### Graphing Functions

- We can graph functions using a table of values. This is the same process that is used for graphing equations.
- To create a table of values, choose several values for the independent variable, then evaluate the function for these values. This will develop of a list of ordered pairs (inputs, outputs) that can be plotted.
- You can determine if a graph is a function by using the vertical line test. If the graph crosses a vertical line in more than one spot, it is not a function.
- To determine the domain and range of a function from its graph, trace points back to the x-axis and y-axis. The points traced back to the x-axis will cover the domain. The points traced back to the y-axis will cover the range.

### Graphing Functions

- y = mx + b

slope = [(y_{2}− y_{1})/(x_{2}− x_{1})] - slope = [(8 − 4)/(0 − 5)] = [4/( − 5)]

- y = mx + b

slope = [(y_{2}− y_{1})/(x_{2}− x_{1})] - slope = [( − 6 − 6)/(0 − ( − 3))]
- slope = [( − 12)/3]
- slope = − 4

- y = mx + b
- m = − 4

b = ? - − 8 = − 4( − 10) + b
- − 8 = 40 + b
- b = − 48

- y = mx + b
- m = − 1

b = ? - 15 = − 1(7) + b
- 15 = − 7 + b
- 22 = b
- y = − 1x + 22

- slope = m = [(y
_{2}− y_{1})/(x_{2}− x_{1})] - m = [(6 − 1)/(0 − 4)]
- m = [5/( − 4)]
- y = mx + b

- y = mx + b
- m = − 7

b = ? - 11 = − 7( − 8) + b
- 11 = 56 + b
- − 45 = b

- m = [(y
_{2}− y_{1})/(x_{2}− x_{1})] - m = [( − 2 − ( − 3))/(4 − ( − 2))]
- m = [1/6]
- y = mx + b
- − 3 = [1/6]( − 2) + b
- − 3 = − [2/6] + b
- − 2[4/6] = b

- m = [(y
_{2}− y_{1})/(x_{2}− x_{1})] - m = [(6 − ( − 4))/( − 1 − 5)]
- m = − [10/6] = − [5/3]
- 6 = − [5/3]( − 1) + b
- 6 = 1[2/3] + b

- Identify slope
- m = 3
- Identify intercept
- b = − 5

- Identify slope
- m = [1/2]
- Identify intercept
- b = 3

*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

### Graphing 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
- Objectives 0:09
- Graphing Functions 0:54
- Using Slope-Intercept Form
- Vertical Line Test
- Determining the Domain
- Determining the Range
- Example 1 6:06
- Example 2 7:18
- Example 3 8:31
- Example 4 11:04

### Algebra 1 Online Course

### Transcription: Graphing Functions

*Welcome to www.educator.com.*0000

*In this lesson, we are going to work more with graphing functions now that we know a little bit more about them.*0002

*What you will see is that when it comes to graphing functions, we use a lot of the tools that we use with graphing just a normal equation.*0012

*We will see first how you can use a chart to plot a whole bunch of different points and graph out an entire function. *0018

*The good news is if you have a linear function, we can often use our tools for lines to shortcut that process. *0027

*Once we can look at the graph of the function, we will be able to determine whether it is truly a function using something like the vertical line test.*0035

*More importantly, once we have the graph of the function we can test out what its domain and range is just from looking at its graph.*0044

*Let us go ahead and take a look.*0052

*Graphing a function is the same process as graphing an equation.*0058

*We want to look at the relationship between the variables.*0063

*In order to do that, we can simply develop a table of values. *0066

*Remember doing this in our graphing of linear equation section, we may pick a lot of different values for x and see what the corresponding y value is. *0071

*The only difference that we might make with function is simply use different notation.*0081

*I would still pick a lot of different values for x but then I would simply see what the corresponding output is for my y values. *0086

*Always keep in mind that when we are dealing with our inputs, those are going to be our x values.*0096

*When we are dealing with our outputs, those are going to be our y values.*0102

*We will still be able to plot them on a coordinate axis.*0107

*Our x and y will correspond to the inputs and outputs of that function. *0109

*Now if a function represents a line then I have some good news for you.*0119

*You can use a lot of your techniques, especially about slope intercept form. *0122

*Here is a function written in slope intercept form, you can see that it still has the n being slope and it still has b being the y intercept.*0127

*The only change that I have done here is instead of writing out the y, I'm using my function notation. *0141

*This just gives us the name of the function and tells us that our independent variable here is x.*0153

*Since it is in slope intercept form, I could simply graph something like this by first using the y intercept as a starting point*0161

