Find the range of `f(x) =x^(2) -5x + 6`

Answer

To find the range of the function \( f(x) = x^2 - 5x + 6 \), we can follow these steps:

### Step 1: Rewrite the quadratic function in vertex form
We start by completing the square for the quadratic expression. The given function is:

\[ f(x) = x^2 - 5x + 6 \]

To complete the square, we take the coefficient of \( x \) (which is -5), divide it by 2, and square it:

\[ \left(\frac{-5}{2}\right)^2 = \frac{25}{4} \]

### Step 2: Add and subtract the square term
Now, we can rewrite the function by adding and subtracting \( \frac{25}{4} \):

\[ f(x) = \left(x^2 - 5x + \frac{25}{4}\right) - \frac{25}{4} + 6 \]

### Step 3: Simplify the expression
Now, we can simplify the expression inside the parentheses:

\[ f(x) = \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + 6 \]

Next, convert 6 into a fraction with a denominator of 4:

\[ 6 = \frac{24}{4} \]

So we can rewrite the function as:

\[ f(x) = \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{24}{4} \]

This simplifies to:

\[ f(x) = \left(x - \frac{5}{2}\right)^2 - \frac{1}{4} \]

### Step 4: Determine the minimum value
The term \( \left(x - \frac{5}{2}\right)^2 \) is always greater than or equal to 0 (since squares are non-negative). The minimum value occurs when \( \left(x - \frac{5}{2}\right)^2 = 0 \):

\[ f(x)_{\text{min}} = 0 - \frac{1}{4} = -\frac{1}{4} \]

### Step 5: Determine the range
As \( x \) increases or decreases without bound, \( f(x) \) will also increase without bound. Therefore, the function can take all values from its minimum value to infinity.

Thus, the range of \( f(x) \) is:

\[ \text{Range of } f(x) = \left[-\frac{1}{4}, \infty\right) \]