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### Analyzing Graphs of Polynomial Functions

- If f(a) < 0 and f(b) > 0, then f has a zero between a and b.
- The graph of a polynomial function of degree n has at most n – 1 local maximums and local minimums.

### Analyzing Graphs of Polynomial Functions

^{3}− 4x

- Find the zeros
- 0 = x(x
^{2}− 4) = x(x − 2)(x + 2) - Using the Zero Product Property you'll have three equations
- x = 0, x − 2 = 0, and x + 2 = 0
- Solve and plot the roots/zeros
- x = 0, x = 2, and x = − 2
- Plot Some Points
x -3 -1 1 3 y x -3 -1 1 3 y -15 3 -3 15 - Sketch graph using end behavior by using the table below
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Odd Degree and Leading Coefficient is positive, the graph will have the following properties:
- As x approaches + ∞ , y approaches + ∞
- As x approaches − ∞ , y approaches − ∞

^{3}− 3x

^{2}− 4x

- Find the zeros
- 0 = x(x
^{2}− 3x − 4) = x(x − 4)(x + 1) - Using the Zero Product Property you'll have three equations
- x = 0, x − 4 = 0, and x + 1 = 0
- Solve and plot the roots/zeros
- x = 0, x = 4, and x = − 1
- Plot Some Points
x -2 2 3 5 y x -2 2 3 5 y -12 -12 -12 30 - Sketch graph using end behavior by using the table below
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Odd Degree and Leading Coesfficient is positive, the graph will have the following properties:
- As x approaches + ∞ , y approaches + ∞
- As x approaches − ∞ , y approaches − ∞

^{3}− x

^{2}+ 6x

- Find the zeros
- 0 = − x(x
^{2}+ x − 6) = x(x − 2)(x + 3) - Using the Zero Product Property you'll have three equations
- x = 0, x − 2 = 0, and x + 3 = 0
- Solve and plot the roots/zeros
- x = 0, x = 2, and x = − 3
- Plot Some Points
x -4 -2 1 3 y x -4 -2 1 3 y 24 -8 4 -18 - Sketch graph using end behavior by using the table below
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Odd Degree and Leading Coesfficient is Negative, the graph will have the following properties:
- As x approaches + ∞ , y approaches − ∞
- As x approaches − ∞ , y approaches + ∞

^{4}− 5x

^{2}+ 4

- Find the zeros
- 0 = (x
^{2})^{2}− 5(x)^{2}+ 4 - We know how to factor x
^{2}− 5x + 4 = 0 - x
^{2}− 5x + 4 = 0 - (x − 4)(x − 1) = 0 Since what we really want is to factor out is the polynomial to the 4th power,
- all we have to do is replace the x with x
^{2} - (x
^{2}− 4)(x^{2}− 1) = 0 - Here we have difference of squares, so we'll use that pattern to factor out completely.
- (x − 2)(x + 2)(x − 1)(x + 1) = 0
- Using the Zero Product Property you'll have 4 equations
- x − 2 = 0, x + 2 = 0, x − 1 = 0, and x + 1 = 0
- Solve and plot the roots/zeros
- x = 2, x = − 2, x = 1, and x = − 1
- Plot Some Points
x -3 −[3/2] 0 [3/2] 3 y x -3 −[3/2] 0 [3/2] 3 y 40 -2.19 4 -2.19 40 - Sketch graph using end behavior by using the table below
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Even Degree and Leading Coesfficient is Positive, the graph will have the following properties:
- As x approaches + ∞ , y approaches + ∞
- As x approaches − ∞ , y approaches + ∞

^{4}+ 3x

^{2}− 2

- Find the zeros
- 0 = − ((x
^{2})^{2}− 3(x)^{2}+ 2) - We know how to factor x
^{2}− 3x + 2 = 0 - x
^{2}− 3x + 2 = 0 - (x − 2)(x − 1) = 0 Since what we really want is to factor out is the polynomial to the 4th power,
- all we have to do is replace the x with x
^{2} - − (x
^{2}− 2)(x^{2}− 1) = 0 - Here we have difference of squares, so we'll use that pattern to factor out completely.
- − (x
^{2}− 2)(x − 1)(x + 1) = 0 - Using the Zero Product Property you'll have 4 equations. For the first equation, solve by taking the square root of both sides.
- x
^{2}− 2 = 0, x − 1 = 0, and x + 1 = 0 - Solve and plot the roots/zeros
- x = + √2 = 1.4, x = − √2 = − 1.4, x = 1, and x = − 1
- Plot Some Points
x -2 −[5/4] 0 [5/4] 2 y x -2 −[5/4] 0 [5/4] 2 y -6 [1/4] -2 [1/4] -6 - Sketch graph using end behavior by using the table below
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Even Degree and Leading Coesfficient is Negative, the graph will have the following properties:
- As x approaches + ∞ , y approaches − ∞
- As x approaches − ∞ , y approaches − ∞

