Question

find the critical points: (attached)

Answer 1

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

Answer 2

correct?

Answer 3

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)

Answer 4

can someone tell me what i did wrong?

Answer 5

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

Answer 6

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

Answer 7

yes

Answer 8

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

Answer 9

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

Answer 10

do we have to use quadratic formula? i just factored

Answer 11

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

Answer 12

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?

Answer 13

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

Answer 14

lol ok

Answer 15

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

Answer 16

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 :)

Answer 17

oh okay thank you @zepdrix :)!

Answer 18

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