM

Post by XOCHITL PITHAWALLA on July 8, 2014

Why do you at 12:15 minute in order to isolate B, you substrated 4/3 from both sides, is that right? i was taught that when you working with fraction (to pass it to the other side of the equation) you should do it by its reciprocal which would be -3/4. WOULD YOU PLEASE HELP ME to clear thing out. THANKS!!!

1 answer

Last reply by: Dr Carleen Eaton
Thu Mar 27, 2014 6:49 PM

Post by edder villegas on February 9, 2014

i am a bit confused about the difference of the y value and the y intercept, why one can not plug in the y value on b

1 answer

Last reply by: Dr Carleen Eaton
Fri Apr 27, 2012 8:50 PM

Post by Kenny Zeng on April 20, 2012

Do we approach the problem the same way if the slope is a fraction while using the point slope method.

Answer EngineGet answers to any question!Ask any question related to Algebra 2

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Writing Linear Functions

The slope-intercept form of a linear equation is y = mx + b, where m is the slope of the line and b is the y-intercept of the line.
The point-slope form of a linear equation is y – y₁ = m(x – x₁), where m is the slope and (x₁, y₁) is a point on the line.
Use information given about a line to write its equation in these forms. Sometimes it is easier to put it in one form than the other one.

Writing Linear Functions

Find the equation in slope intercept form of the line with slope [1/4] and passing through ( − 3,1).

Recall that slope - intercept form is y = mx + b, where m is the slope and b is the y - intercept.
Given m, find b using the point ( − 3,1)
y = mx + b
1 = [1/4]( − 3) + b
1 = [( − 3)/4] + b
1 + [3/4] = b
b = [7/4]
Write the equation

y = [1/4]x + [7/4]

Find the equation in slope intercept form of the line with slope [2/3] and passing through (3,5).

Recall that slope - intercept form is y = mx + b, where m is the slope and b is the y - intercept.
Given m, find b using the point (3,5)
y = mx + b
5 = [2/3](3) + b
5 = [6/3] + b
5 = 2 + b
b = 3
Write the equation

y = [2/3]x + 3

Find the equation in slope intercept form of the line with slope − 2 and passing through ( − 1,7).

Recall that slope - intercept form is y = mx + b, where m is the slope and b is the y - intercept.
Given m, find b using the point ( − 1,7)
y = mx + b
7 = − 2( − 1) + b
7 = 2 + b
5 = b
Write the equation

y = − 2x + 5

Find the equation in slope intercept form of the line passing through ( − 3, − 4) and (6, − 1).

Find the slope
m = [(y₂ − y₁)/(x₂ − x₁)] =
m = [( − 1 − ( − 4))/(6 − ( − 3))] = [( − 1 + 4)/(6 + 3)] = [3/9] = [1/3]
Now that you have m, find b in y = mx + b using one of the points
y = mx + b
− 4 = [1/3]( − 3) + b
− 4 = − 1 + b
b = − 3
Write the equation

y = [1/3]x − 3

Find the equation in slope intercept form of the line passing through ( − 5,4) and (5,2).

Find the slope
m = [(y₂ − y₁)/(x₂ − x₁)] =
m = [(2 − (4))/(5 − ( − 5))] = [( − 2)/(5 + 5)] = [( − 2)/10] = − [1/5]
Now that you have m, find b in y = mx + b using one of the points
y = mx + b
2 = − [1/5](5) + b
2 = − 1 + b
b = 3
Write the equation

y = − [1/5]x + 3

Find the equation in slope intercept form of the line passing through (10, − 1) and parallel to the graph of y = − [1/5]x + 3

Recall that parallel lines have the same slope. Therefore, use slope m = − [1/5] to find the equation .
Use y = mx + b and the point to find b.
y = − [1/5]x + b
− 1 = − [1/5](10) + b
− 1 = − 2 + b
− 1 + 2 = b
b = 1
Write the equation

y = − [1/5]x + 1

Find the equation in slope intercept form of the line passing through ( − 4, − 7) and parallel to the graph of y = [1/2]x − 2