^{6}− 5x

^{4}+ 4x

^{2}

- Find the zeros
- 0 = x
^{2}(x^{4}− 5x^{2}+ 4) = x^{2}((x^{2})^{2}− 5(x)^{2}+ 4) - We know how to factor x
^{2}− 5x + 4 = 0 - x
^{2}− 5x + 4 = 0 - (x − 4)(x − 1) = 0 Since what we really want is to factor out is the polynomial to the 4th power,
- all we have to do is replace the x with x
^{2} - x
^{2}(x^{2}− 4)(x^{2}− 1) = 0 - Here we have difference of squares, so we'll use that pattern to factor out completely.
- x
^{2}(x − 2)(x + 2)(x − 1)(x + 1) = 0 - Using the Zero Product Property you'll have 5 equations.
- x
^{2}= 0, x − 2 = 0, x + 2 = 0, x − 1 = 0, and x + 1 = 0 - Solve and plot the roots/zeros
- x = 0, x = 2, x = − 2, x = 1, and x = − 1
- Plot Some Points
x -2.5 -1.5 -0.5 0.5 1.5 2.5 y x -2.5 -1.5 -0.5 0.5 1.5 2.5 y 73.83 -4.92 0.7 0.7 -4.92 73.83 - Sketch graph using end behavior by using the table below
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Even Degree and Leading Coesfficient is Positive, the graph will have the following properties:
- As x approaches + ∞ , y approaches + ∞
- As x approaches − ∞, y approaches + ∞

^{3}− 2x

^{2}− 2x + 4

- Find the zeros, factor by grouping
- 0 = (x
^{3}− 2x^{2}) + ( − 2x + 4) - 0 = x
^{2}(x − 2) − 2(x − 2) - 0 = (x
^{2}− 2)(x − 2) - Using the Zero Product Property you'll have 3 equations.
- x
^{2}− 2 = 0 and x − 2 = 0 - Solve and plot the roots/zeros. You must use the square root to solve the first equation.
- x = + √2 = 1.41, x = − √2 = − 1.41, and x = 2
- Plot Some Points
x -2 -1 0 1 3 y x -2 -1 0 1 3 y -8 3 4 1 7 - Sketch graph using end behavior by using the table below
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Odd Degree and Leading Coesfficient is Positive, the graph will have the following properties:
- As x approaches + ∞ , y approaches + ∞
- As x approaches − ∞ , y approaches − ∞

^{3}+ x

^{2}+ 7x − 7 by finding the zeros and using the end - behavior only

- Find the zeros, factor by grouping
- 0 = ( − x
^{3}+ x^{2}) + (7x − 7) - 0 = − x
^{2}(x − 1) + 7(x − 1) - 0 = (7 − x
^{2})(x − 1) = − (x^{2}− 7)(x − 1) - Using the Zero Product Property you'll have 3 equations.
- x
^{2}− 7 = 0 and x − 1 = 0 - Solve and plot the roots/zeros. You must use the square root to solve the first equation.
- x = + √7 = 2.65, x = − √7 = − 2.65, and x = 1
- Analyze the end behavior and use it to sketch the graph.
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Odd Degree and Leading Coefficient is Negative, the graph will have the following properties:
- As x approaches + ∞, y approaches − ∞
- As x approaches − ∞, y approaches + ∞
- With your pencil, start somewhere high on Quadrant II and draw a smooth curve going down passing through
- the zero ( − 2.65,0) going down and then start going up going through the zero (1,0) and then start going down and pass through the
- last zero at (2.65,0) until you are in Quadrant IV. Your lines should be smooth and the graph must pass through those three points.
- The rest is just a sketch.
- Compare your graph with the actual graph. How close/far were you?

^{3}− 6x

^{2}− 4x + 24 by finding the zeros, f(0), and end - behavior only.

