1. ## Polynomial Functions

Graph each function given below on a graphing calculator to find a general rule for determining when a graph crosses the x axis at an x intercept or when the graph just touches and turns away from the x axis. State the rule that you find. [3]

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

y = (x - 4)^3(x - 1)^2

y = (x - 3)^2(x + 4)^4

2. Originally Posted by buddah
Graph each function given below on a graphing calculator to find a general rule for determining when a graph crosses the x axis at an x intercept or when the graph just touches and turns away from the x axis. State the rule that you find. [3]

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

y = (x - 4)^3(x - 1)^2

y = (x - 3)^2(x + 4)^4
so ... what have you noticed about the zeros of each graph?

3. there are 3 zeros?

4. Originally Posted by buddah
there are 3 zeros?
Do you know what a "zero" is on a graph?

5. Here's another guider,

To find the x-intercepts of a polynomial, we have to make y = 0

$0 = (x + 1)^2(x - 2)

0 = (x - 4)^3(x - 1)^2

0 = (x - 3)^2(x + 4)^4$

Now have a think... How do you make a product of 2 terms 0?

6. Originally Posted by buddah
Graph each function given below on a graphing calculator to find a general rule for determining when a graph crosses the x axis at an x intercept or when the graph just touches and turns away from the x axis. State the rule that you find. [3]

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

y = (x - 4)^3(x - 1)^2

y = (x - 3)^2(x + 4)^4
Surely you can "Graph each function ... on a graphing calculator ..." and make some observations. What have you tried? Where are you stuck?

7. Originally Posted by mr fantastic
Surely you can "Graph each function ... on a graphing calculator ..." and make some observations. What have you tried? Where are you stuck?
i solved it:

okay well i tried to do it

is this correct?

x intercept y=0

y = (x + 1)^2(x - 2)
0 = (x + 1)^2(x - 2)
(x + 1)^2 = 0 =====> x + 1 = 0 =====> x = -1 (-1 , 0)
x - 2 = 0 =====> x = 2 (2 , 0)

y = (x - 4)^3(x - 1)^2
0 = (x - 4)^3(x - 1)^2
(x - 4)^3 = 0 ====> x - 4 = 0 =====> x = 4 (4 , 0)
(x - 1)^2 = 0 ====> x - 1 = 0 ====> x = 1 (1 , 0)

y = (x - 3)^2(x + 4)^4
0 = (x - 3)^2(x + 4)^4
(x - 3)^2 = 0 ====> x - 3 = 0 ====> x = 3 (3 , 0)
(x + 4)^4 = 0 ===> x + 4 = 0 ====> x = -4 (-4 , 0)

however am having trouble explaining it

8. you've missed the point of the original question ...

Graph each function given below on a graphing calculator to find a general rule for determining when a graph crosses the x axis at an x intercept or when the graph just touches and turns away from the x axis. State the rule that you find.

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

y = (x - 4)^3(x - 1)^2

y = (x - 3)^2(x + 4)^4
did you graph each function? does the graph of each function cross the x-axis at each zero (x-intercept), or does it just "touch" the x-axis and turn away?

which zeros cross? which turn away?

9. y = (x + 1)^2(x - 2)

Crosses each intercept

y = (x - 4)^3(x - 1)^2

Turns away from the x axis

y = (x - 3)^2(x + 4)^4

Turns away from the x axis

10. Originally Posted by buddah
y = (x + 1)^2(x - 2)

Crosses each intercept Mr F says: What? Are you saying the grpah crosses BOTH x-intercepts ....!? There is nothing more we can do for you if you cannot correctly examine a graph that you draw on your calculator (assuming you typed it in correctly, and if you haven't then there's still nothing more we can do).

y = (x - 4)^3(x - 1)^2

Turns away from the x axis

y = (x - 3)^2(x + 4)^4

Turns away from the x axis
There comes a time when you are expected to apply the help you have been given and finish the question yourself. That time is now.

11. mmm.. I'll graph the first one for you

As you can see, there is a x-intercept at -1 and at 2.

However, the line bounces off -1 but goes straight through 2... now compare this observation with the equation $y = (x + 1)^2(x - 2)$

What's special about the $(x+1)$ factor?

12. the square?

13. That's correct. So because (x+1) is squared. If we look on the graph, it bounces off the x axis, it doesn't touch it.

Now have a look at the second graph:

This graph is $y = (x - 4)^3(x - 1)^2$

You already worked out that there's x-intercepts at 4 and 1. Do you notice any trends? Since the x-1 is squared is bounces off the graph....

I'm not sure if you've studied cubics before but a perfect cube is one of the most simple examples of a graph that has what we call an "inflection point". So at x=4 there is an inflection point.

Can you make any conclusions from this?

14. the cubic part is symmetrical? *cantwatchface

15. lol, don't be shy...

Here is a standard cubic graph (you don't have to know this but I'll show you anyway...):

As you can see the "straight" part in the middle is what we call the "Point of Inflection". This is all you need to know, that when there is a shape like that it is called a point of inflection.

Now if you compile all the stuff you've learnt from the 2 graphs I have shown you in my previous 2 posts can you come to any conclusions?

I'll start you off...

If a factor is (x-1) it will cut the x axis
If a factor is $(x-1)^2$ it will....
If a factor is $(x-1)^3$ it will...

If you can't then you should probably get your teacher to help you because it will make it easier if you have some one on one time with a real teacher

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