"

If your solution to stated problem does not match the given solution, you should self-critique per instructions at

 

http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

 

 

Your solution, attempt at solution. If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

 

020. `query 20

Question: `qWhat are the zeros of f(x) = 2x - 6 and g(x) = x + 2?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** f(x) = 2x - 6 is zero when 2x - 6 = 0. This equation is easily solved to yield x = 3.

g(x) = x + 2 is zero when x + 2 = 0. This equation is easily solved to yield x = -2. **

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `qWhat does the quadratic formula give you for the zeros of the quadratic polynomial q(x)?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** We get q(x) = f(x) * g(x) = (2x - 6) ( x + 2) = 2x ( x + 2) - 6 ( x + 2) = 2 x^2 + 4 x - 6 x - 12 = 2 x^2 - 2 x - 12.

 

This polynomial is zero, by the quadratic formula, when and only when

 

x = [ -(-2) +- sqrt( (-2)^2 - 4(2)(-12) ] / (2 * 2) = [ 2 +- sqrt( 100) ] / 4 = [ 2 +- 10 ] / 4.

 

Simplifying we get x = (2+10) / 4 = 3 or x = (2 - 10) / 4 = -2.

 

This agrees with the fact that f(x) = 0 when and only when x = -3, and g(x) = 0 when and only when x = 2.

 

The only was f(x) * g(x) can be zero is for either f(x) or g(x) to be zero. **

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `q2. If z1 and z1 are the zeros of x^2 - x - 6, then what is the evidence that x^2-x + 6=(x - z1) * (x

- z2)?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a**

 

We could factor the polynomial, but the following argument works whether the polynomial can be factored or not (for example it would apply to the expression x^2 - 2 x - 5, which doesn't factor easily, just as well as it does to x^2 - x - 6).

z1 and z2 both give us zero when plugged into x^2 - x - 6, and also when plugged into (x-z1)(x-z2).

 

When expanded, (x-z1)(x-z2) gives us the term x^2, which matches the x^2 term of x^2 - x - 6.

 

Since the zeros and the highest-power term match both functions are obtained from the basic y = x^2 function by the

same vertical stretch, both have parabolic graphs and both have the same zeros. They must therefore be identical. **

Of course we can factor the polynomial into (x - 3) * (x + 2), then use the zero property to determine that the zeroes are z1 = 3 and z2 = -2. 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `q3. Explain why, if the quadratic polynomial f(x) = a x^2 + bx + c has no zeros, that polynomial

cannot be the product of two linear polynomials.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** If f(x) has linear factors, then if any of these linear factors is zero, multiplying it by the other factors will yield zero.

 

Any linear factor can be set equal to zero and solved for x.

 

Thus if f(x) has linear factors, it has zeros.

 

So if f(x) has no zeros, it cannot have linear factors. **

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `q4. Explain why no polynomial of degree 2 can be the product of three or more polynomials of

degree 1.

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** If you have 3 polynomials of degree one then each contains a nonzero multiple of x. Multiplying three such factors

together will therefore yield a term which is a nonzero multiple of x^3. For example (x-2)(x+3)(x-1) = (x^2 + x -

6)(x+1) = x^3 + 2 x^2 - 5 x - 6.

 

Any polynomial containing a nonzero multiple of x^3 has degree at least 3, and so cannot be of degree 2.

 

Therefore a polynomial of degree 2 cannot be a product of three or more polynomials of degree 1. **

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `q5. What then would be the zeros and the large-x behavior of y = (x-7)(x+12)

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** y = 0 when x-7 = 0 or x+12 = 0, i.e., when x = 7 or x = -12.

 

If x is a large positive number then both x-7 and x+12 are large positive numbers so that (x-7)(x+12) is a very large

positive number.

 

If x is a large negative number then both x-7 and x+12 are large negative numbers so that (x-7)(x+12) is again a very

large positive number.

 

So for large positive and negative x the function more and more rapidly approaches infinity.