- Find the zeros, factor by grouping
- 0 = (x
^{3}− 6x^{2}) + ( − 4x + 24) - 0 = x
^{2}(x − 6) − 4(x − 6) - 0 = (x
^{2}− 4)(x − 6) - Using the difference of Squares, factor completely
- 0 = (x − 2)(x + 2)(x − 6)
- Using the Zero Product Property you'll have 3 equations.
- x − 2 = 0, x + 2 = 0, adn x − 6 = 0
- Solve and plot the roots/zeros.
- x = 2, x = − 2, and x = 6
- Find f(0) and plot it
- f(0) = (0)
^{3}− 6(0)^{2}− 4(0) + 24 = 24 - Analyze the end behavior and use it to sketch the graph.
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Odd Degree and Leading Coefficient is Positive, the graph will have the following properties:
- As x approaches + ∞ , y approaches + ∞
- As x approaches − ∞, y approaches − ∞
- With your pencil, start somewhere low on Quadrant III and draw a smooth curve going up passing through
- the zero ( − 2,0) going up and then start going down as you pass (0,24), then continue going down as you pass through (2,0) and continue going down about
- 24 down then start going up as you cross (6,0) and continue going up until you're in Quadrant I. . Your lines should be smooth and the graph must
- pass through those 4 points. The rest is just a sketch.
- Compare your graph with the actual graph. How close/far were you?

^{4}+ 5x

^{2}− 4 by finding the zeros, f(0), and end - behavior only.

- Find the zeros, imagine you're factoring x
^{2}and then replace x with x^{2}as in previous examples - 0 = − (x
^{4}− 5x^{2}+ 4) - 0 = − (x − 4)(x − 1)
- 0 = − (x
^{2}− 4)(x^{2}− 1) - Using the difference of Squares, factor completely
- 0 = − (x − 2)(x + 2)(x − 1)(x + 1)
- Using the Zero Product Property you'll have 4 equations.
- x − 2 = 0, x + 2 = 0, x − 1 = 0, and x + 1 = 0
- Solve and plot the roots/zeros.
- x = 2, x = − 2, x = 1, and x = − 1
- Find f(0) and plot it
- f(0) = − (0)
^{4}+ 5(0)^{2}− 4 = − 4 - Analyze the end behavior and use it to sketch the graph.
End Behavior Chart Even Degree Odd Degree + leading coefficient 1. As x approaches +∞, y approaches +∞ 1. As x approaches +∞, y approaches +∞ 2. As x approaches -∞, y approaches +∞ 2. As x approaches -∞, y approaches -∞ - leading coefficient 1. As x approaches +∞, y approaches -∞ 1. As x approaches +∞, y approaches -∞ 2. As x approaches -∞, y approaches -∞ 2. As x approaches -∞, y approaches +∞ - Since the graph is of Even Degree and Leading Coefficient is Negative, the graph will have the following properties:
- As x approaches + ∞ , y approaches − ∞
- As x approaches − ∞ , y approaches − ∞
- With your pencil, start on Quadrant III and go up and pass through ( − 2,0) and continue going up a little bit then start
- coming down and go through point ( − 1,0). Continue going down and then start going up as you cross (0, − 4) heading towards (1,0) going up.
- Continue going up and then start going down as you cross the point (2,0) and continue going down. You should end on Quadrant IV.
- Compare your graph with the actual graph. How close/far were you?

*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

### Analyzing Graphs of Polynomial 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
- Graphing Polynomial Functions 0:11
- Example: Table and End Behavior
- Location Principle 4:43
- Zero Between Two Points
- Example: Location Principle
- Maximum and Minimum Points 8:40
- Relative Maximum and Relative Minimum
- Example: Number of Relative Max/Min
- Example 1: Graph Polynomial Function 11:57
- Example 2: Graph Polynomial Function 16:19
- Example 3: Graph Polynomial Function 23:27
- Example 4: Graph Polynomial Function 28:35

### Algebra 2

### Transcription: Analyzing Graphs of Polynomial Functions

*Welcome to Educator.com.*0000

*In a previous lecture, we introduced the concept of the graphs of polynomial functions.*0002

*Now, we are going to go further and actually talk about how to develop those graphs.*0006

*OK, in order to obtain the graph of a polynomial function, you make a table of values,*0012

*connect these points with a curve, and then use information about the end behavior of the function.*0017

