The problem asks to find the range of the quadratic function $y = x^2 - 6x + 4$ when $2 \le x \le 6$. The solution shows the graph of the function for the given interval and identifies the maximum and minimum values of $y$ within that interval.

1. Problem Description

The problem asks to find the range of the quadratic function $y = x^2 - 6x + 4$ when $2 \le x \le 6$. The solution shows the graph of the function for the given interval and identifies the maximum and minimum values of $y$ within that interval.

2. Solution Steps

The given function is $y = x^2 - 6x + 4$.

The function is rewritten in vertex form by completing the square:

$y = (x - 3)^2 - 5$.

The vertex of the parabola is at $(3, -5)$.

Since the coefficient of the $x^2$ term is positive, the parabola opens upwards.

The interval for $x$ is $2 \le x \le 6$.

The minimum value of the function occurs at the vertex, which is at $x = 3$, and the minimum value is $y = -5$.

To find the maximum value, we check the endpoints of the interval.

When $x = 2$, $y = (2 - 3)^2 - 5 = (-1)^2 - 5 = 1 - 5 = -4$.

When $x = 6$, $y = (6 - 3)^2 - 5 = (3)^2 - 5 = 9 - 5 = 4$.

The maximum value is $4$ at $x = 6$.

Therefore, the range of the function on the interval $2 \le x \le 6$ is $-5 \le y \le 4$.

3. Final Answer

The range is $-5 \le y \le 4$.