 

The graph will be decreasing, beginning with very large positive values at large negative x, as it passes through its leftmost

zero at x = -12. The rate of decrease will initially be very rapid but will decrease less and less rapidly until the graph

reaches a low point between x = 7 and x = -12, at which point it begins increasing at an increasing rate, passing through

its rightmost zero at x = 7 and continuing with increasing slope as x becomes large. **

 

STUDENT COMMENT

 

I’m still not sure how to describe it.  I used this table:

 

x x-12 x+7 y = (x-12)(x+7)
-100 -112 -93 10 000 approx
-20 -32 -13 286
-12 -24 -5 120
-7 -19 0 0
-1 -13 6 -78
0 -12 7 -84
1 -11 8 -88
7 -5 14 -70
12 0 19 0
100 88 119 88 000



INSTRUCTOR RESPONSE:

 

Your table can serve as a basic for understanding.  Think about these questions:

You don't really need a detailed table to think about patterns of this sort.

 

Try answering the following questions without referring to the table:

As you move to the left along the graph, to the left of all the zeros, and continue moving to the left x becomes a larger and larger negative number.

If x is a large negative number, then is x - 7 positive or negative?
If x is a large negative number, then is x + 12 positive or negative?
What sign do you therefore get when you multiply the two quantities?
Does the result keep getting larger and larger?
Is there a limit to how large the result can get, or does it remain bounded?
How therefore does the graph behave as you move further and further to the left?

As you move to the right along the graph, to the right of all the zeros, and continue moving to the right x becomes a larger and larger positive number.

If x is a large positive number, then is x - 7 positive or negative?
If x is a large positive number, then is x + 12 positive or negative?
What sign do you therefore get when you multiply the two quantities?
Does the result keep getting larger and larger?
Is there a limit to how large the result can get, or does it remain bounded?
How therefore does the graph behave as you move further and further to the right?

Now consider values of x between -12 and 7.

If x is between -12 and 7, is x - 7 positive or negative?
If x is between -12 and 7, is x + 12 positive or negative?
What sign do you therefore get when you multiply the two quantities?
Is there a limit to how large the result can get for x values in this interval, or does it remain bounded?

To get from one zero to the next (in this case from (-7, 0) to (12, 0)), the graph of a polynomial has to move away from the x axis, then back.

 

At some point between the zeros, it will reach a relative maximum distance from the x axis.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `q6. Describe your graph of y = f(x) = (x-3)(x+2)(x+1), describing all intercepts, intervals of

increasing or decreasing behavior, concavity, and large-|x| behavior.

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** The function has zeros at x = 3, x = -2 and x = -1.

 

For large positive x all three factors will be large positive numbers, so that the product will be a very large positive

number.

 

For large negative x all three factors will be large negative numbers, so that the product will be a very large negative

number.

 

The graph will be increasing, beginning with very large negative values at large negative x, as it passes through its leftmost

zero at x = -2. The rate of increase will initially be very rapid but the graph will increase less and less rapidly until the

graph reaches a relative maximum point between x = -2 and x = -1, at which point it begins decreasing.

 

The function will be decreasing as it passes through its zer0 at x = -1. Somewhere between x = -1 and its next zero at x

= 3 the function will reach a relative minimum value after which it will begin to increase more and more rapidly. It will be

increasing as it passes through its zero at x = 3 and will continue to increase faster and faster as x becomes larger. **

 

FURTHER SUGGESTIONS FOR THE PERPLEXED

 

The zeros of this function occur at x = -2, x = -1 and x = 3.  So all the zeros are between x = -2 and x = 3.

 

As we move to the left from x = -2, we encounter larger and larger negative values of x.  As we do so:

As we move to the right from x = 3, we encounter larger and larger positive values of x.  As we do so:

By thinking about these questions, you should see how the graph of our function rises rapidly from negative values toward the zero at (-2, 0).  You should also see how the graph rises rapidly into positive values from its zero at (3, 0).