*Earlier on, when you had to graph linear and quadratic functions, you used the technique of making a table.*0024

*So, it is the same idea here, but with the additional concept of thinking about end behavior.*0032

*For example, if you were to graph a function such as f(x) = 3x ^{3} - 2x^{2} + x,*0039

*you would approach this by finding some x and f(x) values and graphing those out.*0054

*Then, look at the areas of the function that tell you about end behavior.*0063

*I am not going to go ahead and make the table on this one.*0070

*But let's say that you were to graph it out, and it came out something like this.*0072

*OK, using points, and also finding the zeroes, can help you graph.*0080

*In addition to just finding various points, sometimes you can approach making a graph of a polynomial function by also finding the zeroes*0088

*(and recall that the zeroes are where the graph crosses the x-axis, so they are the x-intercepts), and then finding some other points.*0097

*And then, you think about what happens to the graph at the ends.*0109

*Well, recall that, if you have an odd-degree polynomial, such as this one, the ends are going to go in different directions.*0112

*So, one will increase; the other will decrease; and then you have to look at the leading coefficient.*0124

*The leading coefficient: if it is positive (if it is greater than 0), as x increases, y will also increase.*0131

*So, this is odd degree with a positive leading coefficient: x and y are both increasing, and as x gets very, very small,*0140

*y is going to get very, very small, as well; so this is odd degree, where a of n, the leading coefficient, is greater than 0.*0152

*If you have an odd degree where the leading coefficient is negative, then the graph is going to pretty much be the opposite,*0168

*where you are going to have y decreasing (f(x) decreasing) as x gets very large.*0177

*And when x gets very small, f(x) is going to increase; this is odd degree with a negative value for the leading coefficient.*0184

*Let's look at even degrees right here.*0199

*If you are talking about a function with an even degree, such as the largest exponent of the fourth power or the sixth power,*0207

*so it's an even degree, both ends go in the same direction--they both face up, or they both face down.*0222

*And the leading coefficient tells you which direction that is.*0228

*With a leading coefficient that is positive, you are going to get a graph where both ends are facing up.*0231

*So, as x gets very large, or as x gets very small, y is going to get very large; this is even with a leading coefficient that is positive.*0238

*Now, if you have an even degree where the leading coefficient is negative, then both ends will face down.*0249

*So, this could be an example of even with a negative leading coefficient.*0260

*This is review from a previous lecture; but recall that, in addition to finding what is going on*0265

*here in the middle of the graph, and finding your zeroes--finding various points--*0270

*you also need to know about end behavior in order to develop a graph of a polynomial function.*0275

*OK, I mentioned talking about finding the zeroes.*0283

*Well, sometimes it can be difficult or time-consuming to find the exact location of a zero.*0286

*But if you are just trying to sketch out a graph, you can use the location principle*0292

*to estimate where a zero will be--where a graph is going to cross the x-axis.*0296

*So, let's look at what this is saying: this is saying that, if you have a polynomial function f(x),*0300

*suppose that there are numbers a and b, such that f(a) < 0 and f(b) > 0.*0306

*Then, there is a zero, f(x), between these two points, a and b.*0314

*So, let's make this concrete: let's say that this is my point a--that I am going to let a equal 1.*0318

*And this is saying that if f(a) is less than 0 (I find f(a)--let's say it is somewhere down here: f(a) = -3), a is at 1, and f(a) is -3;*0328

*at (1,-3), that is (a,f(a)) on the graph; there is a negative value to the function.*0349

*OK, now let's say I find another value b, and let b equal 3.*0362

*So, here I had a equal 1, and let b equal 3; f(a) = -3; let's let f(b) equal 4.*0367

*Now, what this is telling me is that, somewhere between this point and this point,*0382

*we went from a negative value of the function to a positive value of the function.*0394

*That tells me that there has to be a zero somewhere...actually, it could even be over here; but somewhere between 1 and 3, this is going to cross.*0398

*So, I don't know that it is exactly here; but I know that somewhere between 1 and 3, there is a zero,*0410

*at an x-value somewhere between x = 1 and x = 3.*0420

*And I know that because the function changed from a negative value to a positive value.*0432

*Now, there actually could be more than one zero between those points.*0442

*For example, I could have a graph right here that goes like this.*0446

*And if I just graphed this point out here, I would see that the function is negative at this point, at -4: at f(-4), I have a negative value.*0452

