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INSTRUCTORS Carleen Eaton Grant Fraser Eric Smith
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Lecture Comments (2)

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 − y1)/(x2 − x1)]
   Step 2. m = [(6 − ( − 4))/( − 1 − 5)]
   Step 3. m = − [10/6] = − [5/3]
   Step 4. 6 = − [5/3]( − 1) + b
   Step 5. 6 = 1[2/3] + b

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

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

A line passes through the points (5,4) and (0,8). Find the equation of the line in slope intercept form.
  • y = mx + b
    slope = [(y2 − y1)/(x2 − x1)]
  • slope = [(8 − 4)/(0 − 5)] = [4/( − 5)]
y = − [4/5]x + 8
A line passes through the points ( - 3,6) and (0, - 6). Find the equation of the line in slope intercept form.
  • y = mx + b
    slope = [(y2 − y1)/(x2 − x1)]
  • slope = [( − 6 − 6)/(0 − ( − 3))]
  • slope = [( − 12)/3]
  • slope = − 4
y = − 4x − 6
A line has slope - 4 and passes through ( - 10, - 8). Find its equation in slope intercept form.
  • y = mx + b
  • m = − 4
    b = ?
  • − 8 = − 4( − 10) + b
  • − 8 = 40 + b
  • b = − 48
y = − 4x − 48
A line has slope - 1 and passes through (7,15). Find its equation in slope intercept form.
  • y = mx + b
  • m = − 1
    b = ?
  • 15 = − 1(7) + b
  • 15 = − 7 + b
  • 22 = b
  • y = − 1x + 22
y = − x + 22
A line passes through the points (4,1) and (0,6). Find the equation of the line in slope intercept form.
  • slope = m = [(y2 − y1)/(x2 − x1)]
  • m = [(6 − 1)/(0 − 4)]
  • m = [5/( − 4)]
  • y = mx + b
y = − [5/4]x + 6
A line has slope - 7 and passes through ( - 8,11). Find its equation in slope intercept form.
  • y = mx + b
  • m = − 7
    b = ?
  • 11 = − 7( − 8) + b
  • 11 = 56 + b
  • − 45 = b
y = − 7x − 45
A line passes through the points ( - 2, - 3) and (4, - 2). Find the equation of this line in slope intercept form.
  • m = [(y2 − y1)/(x2 − x1)]
  • m = [( − 2 − ( − 3))/(4 − ( − 2))]
  • m = [1/6]
  • y = mx + b
  • − 3 = [1/6]( − 2) + b
  • − 3 = − [2/6] + b
  • − 2[4/6] = b
y = [1/6]x − 2[4/6]
A line passes through the points (5, - 4) and ( - 1,6). Find the equation of this line in slope intercept form.
  • m = [(y2 − y1)/(x2 − x1)]
  • m = [(6 − ( − 4))/( − 1 − 5)]
  • m = − [10/6] = − [5/3]
  • 6 = − [5/3]( − 1) + b
  • 6 = 1[2/3] + b
4[1/3] = b
Graph y = 3x − 5
  • Identify slope
  • m = 3
  • Identify intercept
  • b = − 5
Graph utilizing slope and intercept
Graph y = [x/2] + 3
  • Identify slope
  • m = [1/2]
  • Identify intercept
  • b = 3
Graph utilizing slope and intercept

*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

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 point0161

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 I’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 y’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 I’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 I’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

I’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 × √4 - 4 or 2 × √0 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 × √5-4 that would be 1, √1 = 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 √4 is 2, 2 × 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 × √9, 2 × 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

I’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 back0959

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

Thanks for watching www.educator.com.0970