 

The figure below shows the graph to the left of x = -2 and to the right of x = 3:

 

           

 

The graph also has a zero at (-1, 0).  So the zeros at x = -2, -1 and 3 divide the x axis into a series of intervals.

 

Between x = -2 and x = -1:

Between x = -1 and x = 3:

Now consider the following questions:

Sketch your graph, then note the following:

On your sketch, at what point does the graph stop increasing?

 

To get to (-1, 0) the graph has to start decreasing (this occurs at the point where it stops increasing), and must continue decreasing for at least awhile past (-1, 0).  It will then have to start increasing once more to reach the point (3, 0).

 

At what point does the graph, as you sketched it, stop decreasing?

 

The graph depicted below is not the same as the graph of this question. 

 

 

Here is a description of the graph:

 

The graph decreases at a decreasing rate, passing through a zero at x = -2 and continuing to decrease at a decreasing rate until it reaches the approximate point (-1.6, -1), which is the lowest value in its vicinity (we call such a point a 'relative minimum').

 

The graph then increases at an increasing rate, and continues to do so as it passes through a zero at x = -1.  It continues increasing at an increasing rate until it reaches a point somewhere around (-0.5, 2).

 

The graph continues increasing, but at a decreasing rate, until it reaches a point near (0.8, 6), which is the peak value in its vicinity (called a 'relative maximum').

 

The graph then begins decreasing at a decreasing rate, passing through its zero at x = 2 and continuing to decrease at a decreasing rate.

 

The graph is concave up until it reaches a point near (-0.5, 2), after which it is concave down.

 

Wherever a graph is decreasing at a decreasing rate, or increasing at an increasing rate, it is concave down.

Wherever a graph is decreasing at an increasing rate, or increasing at a decreasing rate, it is concave up.

Can you now describe your graph of the function (x + 2) (x + 1) (x - 3) in similar language?

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `q1. Give the y = (x-x1)(x-x2)(x-x3) form of a degree 3 polynomial with zeros at x = -3, 1 and 2, as

well as the y = ax^3 + bx^2 + cx + d form.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** The factored form is

 

y=(x+3)(x-1)(x-2)

 

The standard polynomial form is obtained by multiplying these factors to obtain

 

(x+3) ( x^2 - 2x - x + 2) =

 

(x+3)( x^2 - 3x + 2) =

 

(x^3 - 3 x^2 + 2 x) + (3 x^2 - 9 x + 6) =

 

x^3 - 7 x + 6. **

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `q2. Describe how the two graphs of y = (x-1)(x+3)(x-4) and y = (1/12) * (x-1)(x+3)(x-4)

compare.

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** The graphs both have zeros when x - 1 = 0, when x + 3 = 0 and when x - 4 = 0. These zeros therefore occur at x

= 1, x = -3 and x = 4.

 

The only difference is that the graph of y = 1/12 ( x-1)(x+3)(x-4) is everywhere 12 times closer to the x axis than that of

y = (x-1)(x+3)(x-4), with 1/12 the slope at every point. **

 

STUDENT COMMENT

 

I knew the second would be closer, but how did you come up with 12 times closer?

INSTRUCTOR RESPONSE

 

For example, 1/12 * 24 = 2. The result 2 is 12 times closer to zero than 24.

Now the value of y = (1/12) * (x-1)(x+3)(x-4) is 1/12 as great as y = (x-1)(x+3)(x-4).

So for any x value, the y value of one function is 12 times closer to 0 than the y value of the other.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `q4. What function describes the approximate behavior of the graph of y = p(x) = (x-3)(x-3)(x+4) near the point (3,0)?

 

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** If x is close to 3 then x + 4 is close to 7 and is not significantly different for various values near x = 3.

 

However the nature of x - 3 depends greatly on just how close x is to 3, and whether x is greater or less than 3. x - 3 =

0 when x = 3, x - 3 < 0 when x > 3 and x - 3 > - when x < 3.