*and then using the slope to identify another point on that graph.*0167

*I have 2 points that I could connect them and then I'm good to go.*0171

*To determine if the graph is a function or not we can use what is known as the vertical line test.*0182

*The way the vertical line test works is you imagine a vertical line on the graph.*0189

*As long as it only crosses the graph in one spot, then you can consider it a function.*0195

*If the vertical line crosses the graph in only one spot and then we can consider it a function.*0201

*If it crosses in more than one spot that is where we will get into trouble.*0208

*Here I have two little diagrams, this one crosses here and here, we would say that this is not a function.*0212

*This one, it only crosses in one spot and if I was to move that dotted line into a different spot over here, it still crosses only one spot.*0226

*In that situation, I would say that this is a function.*0247

*To determine the domain and range of a function when you are looking at its graph,*0262

*think of tracing back all of the values that we used back to the x values on the x axis and back to the y values on the y axis. *0266

*This looks a little difficult to do at first but it is not that bad.*0276

*I imagine picking out some point out on the graph but if I am looking for the domain, I will trace it back to figure out what value on the x axis it came from.*0281

*If I pick another point, trace that back where did it come from on the x axis.*0293

*I do this for many, many different points Iâ€™m always tracing it back *0299

*What I'm looking to do is trace back essentially every single point on that graph.*0303

*Now what this would end up doing is I will end up plugging back many different points *0310

*and they would end up shading in the domain of all the x values that we have used in the function.*0315

*This one, if I trace this back trace it back, you can see that it creates that entire line. *0327

*In this part of the line out here comes from tracing back values on this side.*0332

*Even the ones that it can not see it, sure enough they traced all the way back.*0338

*In a similar fashion, you can figure out the range by taking these points and going to the y axis.*0343

*It is because the yâ€™s represents the outputs, shading the axis, so you knew what was in your range.*0349

*That way we get a little bit better sense of how to graph functions and things about them.*0365

*Let us practice.*0371

*Let us use the vertical line test to see whether these following relations are functions or not.*0373

*The way a vertical line test works is we imagine a vertical line or test it to see if it crosses in only one spot.*0379

*On this first one over here, let us go ahead and put down a vertical line.*0387

*No matter where we move that vertical line, it looks like it will only end up crossing in one spot.*0394

*Since it only crosses once, we will say that it is a function.*0402

*With this one over here, when I put down a vertical line it is easy to see that it crosses in 2 spots.*0412

*It crosses in two spots, not a function.*0422

*The vertical line test says it must only cross in one spot at the most.*0430

*Let us go ahead and see if we can graph one of our linear functions.*0440

*In this one, it is a special type since it is linear.*0443

*We want to look at the form to see if that will help us out.*0448

*This one is written in slope intercept, I know that the y intercept is the 3 and I have a slope of -1/4.*0451

*Let us start with our first being right at 3 and from that point I will go down 1 to the right, 1, 2, 3, 4.*0464

*I have a second point, so I will connect the two.*0473

*There is my line.*0483

*I can also graph out this line by simply choosing a whole bunch of different values for x and evaluating them one at a time. *0485

*I simply use the slope intercept form because it will be a lot quicker.*0495

*I do want to point out that either way would be fine, just pick out some different things for x, plug them in and see what you will get for y.*0500

* You use the graph to determine the domain and range of the function.*0513

*This is unusual in that with last time we actually looked at the equation and tried to pick out what the natural domain was.*0518

*In here, I just have the graph and I have no idea what is being used in here but I can see the inputs and outputs.*0525

*Remember that is every single point on this graph here.*0531

*To first figure out the domain, I will imagine all the points and trace them back to this x axis.*0536

*What I'm doing is Iâ€™m figuring out what points we get shaded in on that x axis when I start tracing them all back.*0545

*What it looks like it is doing is it is tracing out a lot of different values here.*0555

*In fact, even my little point way out here we get trace back.*0561

*I will end up shading quite a bit of the x axis. *0567

*One thing to note is nothing is over on the side.*0571

*The reason why I have nothing over there, is there is no graph to trace back to the x axis.*0576

*Our domain looks like it would start here at -3 and it will go on and on forever from there.*0581

*We can use the same process to figure out what the range is.*0596

*We will simply take all our points now and trace them back to the y axis.*0600

*We will see what this shades in as we do this, bring that one back and you can see Iâ€™m shading a whole bunch of different values here.*0605

*In some places you might have more than one spot it traces back, but that is okay. *0614

