Analyzing Graphs of Polynomial Functions

Hi! I'm Marielle! (I'm using Mydee's account just because)

Last Monday, March 17, we were assigned to determine the following for Example 1 a and b:

• The least possible degree
• The sign of the leading coefficient
• The x-intercepts and the factors of the function with least possible degree
• The intervals where the function is positive and the intervals where it is negative

*NOTE: This topic is more like a wrap up of the previous lessons from this unit involving graphs

The graphs are:

a.)

 

• Basically, to find the least possible degree of this graph, determine how many x-intercepts there are. In this case, the points are -4, -2, and 2, so the degree of this polynomial function is three.
  
  

• Next, we have to figure the sign of the leading coefficient! For this graph, the sign of the leading coefficient is positive because the function runs down quadrant 3 rises up quadrant 1.
  
  

Moving on~

• Now we have to get the factors of the function. This one should be simple since all you have to do is go back to the zeroes (which are -4, -2, 2) and change them into factors by getting their absolute values [(x+4), (x+2), (x-2)]
  
  

• Lastly, the intervals where the function is positive and the intervals where it is negative.

Positive:

x > 2

-4 < x < -2

Negative:

x < -4

-2 < x < 2

b.)

This function is almost  the same as the first one above, however, this one has different set of values.

The Degree is 3 and sign of the leading coefficient is still positive.

The x-intercepts of this function are -5, -1, 4; while the factors are (x+5) (x+1) (x-4)

The positive intervals will be:

x>4

-5<x<-1

and the negative intervals:

x<-5

-1<x<4

For a different mind exercise, we sketched the given polynomial functions to the graph! And one of the easy ones is the function:

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

Degree	3
Leading Coefficient	+ve
End Behaviour	Goes down to the 3rd quadrant, rises up at 1st quadrant
Zero/x-intercepts	1, -2, -3
y - intercept	-6
Intervals when the function is +ve or -ve	+ve:                              -ve:                 x>1                               x<-3        -3<x<-2                        -2<x<1

This table will be sketched/plotted as:

Again, the degree is 3 because there are three zeroes plotted and the Leading coefficient is positive.

The End Behaviour of the function is it slumps down the third quadrant and rises up the 1st quadrant

The x-intercepts are 1, -2, -3 as presented above

Y-intercept is -6 because we have transform all of the x in every factor [(x-1)(x+2)(x+3)] to zero and solve.

*NOTE: If you want to know how high the function will “bounce” in between points -3 and -2, your best bet is to transform all x’s in the factors again with every value in between the said points.

So, that will be it! I hope this will help you for review purposes and sorry if it won’t and confused you instead.

Either way, hope you will all have a good day ~