(x-3)^2 will be zero when x = 3, and will increase at an increasing rate as x moves away from 3.

 

So the function y = (x-3)(x-3)(x+4) is close to y = 7(x-3)^2. Note that this function describes a parabola with vertex at

(3, 0), the 2d-degree zero of the given polynomial, and basic points (3, 0), (4, 7) and (2, 7).

 

So near x = 3 the graph of p(x) = (x-3)(x-3)(x+4) will be very nearly matched by the parabolic graph of the function y =

7 ( x - 3) ^2. As x moves out of the vicinity of x = 3 the graphs will at first gradually, then more and more rapidly move

apart.

 

In general near z second-degree 0, like 3 in the present example, the graph of a parabola will look like a parabola whose

vertex is at that zero. **

 

STUDENT QUESTION

 

How did you get 7(x-3)^2 ?


I see where the 7 comes from I think. When x = 3, the (x+4) would be 7, but where is the (x-3)^2 It is a little confusing for
me.

INSTRUCTOR RESPONSE

 

Near x = 3 the value of x + 4 is near 7. As long as x doesn't stray too far from 3, the value of x + 4 won't stray far from 7.

However x - 3 changes drastically in any neighborhood of x = 3, taking positive values, negative values and the value 0. On one side of x = 3 this expression is negative, on the other side it's positive. Between x = 2 and x = 4 the expression (x - 3) changes from -1 to 1. On the same interval (x - 3)^2 changes from 1 to 0 and back to 1.

As long as x + 4 stays around 7, give or take a fairly small percent, we can then say that the value of (x + 4) ( x - 3)^2 has to be around 7 * (x - 3)^2. This will be the case anywhere in the near vicinity of x = 3.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `qWhy do we say that near (3,0) the graph of (x-3)(x-3)(x+4) is approximately the same as the

graph of 7(x-3)^2?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`awith the zero of 3, x+4 will equal 7, so that portion of the graph will appear as a quadratic equation or a parabola

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `qDescribe the graph of 7(x-3)^2.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`aThis is a parabola, obtained from the basic y = x^2 parabola by a vertical stretch of 7 and horizontal shift of 3 units. It

will be a steep parabola with vertex (3, 0) and basic points at (2, 7) and (4, 7).

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `qHow do the graphs made on your calculator or computer compare?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`aThe two graphs should match very closely near (3, 0). To the right the graph of the polynomial will gradually move

higher than that of the parabola, and to the left will gradually move lower.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `qWhat does the graph of a polynomial look like near a second-degree zero and why?

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`aSTUDENT ANSWER: parabola, when that portion is factored out it is a quadratic, since that zero is repeated the

graph cannot cross the x axis at that point but must touch it sou appearing as a parabola

 

INSTRUCTOR'S ADDITION: Also because the other factors of the polynomial remain nearly constant close to the

zero.

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

Question: `q5. Sketch graphs of y = (x-2)^2 * (x+3)^2 * (x-1) and y = -.5 * (x-3) (x+2)^3, including

intercepts, the large-| x | behavior for both positive and negative x, concavity, and intervals of increasing and decreasing

behavior.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

`a** The graph of y = (x-2)^2 * (x+3)^2 * (x-1) is nearly parabolic in the vicinity of the zeros at 2 and -3. It only passes

through the x axis at x = 1.

 

Near x = 2 we can approximate all factors except (x-2)^2 by substituting x = 2, which gives us y = (x-2)^2 * (x+3)^2 *

(x-1) = (x-2)^2 * (2+3)^2 * (2-1) = 25 (x-2)^2, an upward-opening parabola with vertex at x = 2.

Near x = -3 we can approximate all factors except (x+3)^2 by substituting x = -3, which gives us y = (x-2)^2 * (x+3)^2

* (x-1) = (-3-2)^2 * (x+3)^2 * (-3-1) = -100 (x+3)^2, a downward-opening parabola with vertex at x = -3.

