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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 – y
_{1}= m(x – x_{1}), where m is the slope and (x_{1}, y_{1}) 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

- 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

- 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

- 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

- Find the slope
- m = [(y
_{2}− y_{1})/(x_{2}− x_{1})] = - 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

- Find the slope
- m = [(y
_{2}− y_{1})/(x_{2}− x_{1})] = - 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

- 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

- 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

- 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

- 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
- m
_{2}= − [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

- 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
- m
_{2}= − [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

*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 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

### 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 - y _{1} = m (x - x_{1}).*0169

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

*So, since the slope is (y _{2} - y_{1})/(x_{2} - x_{1}), you can look and see that that is pretty familiar.*0182

*Now, previously we talked about having two points: (x _{1},y_{1}), and the other (x_{2},y_{2}).*0191

*Well, when we are working with the point-slope form, we have one point (x _{1},y_{1}),*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 - x _{1})--both sides of the equation.*0226

*These cancel out; and that is going to give me (x - x _{1}) times m equals (y - y_{1}).*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 - y _{1} = the slope times (x - x_{1}).*0295

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

*y _{1} is 6; slope is 4; x is another x point on this line that goes along with this y coordinate, minus x_{1}, 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 m _{1}, and the second one has a slope of m_{2}, 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 is*0598

*(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 (x _{1},y_{1}), and this (x_{2},y_{2}).*0813

*OK, so m = y _{2} (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 m _{1}; and m_{2} is the slope of my line, the line I am looking for.*1158

*Let's go ahead and figure out m _{1}: 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 m _{1} equals 2/3.*1217

*I now want to find m _{2}, which is the slope of the perpendicular line, or the line that I am looking for.*1226

*m _{1} times m_{2} equals -1; and I am given m_{1}--I am given that m_{1} is 2/3.*1235

*So, it is 2/3 times m _{2} equals -1.*1243

*If I multiply both sides by 3/2, I can isolate m _{2}, 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 line*1322

*by the equation m _{1} times m_{2} (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: m _{1} times m_{2} equals -1.*1347

*So, that is 2/3 times m _{2} equals -1; m_{2} 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

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.