Recall that parallel lines have the same slope. Therefore, use slope m = [1/2] to find the equation .
Use y = mx + b and the point to find b.
y = [1/2]x + b
− 7 = [1/2]( − 4) + b
− 7 = − 2 + b
− 7 + 2 = b
b = − 5
Write the equation

y = [1/2]x − 5

Find the equation in slope intercept form of the line passing through ( − 8,4) and parallel to the graph of y = [1/4]x + 2

Recall that parallel lines have the same slope. Therefore, use slope m = [1/4] to find the equation .
Use y = mx + b and the point to find b.
y = [1/4]x + b
4 = [1/4]( − 8) + b
4 = − 2 + b
4 + 2 = b
b = 6
Write the equation

y = [1/4]x + 6

Find the equation in slope intercept form of the line passing through (4,0) and perpendicular to the graph of y = 2x − 2

Recall that perpendicular lines have the property that the product of their slopes is − 1. Therefore, the slope of the perpendicular line is always
the negative reciprocal of the first line.
Slope of first line m = 2
Find the negative reciprocal of m
m₂ = − [1/m] = − [1/2]
Use y = mx + b and the point to find b.
y = − [1/2]x + b
0 = − [1/2](4) + b
0 = − 2 + b
2 = b
b = 2
Write the equation

y = − [1/2]x + 2

Find the equation in slope intercept form of the line passing through (12,5) and perpendicular to the graph of y = − [4/3]x + 4

Recall that perpendicular lines have the property that the product of their slopes is − 1. Therefore, the slope of the perpendicular line is always
the negative reciprocal of the first line.
Slope of first line m = − [4/3]
Find the negative reciprocal of m
m₂ = − [1/m] = − ( [1/( − [4/3])] ) = [3/4]
Use y = mx + b and the point to find b.
y = [3/4]x + b
5 = [3/4](12) + b
5 = 9 + b
5 = 9 + b
b = − 4
Write the equation

y = [3/4]x − 4

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*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

Writing Linear Functions

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Intro

Slope Intercept Form

Point Slope Form

Parallel and Perpendicular Lines

Example 1: Slope Intercept Form

Example 2: Slope Intercept Form

Example 3: Parallel

Example 4: Perpendicular

Intro 0:00
Slope Intercept Form 0:11

m and b
Example: Graph Using Slope Intercept

Point Slope Form 2:41

Relation to Slope Formula
Example: Point Slope Form

Parallel and Perpendicular Lines 6:28

Review of Parallel and Perpendicular Lines
Example: Parallel
Example: Perpendicular

Example 1: Slope Intercept Form 11:07
Example 2: Slope Intercept Form 13:07
Example 3: Parallel 15:49
Example 4: Perpendicular 18:42