 

For large positive x the graph is positive and concave up, increasing very rapidly.

For large negative x the graph is negative and concave down, decreasing very rapidly.

 

The graph rises from extremely large negative x values to the zero at x = -3, where it touches the x axis and turns back

toward negative values without ever passing through the x axis. It reaches a minimum somewhere between x = -3 and x

= 1, in the process passing through the y axis at (0, -36). The graph passes through the x axis at x = 1, going from

negative to positive. It turns back toward the x axis at some point between x = 1 and x = 2, touches the x axis moving

along in which is nearly parabolic in the vicinity of that point, and the turns back upward, increasing with a rapidly

increasing slope as x moves to the right.

 

The graph increases at a decreasing rate up to (-3,0), then decreases at an increasing rate until concavity changes from

negative to positive sometime before the function reaches its minimum somewhere between (-3,0) and (1,0). Then it

decreases at an increasing rate and continues to do so until a point between the local minimum and (1,0), probably close

to (1,0), at which concavity again becomes negative. From that point the function increases as a decreasing rate until it

reaches a local maximum somewhere between x=1 and x=2, at which point it begins decreasing at an increasing rate,

remaining concave down until at some point before (2,0) the concavity becomes upward and the function begins

decreasing at a decreasing rate until reaching the local minimum at (2,0). From that point it begins increasing at an

increasing rate, maintaining an upward concavity and rapidly increasing to very large y values.

 

ALTERNATIVE DESCRIPTION:

 

The graph of y = -.5 * (x-3) (x+2)^3 passed thru the x axis at x = 3 and at x = -2.

 

Near x = -2 we can approximate all factors except (x-2)^2 by substituting x = 2, which gives us y = -.5 ( -2 - 3) ( x +

2)^3 = 2.5 (x+2)^3. This function gives us a cubic polynomial with zero at x = -2 and basic points (-2, 0), (-3, -2.5) and

(3, 2.5).

 

For large positive x the graph is negative and concave down, decreasing very rapidly.

For large negative x the graph is negative and concave up, decreasing very rapidly as x moves in the negative direction.

 

The graph rises from extremely large negative x values toward the zero at x = -2, leveling off at (-2, 0) before again

beginning to increase at a increasing rate. Somewhere before the zero at x = 3 the graph turns around and begins

decreasing, passing downward through (3, 0) as it declines faster and faster into negative values.**

 

 

 

 

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique rating:

 

If you understand the assignment and were able to solve the previously given problems from your worksheets, you should be able to complete most of the following problems quickly and easily.  If you experience difficulty with some of these problems, you will be given notes and we will work to resolve difficulties.

Question:  The zeros of a polynomial are x = -3, x = 2 and x = 4. 

What is the minimum possible degree of this polynomial?

Give a possible equation of this degree for the polynomial.

Moderately challenging question:  Can you give an equation of a polynomial of higher degree whose zeros are x = -3, x = 2 and x = 4?

 

Your solution:

 

 

 

Confidence Assessment:

Question:  Sketch a graph of a polynomial of degree 3 whose zeros are at x = -3 and x = 4.

Specify the approximate intervals on which your function is increasing, and the approximate intervals on which it is decreasing.

Specify the large-|x| behavior of your polynomial.

 

Your solution:

 

 

 

Confidence Assessment:

Question:  The value of a certain polynomial of degree 5 approaches -infinity as x approaches +infinity.  The expression (x - 4)^2 occurs in the factored form of the polynomial.  The polynomial has just two zeros, one of which is at x = -1. 

Sketch and describe a graph fitting this description, and give the intervals where the graph is increasing, and where it is decreasing.

Give your best description of its graph in the immediate neighborhood of the point (4, 0), and explain as best you can why the graph has this shape.

 

Your solution:

 

 

 

Confidence Assessment:

 

Self-critique rating:
"