Find the point(s) where the line tangent to the graph of f(x)= 2x^3+6x^2-18x-2
is horizontal

Question

anonymous · Answer

I found the first deriv

anonymous · Answer

the slope, but stuck on trying to find points

tad1 · Answer

If the tangent to the graph is horizontal, the first derivative, which is the slope , = 0.
So the equation will be y = a constant.  So set the first derivative = 0
6x^2 + 12x - 18 = 0
factor out the 6 to get 6(x^2 + 2x - 3) = 0
Factor (x-3)(x+1) = so X = 3 or X = -1
Substitute into f(x) to find y.

anonymous · Answer

oh of course! horizontal line has not slope! Thank you!

tad1 · Answer

It does have a slope; the slope is zero.  a vertical line has no slope.

anonymous · Answer

plugging in 3 though gives me the wrong y coordinate - i think it might be -3

anonymous · Answer

unless, i did something wrong

anonymous · Answer

nope should be x+3 as a factor therefore x=-3

tad1 · Answer

You're right.  Sorry for the error.
Tad

anonymous · Answer

no worries, thanks again for the help

tad1 · Answer

Thanks for the medal.

radar · Answer

I got two points for y(-3)=52  for a point (-3, 52) and
                              Y(1)=-12  for a point (1, -12)

Now which point is correct or is there two points?

radar · Answer

I noticed that the problem asked for point(s), leaving that as a possibility.

tad1 · Answer

There are two points. where the slope of the tangent line = 0.  the original function f(x) is cubic.  It would have three roots, though two of them may be complex.

tad1 · Answer

The degree of the equation tells you how many roots it has, though sometimes the roots are repeated as in a perfect square quadratic.

radar · Answer

OK. so I guess the horizontal lines would be:    y=-12 and y=52 is that what you came up with?