To find the minimum value of the function $y = 2x^2 + 4x$, we can follow these steps:

1. Find the derivative of the function

First, we differentiate the function $y = 2x^2 + 4x$ with respect to $x$:

$\frac{dy}{dx} = \frac{d}{dx}(2x^2) + \frac{d}{dx}(4x)$

$\frac{dy}{dx} = 4x + 4$

2. Set the derivative equal to zero

To find the critical points, set the derivative equal to zero:

$4x + 4 = 0$

3. Solve for $x$

Solve the equation to find the value of $x$:

$4x = -4$ $x = -1$

4. Determine if this point is a minimum

To confirm that this point is a minimum, we can check the second derivative. The second derivative of $y$ is:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(4x + 4) = 4$

Since the second derivative is positive ($4 > 0$), the function is concave up, and the critical point $x = -1$ is indeed a minimum.

5. Find the minimum value of the function

Substitute $x = -1$ back into the original function to find the minimum value of $y$:

$y = 2(-1)^2 + 4(-1)$ $y = 2(1) - 4$ $y = 2 - 4$ $y = -2$

Conclusion

The minimum value of the function $y = 2x^2 + 4x$ is $y = -2$, which occurs at $x = -1$.

Would you like any more details or have any questions?

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Here are 5 related questions to explore further:

1. How would the minimum value change if the coefficient of $x^2$ were different?
2. What happens to the minimum value if a constant is added to the function, say $y = 2x^2 + 4x + 3$?
3. Can you determine the vertex of the parabola represented by $y = 2x^2 + 4x$ without using calculus?
4. How would the function change if we instead maximized the function?
5. How do you interpret the concavity of the function based on the second derivative?

Tip: Always verify the nature of critical points by using the second derivative test; this helps confirm if the point is a minimum, maximum, or a saddle point.