*And then I say, "OK, let's say over here, at f(-2), I have a positive value."*0464

*I know that there is a zero in here; but as it turns out, there is actually more than one zero.*0474

*I know that there is at least one zero here, but there also could be more than one.*0476

*So, when you go ahead and make your table of values, x and f(x), if you see that you have,*0480

*say, positive, positive, and then it goes to negative, I know that at an x-value somewhere between here and here, there is a zero.*0487

*And then, I might have negative and negative, and then switch to positive again.*0498

*And I know that, in here somewhere, there is a zero.*0501

*So again, the idea is that, if you see the value of the function switch from positive to negative,*0508

*or negative to positive, you know that the graph has to pass through the x-axis, and that there is a zero in that location.*0512

*OK, maximum and minimum points: recall that, when we worked with quadratic functions, we talked about maximums and minimums.*0521

*And here, today, with polynomials, we are talking about relative maximums.*0528

*When we talked about quadratics that were downward-facing parabolas, we said, "OK, the vertex is a maximum."*0533

*We didn't say relative; we just said maximum, because this is the highest value that the function could achieve.*0540

*If we were dealing with a minimum value, we didn't say relative; we just said, "OK, this is the minimum,"*0546

*because this is the smallest value that the function could achieve.*0552

*However, when we talk about more complicated, or higher-degree, polynomials, we talk about relative maximums and minimums.*0556

*And what we mean is relative to the x-values around those.*0564

*For example, if I have a graph that looks like this, I could say that this right here--this point (we will call that a,*0569

*and then, right here, the y-value is going to be f(a)--f(a) would be right there,*0581

*so this point is going to be (a,f(a)))--this is a relative maximum.*0589

*And the reason it is a relative maximum is: if I look at any value of x around in this region, the function value is lower than it is right here.*0597

*So, it is the maximum, relative to the values of x near this point.*0607

*A relative minimum is the same idea: if I look at this point here, this gives me a relative minimum.*0613

*And that means that this value of the function (I will call this f(b), so this would be b, and then the y-value would be f(b),*0621

*so this is at (b,f(b)))--this is the smallest value for the function in that region, for these x's.*0636

*But you see here: the graph actually goes lower way over here.*0645

*So, it is not an absolute minimum; it is just relative to that region of the graph.*0648

*And there might be a maximum that is higher up at some area in the graph; but this is a relative maximum for that region.*0652

*The other important point is that the graph of a polynomial function of a certain degree,*0660

*degree n, can have at most n - 1 points that are relative maximums or relative minimums.*0665

*So, if I have a function--let's say f(x) = 4x ^{5} + 3x^{3} - 2x + 6,*0672

*then here, the degree, n is 5 in this case, so the maximum number of relative maximums and relative minimums is 5 - 1, which equals 4.*0682

*So, in a graph like this, degree 5, I can have at most 4 of these maximums and minimums.*0707

*OK, putting together what we have discussed about finding points, the zeroes, using end behavior,*0718

*and relative maximums and minimums, we can develop a graph of some polynomials.*0724

*OK, looking at this first one, this is x ^{3} - 2x; and I am just going to start this one by plotting out some points.*0729

*OK, so let's let x equal -2; this would equal -8; and then, -2 times -2 is 4; -8 + 4 is going to give me -4.*0739

*Here I have -1 for my x-value, and that is going to give me a -1 here, and then -1 times -2 is going to give me 2, so this will be 1.*0756

*Now, notice something: my value for the function switched from negative to positive.*0767

*That means that somewhere between -2 and -1, there is a zero.*0774

*Using the location principle, I know that, somewhere between x = -2 and x = -1, this graph of this function crosses the x-axis.*0781

*When x is 0, f(x) is 0; well, if f(x) is 0, then this point lies on the x-axis, and I actually found a zero--this is a zero.*0793

*Now, when x is 1, that gives me 1, minus 2; that is -1.*0808

*When x is 2, 2 ^{3} is 8, minus 2 times 2 (that is minus 4)--that is going to give me 4.*0815

*Oh, and notice also, here, another zero.*0823

*There is a zero between these values (between the corresponding x-values); somewhere between 1 and 2, there is a zero.*0835

*I have a zero between -2 and -1; I know I have a zero right here; and then, I have another zero between 1 and 2.*0841

