# Quick question of polynomial function

1. Sep 18, 2011

### Nelo

1. The problem statement, all variables and given/known data

Determien an euation for each polynomial function described below state whether the function is even, odd or neither. Sketch a graph of each.

2. Relevant equations

3. The attempt at a solution

My question is.. If youre given information like this ::

a) A quintic function with zeroes at -2(o3) and 3(o2) and that has a y intercept at 70.

From this statement, are you able to determine the value of the leading coefficiant ( ALWAYS?)

So a) would be.. y = (x+2)^3 (x-3)^2 + 70

If i set X = 0, then y= (0+2)^3 (0-3)^2 + 70

Which gives me (2)^3 * (-3)^2 which is 72.

Do i then do the y intercept value over the value of setting the f(0) ?

Like The leading coefficiant for this one would be 70/72(x+2)^3(x-3)^2 + 70

Is that correct?

In the back of the book the "+70" is gone.. So that represents a vertical stretch of that value instead of the y intercept? Im confused. Why is that the coefficiant

2. Sep 18, 2011

### Dick

y = (x+2)^3 (x-3)^2 + 70 doesn't have zeros at x=(-2) or x=3. Does it? Get rid of the 70 and do it again. You've got the leading coefficient right.

3. Sep 18, 2011

### Nelo

What do you mean..? It says in the question "With a y intercept of 70" , why dont that work..? i get rid of the 70 and do wat? , i got that leading coefficiant by dividing 70 by the value of plugging in 0 into x. How else can i do it?

4. Sep 18, 2011

### Dick

I mean if you put x=(-2) or x=3 into y = (x+2)^3 (x-3)^2 + 70 you don't get zero for y. They aren't roots. Just start with y = A (x+2)^3 (x-3)^2 and determine A so you get y=70 at x=0.

5. Sep 18, 2011

### Nelo

ohh okay, and i do that anytime i get a Y intercept value but no leading coefficiant value right?

So if it told me y int was -14 then y = (x+2)^3 (x-3)^2 -14

y = a(0+2)^3 (0-3)^2 -14

Then solve that, and it becomes a part of a, move one to the other side and divide, right?

Does the -14 join with the a?

Like.. 0= 72a -14 14 = 72a 14/72 . ?

6. Sep 18, 2011

### Dick

There is NO REASON to put the -14 into y = (x+2)^3 (x-3)^2 -14 to begin with. I don't know why you are doing it. It makes the y value at x=(-2) and x=3 equal to -14, not the value at x=0, y=-14. The only part of y = (x+2)^3 (x-3)^2 you can adjust and still keep the roots is the leading coefficient.

7. Sep 19, 2011

### Nelo

ok... but the value of the y intercept(when stated) divided by the value of setting f(0) gives me the leading coefficiant right?

8. Sep 19, 2011

### Nelo

Another question.. If i have a grah like (1-x)^4 , then i can remove the x..

[-(x-1)]^4

Does that mean its a reflection on the y axis? Why does the graph still go from q2 to q1 with a upward opening.. It doesnt even reflect. Or am i jus removing a negetive to simplify it, so it doesnt mean y axis reflection?

-x+1

9. Sep 19, 2011

### Staff: Mentor

(1 - x)4 = (-(1 - x))4 = (x - 1)4, for all real x. Note that 1 - x $\neq$, because these quanties have opposite signs.

The graph of y = (1 - x)4 is its own reflection across the vertical line x = 1, but not across the y-axis (the line x = 0).