Find the range of `f(x)=x^2-x-3.`

Explanation

To find the range of the function \( f(x) = x^2 - x - 3 \), we will follow these steps:

### Step 1: Identify the coefficients
The function is a quadratic function of the form \( ax^2 + bx + c \). Here, we have:
- \( a = 1 \)
- \( b = -1 \)
- \( c = -3 \)

### Step 2: Calculate the discriminant
The discriminant \( D \) of a quadratic equation is given by the formula:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-1)^2 - 4 \cdot 1 \cdot (-3) = 1 + 12 = 13 \]

### Step 3: Determine the vertex
The vertex of the quadratic function \( f(x) \) can be found using the formula for the x-coordinate of the vertex:
\[ x = -\frac{b}{2a} \]
Substituting the values of \( b \) and \( a \):
\[ x = -\frac{-1}{2 \cdot 1} = \frac{1}{2} \]

### Step 4: Calculate the minimum value of \( f(x) \)
To find the minimum value of the function, substitute \( x = \frac{1}{2} \) back into the function:
\[ f\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right) - 3 \]
Calculating this step-by-step:
\[ = \frac{1}{4} - \frac{1}{2} - 3 \]
Convert \( \frac{1}{2} \) and \( 3 \) to have a common denominator of 4:
\[ = \frac{1}{4} - \frac{2}{4} - \frac{12}{4} = \frac{1 - 2 - 12}{4} = \frac{-13}{4} \]

### Step 5: Determine the range
Since the coefficient \( a = 1 \) is positive, the parabola opens upwards. Therefore, the range of the function is:
\[ \left[-\frac{13}{4}, \infty\right) \]

### Final Answer
The range of \( f(x) = x^2 - x - 3 \) is:
\[ \boxed{\left[-\frac{13}{4}, \infty\right)} \]