*Let us see, keep shading going to the y axis, this guy will go back to -2.*0620

*Notice how below that I'm not going to shade anything on that part because there is no graph to trace back to the y-axis.*0630

*What do we have for our range? Well, the lowest value I have here is that -2.*0642

*We will start there and then it keeps going on from there since the rest of it is shaded.*0648

*Let us do one more domain and range.*0660

*Graph the relation and determine if it is a function then state its domain and range.*0665

*We got a little bit to do with this one, let us first just develop a graph in it.*0670

*This one is not a linear equation so I do not have too many shortcuts at my disposal.*0675

*Iâ€™m just going to end up creating a table of values to help me out.*0680

*Let us choose some different values like 4, 5, 8, and 13.*0687

*These values will make it a little bit easier to evaluate this.*0698

*If I was to use 4, I would end up with 2 Ã— âˆš4 - 4 or 2 Ã— âˆš0 which is 0.*0702

*That is one point I know is on my graph, at 4, 0.*0721

*Let us go ahead and put in our next value and that would be 5.*0731

*2 Ã— âˆš5-4 that would be 1, âˆš1 = 1 so I have 2 as this value.*0736

*I will plug in 5 and I have 2 as my output.*0748

*Looking good, let us try some more.*0752

*Let us go ahead and put in the 8.*0756

*8 - 4 would be 4 and the âˆš4 is 2, 2 Ã— 2 =4.*0762

*This shows that when I put in 8 my output is 4.*0773

*Let us put in 1, 2, 3, 4, 5, 6, 7, 8, 4, there you go.*0792

*It looks like 13 is going to be off my-*0803

*13 - 4 would be 9, 2 Ã— âˆš9, 2 Ã— 3 or 6.*0815

*That is another point on our graph.*0823

*Looking at our points, I might as well start putting them together so we can have a nice little curve right here.*0827

*Onto our first question, I was able to graph the relation, but is it a function or not?*0835

*Does it pass the vertical line test? *0844

*That is our question that we should be asking. *0846

*If we imagine a vertical line on here, does it cross once, more than once? What is happening?*0849

*What was this vertical line? I can see that no matter where I put it, it is only going to cross this graph in one spot.*0856

* I will say yes it is a function since it passes the vertical line test.*0862

*Now we have to figure out what is its domain and range.*0877

*What are all the inputs we could use and what are all the outputs?*0880

*In making this chart, I can already see some of the inputs that I used that you could potentially use even more than that.*0884

*If we trace back all these values, it also includes all the numbers between the ones we used.*0890

*Iâ€™m tracing things back to the x axis and it looks like I would shade in all of this.*0896

*This starts at 4 and continues on from there.*0903

*The domain would start at 4 and just go on from there, 4 to infinity.*0907

*If I take all of the same values then I start to trace them back to the y axis it will shade in a lot of other values but it looks like nothing less than 0.*0917

*We would shade in all that and now we have our range from 0 up to infinity.*0933

*You can see that graphing functions are the same process as just graphing any type of relation. *0944

*Keep track of your inputs and outputs. *0949

*If it is a special type of function like a linear function, then use your tools for graphing lines.*0953

*When it comes to the domain and range, look at your inputs and outputs by tracing all of the values back*0959

*and then show the intervals of all the numbers that should be included. *0967

*Thanks for watching www.educator.com.*0970

1 answer

Last reply by: Professor Eric Smith

Tue Apr 22, 2014 8:33 PM

Post by Wanda Thomas on April 1, 2014

Hello Professor,

Enjoying your lectures. Glad I found this site to assist with my preparation for a CLEP exam.

I have some questions.

Will you please expound on when the rule for using y2 or y1 for the y factor? I notice some problems use on or the other.

Also, please the answer to this problem:

. A line passes through the points (5, - 4) and ( - 1,6). Find the equation of this line in slope intercept form.

*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.

Step 1. m = [(y2 Ã¢Ë†â€™ y1)/(x2 Ã¢Ë†â€™ x1)]

Step 2. m = [(6 Ã¢Ë†â€™ ( Ã¢Ë†â€™ 4))/( Ã¢Ë†â€™ 1 Ã¢Ë†â€™ 5)]

Step 3. m = Ã¢Ë†â€™ [10/6] = Ã¢Ë†â€™ [5/3]

Step 4. 6 = Ã¢Ë†â€™ [5/3]( Ã¢Ë†â€™ 1) + b

Step 5. 6 = 1[2/3] + b

Answer

4[1/3] = b ***Where did the 4 come from?***