Question

Find the points on the curve y= 2x^3+3x^2-12x+1
where the tangent is horizontal.

Answer 1

@Photon336

Answer 2

let's first get the derivative first

Answer 3

\[\frac{ d }{ dx } = (6x^{2}+6x-12)\]

Answer 4

hmm im thinking it would look like this
y' = 12x+6

Answer 5

so you took the second derivative?

Answer 6

oh shoot i messed up lol okay so we would equal it to 0 for first derivative right ?

Answer 7

let me see

Answer 8

6x ^2+6x-12= 0

6( x^2+6x-12)= 0

(x-2 ) (x+1 ) =0

x= -1 and x=2

Answer 9

yeah that's what I got. now let's find the equation for that.

Answer 10

actually I got m = -2 and 1

Answer 11

when x=1 find y
again when x=-2 find y

Answer 12

oh riight sorry lool hmm so we would plug these points back to the original equation right ?

Answer 13

Yeah okay @marcelie I think I got it

Answer 14

Remember when we took the derivative and set it equal to zero?

Answer 15

\[2(-2)^{3}+3(-2)^{2}-12(-2)+1 = 21\]

\[2(1)^{3}+3(1)^{2}-12(1)+1 = -6\]

Answer 16

these are the two points where the derivative = 0

\[m(x-x_{0}) = (y-y_{0})\]

so by definition

m = 0 when x = -2 or x = 1

so this means

\[0*(x-x_{0}) = y-21\]

and

0*(x-x_{0}) = y+6

re arranging we get

y = 21

and

y = -6

Answer 17

@marcelie using the site desmos online graphing calculator

graph the original function and

y = -6
and
y = 21

Answer 18

oh okay so our answers would be y= -6 and 21 ?

Answer 19

yes

Answer 20

oh okay x)

Answer 21

points are (1,-6) and (-2,21)