## mathcalculus Group Title find the critical points: (attached) 11 months ago 11 months ago

1. mathcalculus Group Title

2. mathcalculus Group Title

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

3. mathcalculus Group Title

correct?

4. mathcalculus Group Title

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 Group Title

can someone tell me what i did wrong?

6. mathcalculus Group Title

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

7. zepdrix Group Title

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

8. mathcalculus Group Title

yes

9. zepdrix Group Title

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 Group Title

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

11. mathcalculus Group Title

do we have to use quadratic formula? i just factored

12. zepdrix Group Title

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

13. mathcalculus Group Title

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 Group Title

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

15. mathcalculus Group Title

lol ok

16. mathcalculus Group Title

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

17. zepdrix Group Title

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 Group Title

oh okay thank you @zepdrix :)!

19. mathcalculus Group Title

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