*So, I am going to go ahead and plot these out; when x is -2, the function is -4.*0848

*When x is -1, the function is 1; when x is 0, the value of the function is 0;*0854

*when x is 1, we have a -1 right here; and when x is 2, the function value is 2.*0862

*I can see, between here and here: this had to cross the x-axis in between here and here.*0870

*Now, I connect these with a smooth line; I found my points; I know that there is a zero*0875

*somewhere in here; that there is one here; and that there is one here.*0889

*Now, I am going to think about end behavior; and the degree equals 3, and that is odd.*0894

*The leading coefficient, a of n, is 1, so that is positive.*0902

*And what that tells me is: it confirms that my graph is correct, because if it is an odd degree, the two ends are going to face in different directions.*0910

*And since it is a positive leading coefficient, that means that, as x gets very large, y gets very large.*0920

*As x gets very small, y gets very small.*0927

*The other thing I know is that the number of relative maximums and relative minimums is going to be one less than the degree of the polynomial.*0930

*And since the degree is 3, I can have, at most, 2 relative maximums and relative minimums.*0939

*And I see right here: I do have a relative maximum--this is the largest function value for the x-values in this region.*0952

*And right here, I have a relative minimum; so I found both relative maximums and minimums.*0962

*So, this is a pretty good graph of this polynomial, just based on using knowledge that I had and finding a few points on the graph.*0969

*In the second example, again, we have a polynomial of degree 3; but notice that it has a negative leading coefficient.*0980

*That is going to give the graph a different shape.*0988

*In the previous example, I just went ahead and graphed some points; I am going to take a slightly different approach here.*0991

*And I am going to find the zeroes; last time, I found one of the zeroes by luck at point (0,0),*0997

*and I also used the location principle to find the approximate location of two more zeroes.*1003

*Here, I am just going to go ahead and find them directly.*1008

*So, what I am going to do is look at the corresponding equation, -x ^{3} + x^{2} + 6x.*1011

*And I am going to set that equal to 0 to find values of x for which the function's value is 0; that will give me the zeroes.*1019

*So, I am going to approach this by factoring; and I see that I have a common factor of x in all three terms.*1028

*And I am actually going to factor out -x, so what I have left behind is simpler to factor.*1035

*If I pull a -x here, I am going to have an x ^{2} left.*1040

*If I pull a -x from x ^{2}, that will leave me -x, because -x times -x is positive x^{2}.*1043

*Here, I am also going to pull out a negative x, leaving behind -6, and checking that -x and -6 is + 6x.*1052

*OK, now I can look here and see that I can factor this further.*1062

*So, I know that this is going to have the general form (x + something) (x - something).*1070

*And I know that, because I have a negative sign here; so a positive and a negative is going to give me a negative.*1077

*Factors of 6 are 1 and 6, and 2 and 3; and I need those factors when one is positive and one is negative.*1084

*I am going to make one positive and one negative; I want them to add up to the middle term of -1.*1096

*Well, these two are too far apart; now, if I take -3 + 2, that is going to equal -1.*1102

*So, I know that this is the right set of factors, and that I want 3 to be negative and 2 to be positive.*1110

*If I use FOIL to check this, I get x ^{2} - 3x + 2x (is going to give me -x) + 2 times -3 (is going to give me -6).*1116

*So, this is factored out as far as it can go.*1133

*And then, recall the zero product property that says that, if, say, a times b equals 0, then either a equals 0 or b equals 0, or they both equal 0.*1135

*So, any of these factors could equal 0, and that would give me a total value of 0.*1147

*And don't forget about this -x; that is part of it; x + 2 could equal 0, or x - 3 could equal 0, to make this equation true.*1154

*Dividing both sides by -1 just gives me x = 0; here, subtracting 2 from both sides gives me x = -2.*1166

*And adding 3 to both sides, I get x = 3.*1176

*These are the zeroes; these are the points at which the graph crosses the x-intercept.*1179

*So, x equals 0; x equals -2; and x equals 3.*1185

*OK, now, to flesh out this graph some more, I am going to find some other points.*1192

*I found three points--I found my zeroes; now I am just going to find some other points on the graph in this region.*1195

*When x is -1, if you work this out, it comes out to f(x) is -4.*1203

*When x is 1, this will give you -1 + 1 + 6, so that would give you 6.*1212

*When x is 2, working this out, it will give you f(x) is 8.*1221

*All right, so when x is -1, f(x) is 4; when x is 1, f(x) is 6 (about up here).*1229

