1. ## Horizontal tangent?

given function: f(x)=3x^2+6x-10

how can i find horizontal tangent of function f(x)=3x^2+6x-10 ?

2. 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

3. 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 $y=-13$ ?!!

4. why y = -13 ?
hm...

5. y = f '(-1)(x+1) +f(1) = f(1)

My mistake
should be f(-1) not f(1)

6. 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 !!