Let `y=x^2+ x` , the minimum value of y is

Options

A. `-1/4`
B. `1/2`
C. 1/4`
D. `-1/2`

Answer

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

### Step 1: Differentiate the function
We start by differentiating \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = 2x + 1 \]

### Step 2: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ 2x + 1 = 0 \]

### Step 3: Solve for \( x \)
Now, we solve for \( x \):
\[ 2x = -1 \\ x = -\frac{1}{2} \]

### Step 4: Determine if it is a minimum or maximum
Next, we need to determine whether this critical point is a minimum or maximum by using the second derivative test. We differentiate \( \frac{dy}{dx} \) again:
\[ \frac{d^2y}{dx^2} = 2 \]
Since \( \frac{d^2y}{dx^2} = 2 \) is greater than 0, this indicates that the function has a local minimum at \( x = -\frac{1}{2} \).

### Step 5: Find the minimum value of \( y \)
Now we substitute \( x = -\frac{1}{2} \) back into the original function to find the minimum value of \( y \):
\[ y = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) \\ y = \frac{1}{4} - \frac{1}{2} \\ y = \frac{1}{4} - \frac{2}{4} \\ y = -\frac{1}{4} \]

### Conclusion
Thus, the minimum value of \( y \) is:
\[ \boxed{-\frac{1}{4}} \]