## mathcalculus 2 years ago find the critical points: (attached)

1. mathcalculus

2. mathcalculus

i found the derivative then i set it to zero. i factored then found the critical points: x=-8/3

3. mathcalculus

correct?

4. mathcalculus

i tried to find y by plugging in the x into original function.... then got this: (-8/3,-419.92) and (4/5,-58.16)

5. mathcalculus

can someone tell me what i did wrong?

6. mathcalculus

critical points *** x=-8/3 and x=4/5

7. zepdrix

So did you get this for your derivative function?$\large h'(x)=15x^2-116x-32$

8. mathcalculus

yes

9. zepdrix

Setting it equal to zero, tossing it into the Quadratic Formula gives us,$\large x=\frac{116 \pm \sqrt{116^2+4\cdot15\cdot32}}{2\cdot15}$

10. zepdrix

I'm coming up with different solutions, lemme know if that setup looks like what you have.

11. mathcalculus

do we have to use quadratic formula? i just factored

12. zepdrix

You were able to factor it into fractions? lol... You must've factored something incorrectly :[

13. mathcalculus

i don't know i was told you can break it down bby multiplying the coefficient and constant= # and finding the factors and sums of that number. right?

14. zepdrix

Mmm yah something like that, lemme try to factor it real quick. I'm not very good with factoring myself :) lol.

15. mathcalculus

lol ok

16. mathcalculus

i got: x=8 and x=-4/15 by using the quadratic formula

17. zepdrix

Yah me too. $\large 15=5\cdot3 \qquad \qquad -32=-4\cdot8$ That relates to your factoring method somehow..$\large \frac{-4}{15}\qquad=\qquad \frac{-4}{3 \cdot 5}$ I just can't seem to figure out how :)

18. mathcalculus

oh okay thank you @zepdrix :)!

19. mathcalculus

i think we can't factor so we use quadratic right away..