Algebra 2

Section 1: Equations and Inequalities
	Expressions and Formulas	22:23
	Properties of Real Numbers	20:15
	Solving Equations	19:10
	Solving Absolute Value Equations	17:31
	Solving Inequalities	17:14
	Solving Compound and Absolute Value Inequalities	25:00
Section 2: Linear Relations and Functions
	Relations and Functions	32:05
	Linear Equations	14:46
	Slope	23:07
	Writing Linear Functions	23:05
	Special Functions	31:05
	Graphing Inequalities	21:42
Section 3: Systems of Equations and Inequalities
	Solving Systems of Equations by Graphing	17:13
	Solving Systems of Equations Algebraically	23:53
	Solving Systems of Inequalities By Graphing	27:12
	Solving Systems of Equations in Three Variables	28:53
Section 4: Matrices
	Basic Matrix Concepts	11:34
	Matrix Operations	21:36
	Matrix Multiplication	29:36
	Determinants	33:13
	Cramer's Rule	28:25
	Identity and Inverse Matrices	22:25
	Solving Systems of Equations Using Matrices	22:32
Section 5: Quadratic Functions and Inequalities
	Graphing Quadratic Functions	31:48
	Solving Quadratic Equations by Graphing	27:03
	Solving Quadratic Equations by Factoring	19:53
	Imaginary and Complex Numbers	35:45
	Completing the Square	27:11
	Quadratic Formula and the Discriminant	22:48
	Analyzing the Graphs of Quadratic Functions	30:07
	Graphing and Solving Quadratic Inequalities	27:05
Section 6: Polynomial Functions
	Properties of Exponents	19:29
	Operations on Polynomials	13:27
	Dividing Polynomials	31:11
	Polynomial Functions	22:30
	Analyzing Graphs of Polynomial Functions	33:29
	Solving Polynomial Functions	21:10
	Remainder and Factor Theorems	31:21
	Roots and Zeros	31:27
	Rational Zero Theorem	31:16
Section 7: Radical Expressions and Inequalities
	Operations on Functions	34:30
	Inverse Functions and Relations	22:42
	Square Root Functions and Inequalities	30:04
	nth Roots	20:46
	Operations with Radical Expressions	41:11
	Rational Exponents	30:45
	Solving Radical Equations and Inequalities	31:27
Section 8: Rational Equations and Inequalities
	Multiplying and Dividing Rational Expressions	40:54
	Adding and Subtracting Rational Expressions	55:04
	Graphing Rational Functions	57:13
	Direct, Joint, and Inverse Variation	20:21
	Solving Rational Equations and Inequalities	55:14
Section 9: Exponential and Logarithmic Relations
	Exponential Functions	35:58
	Logarithms and Logarithmic Functions	45:54
	Properties of Logarithms	28:43
	Common Logarithms	25:23
	Base e and Natural Logarithms	21:14
	Exponential Growth and Decay	24:30
Section 10: Conic Sections
	Midpoint and Distance Formulas	32:42
	Parabolas	41:27
	Circles	21:03
	Ellipses	46:51
	Hyperbolas	38:15
	Conic Sections	18:43
	Solving Quadratic Systems	47:04
Section 11: Sequences and Series
	Arithmetic Sequences	21:16
	Arithmetic Series	21:36
	Geometric Sequences	23:03
	Geometric Series	22:43
	Infinite Geometric Series	18:32
	Recursion and Special Sequences	14:34
	Binomial Theorem	48:30

Transcription: Writing Linear Functions

Welcome to Educator.com.0000

In today's lesson, we are going to talk about writing linear functions.0002

And in particular, we are going to discuss two forms of these functions.0006

The first form is the slope-intercept form; and this is a very useful form of the equation, because it can help you to graph a linear equation.0012

The slope-intercept form of a line is y = mx + b, where m is the slope, and b is the y-intercept.0022

Recall that the y-intercept is the point at which the graph of the equation (the line) intersects the y-axis.0033

For example, if we look at an equation in this form, and it is given as y = -2x + 1, then the slope is -2, and the y-intercept, or b, is 1.0043

This information alone allows me to graph the line.0060

Before, we talked about graphing the line by finding a couple of points.0063

And we, in particular, used the intercept method, where we found the x-intercept and the y-intercept, and graphed that.0067

This time, I am going to use a slightly different method.0073

So, here I have the y-intercepts: this is the y-coordinate where the graph of the line is going to cross the y-axis.0076

That is going to be at y = 1; x will be 0, and y is 1.0088

Now, I also have the slope; the slope is the change in y, over the change in x.0094

And this is written as -2; but you can think of it in your mind as -2/1.0100

For every 2 that y is decreased, x is increased by 1.0105

And I am already thinking of what I expect this graph to look like.0110

Recall that, if the slope is negative, the line is going to decrease going from left to right--it is going to go this way.0113

If the slope is positive, the line is going to increase going from left to right.0120

