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y = 2x^3 + 3x^2 − 12x + 9
y' = 6x^2 + 6x − 12

tangent line is horizontal y'=0
6x^2 + 6x − 12=0
x^2+x-2=0
(x+2)(x-1)=0
x=-2 or x=1
when x=-2, y=2x^3 + 3x^2 − 12x + 9 = -16 + 12 + 24 + 9 =29
when x=1, y=2x^3 + 3x^2 − 12x + 9 = 2 +3 -12 +9 =2
two points (-2,29),(1,2) 



2).  y = 2x^3 + 3x^2 − 12x + 9

slope of horizontal-tangent-line, y' = 6x^2+6x-12 = 0
SO,
6(x+2)(x-1) = 0,
x=1, and x=-2,
SO, for, x= 1, y= 2+3-12+9 = 2,
Hence,
the  1st. point is : Answer (1, 2)  

and, for, x=-2,
y = 2(-2)^3 +3(-2)^2 -12(-2) +9 = -16+12+24+9 = 29
Hence,
the 2nd-point is : Answer (-2, 29) 

the points on the curve [tex]y = 2x^3 + 3x^2 - 12x + 9 are (-2,29) \; and \; (1,2)[/tex]where the tangent line is horizontal.

Given :

The equation of the curve is [tex]y=2x^3\:+\:3x^2\:-\:12x\:+\:9[/tex]

Given tangent line is horizontal . Tangent line is horizontal when slope =0

Slope is nothing but the derivative.

So we find out x values where derivative =0

Lets take derivative for the given curve y

[tex]y=2x^3+3x^2-12x+9\\y'=2(3x^2)+3(2x)-12\\y'=6x^2+6x-12[/tex]

Now we set the derivative =0 and solve for

[tex]6x^2+6x-12=0\\Divide\; whole \; equation \; by \; 6\\x^2+x-2=0\\(x+2)(x-1)=0\\x+2=0, x=-2\\\\x-1=0, x=1[/tex]

So , the slope =0 when x=1 and x=-2

Now we find out the points . Use the original function

[tex]y=2x^3\:+\:3x^2\:-\:12x\:+\:9\\x=-2\\2\left(-2\right)^3+3\left(-2\right)^2-12\left(-2\right)+9=29\\(-2,29)\\\\x=1\\2\left(1\right)^3+3\left(1\right)^2-12\left(1\right)+9=2\\(1,2)[/tex]

the points on the curve [tex]y = 2x^3 + 3x^2 - 12x + 9 are (-2,29) \; and \; (1,2)[/tex]where the tangent line is horizontal.

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