what are the critical points of 4y^3+2y-6? cant figure it out .. tried a lot of things. .

Question

amistre64 · Answer

is it 4y^3... ??

anonymous · Answer

how did you get that >?

amistre64 · Answer

i looked at your probelm and then decided to ask you if 4y^3  was what you meanrt to type....then waited for you to answer :)

amistre64 · Answer

if there are no typos in your problem than I would begin by factoring out a "2" tha they all have in common

anonymous · Answer

yes its 4y^3

amistre64 · Answer

is the equation of the curve "4y^3 +2y -6" ?  or is that the 1st derivative of it?

anonymous · Answer

thats the first derivative..the eqauation of the porabola is x-y^2=0

amistre64 · Answer

..... if all you are looking to do is factor "4y^3+2y-6" then I can help.... but as far as it fitting in with all the information that you are leaving out, thatll be up to you :)

anonymous · Answer

yes i can figure the rest out .. its the factoring that for some reason is hurting me ...

amistre64 · Answer

factor a 2:

2(2y^3 +y -3) = 0

2y^3 +y -3 = 0  is easier to play with

amistre64 · Answer

lets find a "pool" of options to narrow our trail and error with ok"

factors of the last #
--------------
factors of the first #

1,3
---  now we can get dirty with it :)
1,2

amistre64 · Answer

we could use long division to try it, but synthetic division is more compact....

amistre64 · Answer

lets try y =1


1 |  2   0  1  -3
      0   2  2    3
  ------------- 
      2   2  3    0 <- remainder is zero, this is a factor

(y-1)

amistre64 · Answer

(y-1) (2y^2 +2y +3)

amistre64 · Answer

the quad facotrs easily now right?

amistre64 · Answer

or not at all ;)

sqrt(2^2 - 4(3)(2)) = sqrt(4 - 24) = sqrt(-20)  doesnt factor...

your only root is y = 1

amistre64 · Answer

which means, there there is only one critical point; and it is at y=1

amistre64 · Answer

any of that make sense?

anonymous · Answer

makes perfect sense.. thanks you much ...

amistre64 · Answer

youre welcome :)