I am starting right here, and I am going to decrease y by 2--1, 2--for every increase in x by 1.0125

So now, I have another point: 1, 2, and then increase x by 1.0133

And I now have plenty to go ahead and graph this with.0139

So, you can see how this form of the equation is very helpful in getting information about the line and actually graphing the line.0152

The second form of these linear equations that we are going to look at today is point-slope form.0161

The point-slope form of the equation of a line is y - y₁ = m (x - x₁).0169

And you will see that this is related to slope and the slope formula.0177

So, since the slope is (y₂ - y₁)/(x₂ - x₁), you can look and see that that is pretty familiar.0182

Now, previously we talked about having two points: (x₁,y₁), and the other (x₂,y₂).0191

Well, when we are working with the point-slope form, we have one point (x₁,y₁),0199

and then the other point could be anywhere on the line; it is just another point (x,y) that isn't specified.0205

So, look at how you could manipulate this to be similar to this,0212

if I said, "OK, the slope is some point on the line, minus a given point, over some point on the line, minus the given point."0215

Now, I am going to multiply both sides of this by (x - x₁)--both sides of the equation.0226

These cancel out; and that is going to give me (x - x₁) times m equals (y - y₁).0243

A little bit of rearranging: I am going to put the y portions on the left side of the equation, and the x and the slope on the right side of the equation.0255

And you see, that gives me the point-slope form.0267

And this is useful, because imagine if I am given some facts about a line, and I am told that the slope equals 4, and that this line passes through a certain point.0271

And I am told that it passes through the point (-2,6).0287

Knowing this, I can write the equation for this line in point-slope form.0290

So, point-slope form: y - y₁ = the slope times (x - x₁).0295

So, y is any other point on this line.0301

y₁ is 6; slope is 4; x is another x point on this line that goes along with this y coordinate, minus x₁, which is -2.0306

Simplifying this: a negative and a negative is a positive.0319

What is helpful about this point-slope form is: with a little bit of work, I can put this into the slope-intercept form.0329

Recall that the slope-intercept form (this is the point-slope form of the equation) is y = mx + b.0338

I am going to multiply this out; this is 4x + 8; and I want to isolate y by adding 6 to both sides.0358

So now, I have it in slope-intercept form; this is a very useful form of the equation.0374

And again, this can allow us to graph the equation.0384

Parallel and perpendicular lines: first, recall that parallel lines have the same slope.0388

So, if I have two parallel lines, line 1 and line 2, and the first one has a slope of m₁, and the second one has a slope of m₂, those are equal.0402

Perpendicular lines: the product of the slope of two perpendicular lines is equal to -1.0415

The slope-intercept form and point-slope form can be used to solve problems involving parallel and perpendicular lines.0429

And the reason is: if I am told that a line is parallel to another line, I have the slope.0435

With a little bit more information, I can write the equation in slope-intercept form.0442

Or I may be given the equation in slope-intercept form and told that a line is parallel to that line.0446

The same with perpendicular lines: by having these forms of the equation, which involve slope,0453

and knowing the relationship between two lines and their slopes, I can write the equation for the second line.0458

I can graph the lines, and it is all about knowing the relationships between these two lines.0464

For example, if I am told that the graph of a line is parallel to the line described by the equation y = 1/2x - 4,0471

and I am also told that the line I am looking for passes through a point (4,-3),0499

I can graph the line I am looking for; I can also write an equation for the line I am looking for.0514

So, the graph of a line is parallel to the line y = 1/2x -4, and the line I am looking for passes through a certain point.0521

Well, since parallel lines have the same slope, and I am looking at this, and it is in slope-intercept form, which is y = mx + b, I now have the slope.0529

So, the slope equals 1/2; since these lines are parallel, I also have the slope of the line I am looking for, and it is 1/2.0540

The line I am looking for passes through (4,-3); so that is (4,-3), and I am going to plot that out: (4,-3).0549

And I have the slope, so I know that when I increase y by 1, I increase x by 2; therefore, I can plot the line.0564