*And notice here that this tells me that there is a zero in here--that between x is -1 and x is 1, there is a zero right in here.*1244

*And I defined that zero between x is -1 and x is 1; there is a zero that lies in here.*1254

*Let's see, when x is 2, f(x) is way up here; it is going to be way up here somewhere--it is at 8, so we will just put it right there.*1264

*All right--oh, and when x is...actually, I wrote that incorrectly; when x is -1, this is going to be -4.*1282

*And that is why, when you get the switch from negative to positive, there is a zero right in here (in the switch from here to here).*1294

*So, I have some points graphed, and I have my zeroes; so I am going to go ahead and connect these points.*1305

*I am also going to use end behavior; and since this is an odd degree, one end is going to go up, and the other is going to go down.*1324

*And since it is a leading coefficient that is negative, then I am going to have--as x becomes very large, y would be very small;*1336

*and as x becomes very small, over here, y becomes very large.*1349

*So, end behavior helped me figure out these portions of the graph.*1356

*Also, I know that, because my degree is 3, the greatest number of relative maximums and minimums I can have is 2.*1360

*2 relative maximums and minimums at most--that is the greatest number I can have.*1369

*And I do have a relative minimum here and a relative maximum there.*1376

*So, graphing is based on finding the zeroes, graphing a few points, and then thinking about end behavior.*1384

*And notice, also, that here, when my graph switched from a value of -4 (a negative value for the function)*1391

*to a positive value, up here at 6, I knew that there had to be a zero in between those; and I did find that zero.*1399

*OK, I am going to approach this, also, by first finding the zeroes.*1406

*Let's work with the corresponding equation and set this equal to 0.*1411

*Now, looking at this, there are no common factors to pull out; and I have four terms, so I am going to factor by grouping.*1417

*Grouping means that we are going to put x ^{3} - x^{2} together, and add that to -3x + 3.*1424

*Now, looking at this, I have a common factor of x ^{2}; that leaves behind an x and a -1.*1434

*Here, I have a common factor of -3; I pull that out--that leaves an x behind here, and a -1, because -3 and -1 would give me the 3 back.*1442

*Now, I see that I have a common binomial factor of x - 1; I am going to pull that out in front.*1455

*When that is pulled out, that leaves behind x ^{2} - 3.*1462

*OK, using the zero product property, I know that x - 1 = 0 or x ^{2} - 3 = 0.*1467

*If either of these expressions is 0, then the equation will hold true; this left side will be equal to 0.*1476

*So, this is easy: just add 1 to both sides--that gives me x = 1; and then, over here,*1485

*I am going to add 3 to both sides, and that is going to give me x ^{2} = 3.*1490

*I am going to take the square root of both sides, and that is going to give me x = ±√3.*1496

*OK, so I have zeroes at x = 1, x = √3, and x = -√3.*1504

*Well, the square root of 3 can be estimated at about 1.7; so let's rewrite this as 1.7 and -1.7, so that we can graph it.*1517

*OK, I have a zero here at 1; I have a zero here at 1.7 (which is about there) and at -1.7 (which is right about there).*1528

*All right, again, I need to just find a few more points to get a complete graph.*1539

*I am going to let x equal -2; and if I plug that into here, I will get out a value of a function that is -3.*1547

*I am going to let x equal -1, and if you do the calculation on that and substitute -1 in here, you are going to find that y is 4.*1559

*Plotting that out: when x is -2, f(x), or y, is -3, right here.*1570

*When x is -1, f(x) is 4; and you see the location principle at work--that, since we switched from negative to positive,*1580

*that means that there is a zero somewhere in here, somewhere between x is -2 and x is -1.*1588

*And we already know that there is a zero between those two, and it is at -1.7.*1597

*OK, finding a few more points: let's let x equal 0; f(x) is going to be 3.*1605

*When x is 2, plugging that in here, we get a value for the function of 1.*1612

*So, when x is 0, f(x) is 3; when x is 2, f(x) is 1.*1621

*So now, we have enough points to get some sense of what is happening.*1633

*And I am going to start here and connect these.*1639

*Now, using end behavior, I have an odd degree (this is degree 3), which means the two ends are going to go in opposite directions.*1653

