find f'(x)
find the slope of the graph of f a x=2 and x=4
find the equations of the tangent lines at x=2 and x=4
and find the value(s) of x where the tangent line is horizontal.
f(x)=2x^2+8x
I found that f'(x)=4x+8
and that the slopes were 16 and 24
and the equations were y=16x-8 and y=24x-46 but i couldnt find out how to do the last part.
horizontal means zero slope
set f' equal to zero and solve for x
what, you can't solve
4x+8 = 0
??
To find the value(s) of x where the tangent line is horizontal, we need to find the x-values that will make the derivative, f'(x), equal to zero.
Given that f'(x) = 4x + 8, we can set this equation equal to zero:
4x + 8 = 0
Now, solve for x:
4x = -8
x = -8/4
x = -2
Therefore, when the tangent line is horizontal, it occurs at x = -2.
So, the value of x where the tangent line is horizontal is x = -2.