If t is a constant and the curve y=2x^3+3tx^2+6x+131 has NO horizontal tangents, what conditions must t satisfy?

Question

(a) t >-2
(b) t < 2
(c) t > 2
(d) t < -2
(e) none of these

Apparently, the correct answer is (e).

So for the curve to have no horizontal tangents, I know it's derivative cannot equal to 0.

If y=2x^3+3tx^2+6x+131, then
y'=6x^2+6tx+6 cannot equal 0
y'=6(x^2+tx+1) cannot equal 0

What are the following steps that what would bring me to answer (e) none of these?

Answer 1

6(x^2+tx+1) cannot equal 0
so, the discriminant (b^2-4ac) must be negative
t^2 - 4 < 0
|t| < 2

Answer 2

Your explanation makes sense. However, if I graph f(x)=6(x^2+x+1), for example, on desmos I can see there is a horizontal tangent line at (-0.5, 4.5). This observation contradicts your answer since |1| < 2. Why is this?

Answer 3

6(x^2+x+1) is the derivative. It does not matter where it has a horizontal tangent.

We want to find where y=2x^3+3tx^2+6x+131 has a horizontal tangent. It does not, because 6(x^2+x+1) is never zero.