*And I have a positive leading coefficient, meaning that, over here on the right, as x gets very large, y is going to get large.*1663

*Over here on the left, as x gets very small, f(x), or y, is going to get very small.*1670

*I am also going to look and see that my degree is 3; so I have, at most, 2 relative maximums and relative minimums.*1677

*And I have a relative maximum here, and I have a relative minimum here; so I have them both.*1690

*So again, I am handling this by finding the zeroes to get a few points on the x-axis,*1699

*finding some more points to find the shape of the rest of this area of the graph,*1704

*and using end behavior to predict how the graph will look at large and small values.*1709

*So, let's take this and set it equal to 0.*1715

*All right, as far as factoring this, this is a trinomial; and we are used to working with these when they are quadratic equations, and the first term is x ^{2}.*1722

*But it is really not that much different with x ^{4}, because x^{2} times x^{2} is x^{4}.*1730

*So, I am looking here, and I see that I have a positive term here, and a negative here.*1740

*And the only way to get that is if I am multiplying a negative times a negative; that gives me a positive.*1745

*And yet, when I sum up the outer and inner terms after multiplying, I will get a negative term here.*1751

*Factors of 8 are 1 and 8, and 2 and 4; and I need them to add up to 6.*1757

*So, if I added -2 and -4, I am going to get -6; so I know that these are the factors that I want to use.*1764

*OK, I can't do anything more with this, so I am just going to leave this as it is.*1774

*However, this is the difference of two squares, so I can factor that out a bit more to this.*1777

*OK, so using the zero product property: x ^{2} - 2 = 0 or x - 2 = 0 or x + 2 = 0.*1785

*So, starting with these easier ones: x = 2, x = -2; I found two zeroes.*1794

*Now, I am going to look over here: x ^{2} = 2--taking the square root of both sides gives me x = ±√2.*1803

*All right, so I have four zeroes; I have zeroes at x = 2, -2, and then √2 is approximately 1.4...*1813

*so let's rewrite that as 1.4 and -1.4 to help me with the graph.*1824

*So, plotting those out: I have a zero here, at -2, and then another one at 2.*1830

*I have a zero here at 1.4 and -1.4.*1836

*I need a few more points to complete my graph; so right up here, let's find x and f(x).*1843

*When x is -1, if you work this out, you get 1; you get -6, because this x ^{2} would just be 1, plus 8; so it is going to give you 9 - 6, or 3.*1852

*When x is 0, that cancels; that cancels; that becomes 0; that leaves me with 8 right here.*1866

*And then finally, when x is 1, this will be 1, minus 6, plus 8; so again, I will have 3.*1873

*All right, so when x is -1, f(x) is 3; when x is 0, f(x) is way up here, about here, at 8 (8 would be right about here); and then, when x is 1, f(x) is 3.*1882

*I am going to connect these points; and I am also going to use my knowledge of end behavior.*1905

*And since this is an even degree, both ends are either going to point up or down.*1909

*And since this is a positive leading coefficient, both ends will actually be up.*1915

*So, when x is very small, y is going to be large; and then, I know that after this point, I am going to go up,*1921

*because when x is very large, y is going to be very large.*1938

*After this point, I am going to continue on up, because when x is small, f(x) is going to be very large,*1942

*because it is an even degree with a positive leading coefficient.*1948

*Now, I am checking that I have degree 4; since this is degree 4, the greatest number of relative maximums and minimums I am going to have are 3.*1951

*And I have a relative minimum there; I have a relative maximum here; and I have a relative minimum here.*1964

*So, I found all three relative maximums and minimums.*1973

*So again, finding the zeroes, I found 1, 2, 3, 4 zeroes.*1976

*Then, I plotted a few other points to give me the shape of the graph.*1983

*I used my knowledge of end behavior to figure out what is going on out here and here.*1987

*And then, I just verified that I had the generally correct shape by seeing that I would expect, at most, three relative maximums and minimums.*1991

*And I found that I had all three.*1999

*That concludes this session of Educator.com on analyzing the graphs of polynomials.*2003

*And I will see you again next lesson!*2008

1 answer

Last reply by: Dr Carleen Eaton

Thu Nov 1, 2012 10:59 PM

Post by Daniel Cuellar on October 30, 2012

by looking at a polynomial of the degree of n, can we have "at most" n-1 relative max & mins? or is it that we WILL have n-1 relative max & mins. Thanks