Q: Find the point where the tangent is parrallel to x-axis, y=2x^3-3x^2-12x+6?
All I need to get the first derivative so I can get (x,y) and I'll do the rest.
Idk what to do if my y'= 6x^2-3x-6=0, should I transpost -6? Facotor it?
#Derivatives of application
tough crater
storm yarrow
Q: Find the point where the tangent is parrallel to x-axis, y=2x^3-3x^2-12x+6?
...
solve for x, then plug it in to the equation to get y
solve like usual quadratic equation
divide both sides by 3 if you want to simplify it
quiet meadow
Q: Find the point where the tangent is parrallel to x-axis, y=2x^3-3x^2-12x+6?
...
Wait, how'd you get y' = 6x^2 - 3x - 6?
That's not correct.
tough crater
Wait, how'd you get `y' = 6x^2 - 3x - 6`?
Ops yea I just saw that forgot to multiply* the exponent.
tough crater
divide both sides by 3 if you want to simplify it
Also thank you already got the two X, :)
hollow hollowBOT
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storm yarrow
Also thank you already got the two X, :)
urwelcome