OK, and this is as expected, because it has a positive slope, so it is increasing as it goes to the right.0577

The same holds true for perpendicular lines: for example, if I am told that the graph of a line is0598

(this stands for perpendicular) perpendicular to the graph of the line defined by the equation y = 1/4x + 6,0612

and the line passes through some point (say (1,2)), I can graph this line.0627

And the reason I can graph it is that I know that this slope is 1/4, and I recall that the two slopes are related by this formula.0638

So, if I am given a point on a line, as well as the knowledge and relationship between that line and a parallel or perpendicular line,0648

I have the point; I can find the slope; I can graph the line.0659

OK, first example: Find the equation in slope-intercept form of the line with the slope 2/3 and passing through (2,-4).0665

Slope-intercept form is y = mx + b; and I am given slope, so I am given m = 2/3; so let's start from there: y = 2/3 x + b.0678

Well, in order to write this out, I also need to find b; and b is unknown.0694

However, I am given an x value and a y value; and since I have m, x, and y, I can solve for b.0699

So, substituting in -4 for y, and 2 for x, now I can solve for b.0707

So, -4 = 2 times 2...that is 4, so that is 4/3, plus b.0718

Subtract 4/3 from both sides...equals b...and you could really think of this as -1 and 1/3; it might be easier to look at it that way.0734

This is -5 and 1/3...equals b.0746

OK, going back to the beginning here: y = mx + b, so y equals...m is given as 2/3; x; and then b is -5 and 1/3.0749

Using the facts that I was given, I am able to write this in slope-intercept form.0765

I was given the slope, and I was given a point on a line.0773

The point on the line, and the slope: by plugging those into this equation, I could find the y-intercept; and therefore, I could write this out in slope-intercept form.0776

Example 2: find the equation in slope-intercept form of the line passing through these two points.0786

Slope-intercept form is y = mx + b, so I need to have the slope, and I need to have the y-intercept, in order to write this.0793

The slope is the change in y, over the change in x; and I am given two points, so I can find the change in y over the change in x.0802

I am going to call this (x₁,y₁), and this (x₂,y₂).0813

OK, so m = y₂ (which is -3) minus -7, over -6 minus -2.0819

-3...and that negative and negative becomes a positive, so plus 7, minus 6--a negative and a negative--that is plus 2.0833

-3 + 7 is 4; -6 + 2 is -4; I have the slope now--the slope is -1.0841

My next thing is to find b, which is the y-intercept; so, I have y = mx (-1x; we usually just write this as -x,0853

but we are writing it out right now as -1x) + b.0864

I need to solve for b, and I can do that, because I have some x and y values I can substitute in.0868

You could choose either set of coordinates, either ordered pair.0874

I am going to go ahead and choose this first one: y is -7; it equals (instead of writing -1, I am just going to write) -x (which is -2) + b.0880

Solving for b: -7 equals...a negative and a negative is a positive, so that is 2 + b.0898

Subtracting two from both sides, b equals -9.0905

In order to write this equation, I needed to have my slope, which I do.0910

And so, I have that m equals -1, and b equals -9.0916

So now, I can go ahead and write this as y = -x (that is mx, where m is -1) + b (and that is -9): y = -x - 9.0922

This is the equation for this line, written in slope-intercept form.0942

Example 3: Find the equation in slope-intercept form of the line passing through the point (-2,-3), and parallel to the graph of y = 3x - 7.0951

Slope-intercept form is y = mx + b: the first thing I need to do is to find the slope.0964

I am not directly given the slope; however, I am told that this line is parallel to the line described by this equation.0972

Parallel lines have the same slope, so if I know this slope, I know the slope for the line I am looking for an equation of.0981

Well, this is in slope-intercept form; therefore, y = mx + b, so I have the slope.0994

The slope of this line is 3, and the slope of the parallel line (which is my line) is also 3.1005

