How can I determine the points on the curve defined by the equation y = 2x^3 + 3x^2 + 12x + 5 where the tangent line is horizontal?

To find the points on the curve y = 2x^3 + 3x^2 + 12x + 5 where the tangent is horizontal, we need to follow these steps:

Understand the Condition for Horizontal Tangents: A tangent line is horizontal at points where the derivative of the function, dy/dx, equals zero. This means we need to find the derivative of the function and set it to zero.

Calculate the Derivative: First, we differentiate the function:

y = 2x^3 + 3x^2 + 12x + 5

dy/dx = d(2x^3)/dx + d(3x^2)/dx + d(12x)/dx + d(5)/dx

dy/dx = 6x^2 + 6x + 12

Set the Derivative to Zero: Now, we need to solve for x by setting the derivative equal to zero:
```
6x^2 + 6x + 12 = 0
```
Simplifying this equation yields:
```
x^2 + x + 2 = 0
```
Finding the Roots: To solve the quadratic equation, we can use the quadratic formula:
```
x = (-b ± √(b² - 4ac)) / 2a
```
where a = 1, b = 1, and c = 2. Plugging in the values, we have:
```
x = (-1 ± √(1² - 4(1)(2))) / 2(1)
   = (-1 ± √(1 - 8)) / 2
   = (-1 ± √(-7)) / 2
```
Since the discriminant (1 - 8 = -7) is negative, there are no real solutions to this equation.
Conclusion: Therefore, there are no points on the curve y = 2x^3 + 3x^2 + 12x + 5 where the tangent line is horizontal. This means that the curve does not have any maxima or minima within real values of x.