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 the function with respect to \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx}(x^2 + x) = 2x + 1 \]

### Step 2: Set the derivative equal to zero
To find the critical points, we set the first derivative equal to zero:

\[ 2x + 1 = 0 \]

### Step 3: Solve for \( x \)
Now, we solve for \( x \):

\[ 2x = -1 \implies x = -\frac{1}{2} \]

### Step 4: Second derivative test
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:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2x + 1) = 2 \]

### Step 5: Analyze the second derivative
Since \( \frac{d^2y}{dx^2} = 2 > 0 \), this indicates that the function has a local minimum at \( x = -\frac{1}{2} \).

### Step 6: Calculate the minimum value of \( y \)
Now, we substitute \( x = -\frac{1}{2} \) back into the original function to find the minimum value of \( y \):

\[ y = \left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right) = \frac{1}{4} - \frac{1}{2} = \frac{1}{4} - \frac{2}{4} = -\frac{1}{4} \]

### Conclusion
Thus, the minimum value of \( y \) is:

\[ \boxed{-\frac{1}{4}} \]

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