So now, I have y = 3x + b; and in order to fully have this in slope-intercept form, I need to have b.1012

Well, I have a point on this line, (-2,-3); so I am going to substitute in; I am going to let x equal -2 and y equal -3, and then solve for b.1021

OK, so y is -3; x is -2; and I have the slope: so 3 times -2 plus b; that gives me -3; 3 times -2 is -6; plus b.1041

Adding 6 to both sides, b equals 3.1061

I have the slope; I have the y-intercept; I can write this out in slope-intercept form: y = (slope is 3) 3x...and b is 3.1068

OK, so this one was a little bit more complicated than before, because they didn't directly give us the slope or two points on the line.1083

But what they did give us is the fact that this line is parallel to the line described by this equation.1089

Knowing that means that I have this slope (which is 3), and since my line is parallel, it is the same slope.1096

Once I have the slope, I just take the point on that line, substitute that in for x and y, and solve for b.1104

So, to write it in this form, I am essentially given m; I can figure out b; and this is the equation in slope-intercept form.1111

Find the equation in slope-intercept form of the line passing through (2,-1) and perpendicular to the graph of 2x - 3y = 6.1122

In order to write this in slope-intercept form, y = mx + b, I need to have the slope of this line.1132

I am not given the slope directly; however, I am told that it is perpendicular to the graph of this line.1139

Recall that the slopes of perpendicular lines are related by this equation.1146

So, if you have the slope of two lines that are perpendicular, and you take their product, it is equal to -1.1152

So, let's let the slope of this line equal m₁; and m₂ is the slope of my line, the line I am looking for.1158

Let's go ahead and figure out m₁: what I need to do is write this equation in slope-intercept form,1172

and that will give me the slope of this line, which in turn will give me the slope of the line I am looking for.1180

First, I am going to subtract 2x from both sides; then, I am going to divide both sides by -3.1190

That is going to give me y = -2x/-3 + 6/-3.1201

Simplify: the negatives cancel out, and I will get 2/3x - 2.1209

OK, since this is in slope-intercept form, this is m; this is the slope; so m₁ equals 2/3.1217

I now want to find m₂, which is the slope of the perpendicular line, or the line that I am looking for.1226

m₁ times m₂ equals -1; and I am given m₁--I am given that m₁ is 2/3.1235

So, it is 2/3 times m₂ equals -1.1243

If I multiply both sides by 3/2, I can isolate m₂, and it is -1 times 3/2.1248

Therefore, the slope of the line that I am looking for is -3/2.1257

So now, I have the slope: I have that, for my line, y = -3/2x + b.1265

I have slope; I need the y-intercept; well, I am also given a point on this line, which means I have an x and y value to substitute here.1274

So, y is -1; x is 2; the 2's cancel out, and that is going to give me -3 plus b.1282

I am going to add 3 to both sides, which is going to give me b = 2.1298

So now, I have the slope; I have the y-intercept; so I can write the equation for this line in slope-intercept form: slope is -3/2; b is 2.1304

Again, we approached this by realizing that the slope of the line we are looking for is related to the perpendicular line1322

by the equation m₁ times m₂ (the product of the slopes of the perpendicular lines) = -1.1331

So then, I went about looking for the slope of this line; rewriting it in slope-intercept form gave me y = 2/3x - 2.1338

So, I know that this slope is 2/3; once I have this slope, I can find my slope: m₁ times m₂ equals -1.1347

So, that is 2/3 times m₂ equals -1; m₂ equals -3/2.1357

I go back to this form y = mx + b, now knowing the slope of my line.1363

Substitute in x and y values and the slope to solve for b; b equals 2.1369

I have b; I have the slope; that allows me to write this equation in slope-intercept form.1376

That concludes this lesson of Educator.com.1383

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Dr. Carleen Eaton

Writing Linear Functions

Slide Duration:

Table of Contents

Section 1: Equations and Inequalities

Expressions and Formulas

22m 23s

Intro