given function: f(x)=3x^2+6x-10 how can i find horizontal tangent of function f(x)=3x^2+6x-10 ?
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the horizontal tangent is where f ' = 0 f ' = 6x +6 = 0 at x= -1 y = f '(-1)(x+1) +f(1) = f(1) y = -1
Originally Posted by Calculus26 the horizontal tangent is where f ' = 0 f ' = 6x +6 = 0 at x= -1 y = f '(-1)(x+1) +f(1) = f(1) y = -1 Sorry, maybe ?!!
why y = -13 ? hm...
y = f '(-1)(x+1) +f(1) = f(1) My mistake should be f(-1) not f(1)
so it should be f ' = 6x +6 = 0 at x= -1 y = f '(-1)(x+1) + f(-1) y = f '(-1)(-1+1) + f(-1) y = f(-1) f(x)=3x^2+6x-10 y = 3(-1)^2 + 6(-1) - 10 = 3 - 6 - 10 = -13 ? thank you !!
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