algebra 2 problem

Question

I'm Lost!!!

 

Given the following polynomial, find
a. The zeros and multiplicity of each
b. Where the graph crosses or touches the axis
c. Number of turning points
d. End behavior
f (x) = - (x-3)2 (x +1)3 (x-7)

Answer 1

Hi Diane, I tried to solve or find out what you are asking for and I think I should give this information first:

 

1.  You already know the "real-zeros" of this function: 3, -1, 7.

2.  How powers (exponents) behaves: odd power passes through and even power touches

 

This graph will look like something like this:

 

 

 

Here is another way to look at it: x y chart.

 

Let's start from -2. and finish at +8

 

f(x) = -(x-3)² (x+1)³ (x-7)

 

 x         y

-2     -1125                -(25) (-1) (-9)

-1        0                    -(16) (0) (-8)

 0       63                     -(9) (1) (-7)

 1      216                    -(4) (8) (-6)

 2      135                    -(1) (27) (-5)

 3        0                      (0) (63) (-4)

 4      375                   -(1) (125) (-3)

 5     1728                   -(4) (216) (-2)

 6     9261                   -(27) (343) (-1)

 7       0                      -(16) (512) (0)

 8     -18225                 -(25) (729) (1)                     

 

These values will tell you how your graph behaves along the Real Zeros.  Since it is even function (degree of 6), both end will go same direction.  In this case, down (because of - sign front of the function).

 

From -2 to 1 (-∞ to 1) (-->  left to right), it is going upward as it passes/crosses x-axis at (-1, 0) [when x < 1 (<---) go downward because it will be all (-) after it passes/crosses x-axis at (-1,0)]. 

 

From 1 to 3, it is going downward and it touches "0" at 3 but it doesn't go through or passes 3. 

 

From 3 to 6, it is going upward again.  From 6 and beyond (6 to ∞), it is going downward as it passes/crosses x-axis (7,0) and it continues downward. 

 

So, from left to right, both ends going downward.

 

I hope you understand what I represented.  Good luck to you.

 

 

Reminder:

 

even function (same behavior on both ends) (both up or both down)    

odd function   (different behavior on ends) (one-side up and one-side down)

 

Answer 2

Hi Diane,

 

We'll go slow and consider each part of this problem.

 

Part A (zeroes and multiplicity):

    1.  Let's write our polynomial function:

             f(x) = -(x-3)²(x+1)³(x-7)

    2.  The zeroes would be when any one of the above three factors would have a value of zero and, thus, causing the value of the function to be zero.

    3.  Let's set each factor to zero and then solve each one for x:

             (a)  x - 3 = 0; x = 3 

             (b)  x + 1 = 0; x = -1

             (c)  x - 7 = 0; x = 7

    4.  Now we'll determine the multiplicity of each factor - it is simply how many times the factor is multiplied by itself, i.e. it's the exponent of each factor:

             (a)  (x - 3)² - multiplicity of 2

             (b)  (x + 1)³ - multiplicity of 3

             (c)  (x - 7) - multiplicity of 1, because of implied exponent of 1

 

Part B (where the graph crosses the x axis):

    1.  The points at which the graph crosses the x axis is where the value of y (the function) is equal to zero.

    2.  Hence the values are the zeroes of the function which we determined in Part A:

             (a)  x = 3

             (b)  x = -1

             (c)  x = 7 

 

Part C (number of turning points):

    1.  To determine the number of turning points we must first determine the order (highest variable exponent) of the function.

    2.  We can determine this by simply adding the exponents of all the factors together:

             (a)  (x-3)²

             (b)  (x+1)³

             (c)  (x+7) - implied exponent of 1

             (d)  Hence, adding all our exponents together, we have:  2 + 3 + 1 = 6

    3.  So if we were to multiply all these factors together to get all the terms of the polynomial, we would see that the highest exponent value for our variable (x) would be 6 (the result of adding the exponents of all the multiplied factors).

    4.  Now we can determine the maximum number of turning points for this function - it is always one less than the order of the polynomial function.

    5.  In this case, since the order of our polynomial function is 6 - the maximum number of turning points is 5.

 

Part D (End Behavior):

    1.  Two attributes of a polynomial function determine it's end behavior:

             (a)  Order of Polynomial (even or odd).

             (b)  Sign of leading term (the sign of the term with the highest variable exponent).

    2.  Let's examine the four possibilities concerning these two attributes and then we'll be able to determine which one applies to our case:

             (a)  If order is odd and sign is positive then graph falls to the left and rises to the right.

             (b)  If order is odd and sign is negative then graph rises to the left and falls to the right.

             (c)  If order is even and sign is positive then graph rises to the left and rises to the right.

             (d) If order is even and sign is negative then graph falls to the left and falls to the right.

    3.  In the case of our polynomial function, we have an order that is even (exponent of 6) and we have a leading minus sign (negative), so according to our above scenarios - it's the last one that applies.

     4.  End behavior of the graph of our polynomial function falls to the left and falls to the right.

 

Kudos for submitting this very intriguing problem !! It's solution allows us to gain a good understanding on function behavior.

 

God bless, Jordan.   

                       

Answer 3

a.

3, with multiplicity of 2

-1, with multiplicity of 3

7, with multiplicity of 1

 

b. 

The graph crosses the axis at -1 and 7. It touches the graph at 3

 

c.

Three

 

d.

x →∞, f(x) → -∞