The problem asks us to draw the graph of the quadratic function $y = x^2 - 6x + 4$ for $0 \le x \le 5$. The function is already given in vertex form as $y = (x-3)^2 - 5$. We are given a coordinate plane with a partially drawn parabola.

1. Problem Description

The problem asks us to draw the graph of the quadratic function $y = x^2 - 6x + 4$ for $0 \le x \le 5$. The function is already given in vertex form as $y = (x-3)^2 - 5$. We are given a coordinate plane with a partially drawn parabola.

2. Solution Steps

The given equation is $y = (x-3)^2 - 5$. This is a parabola with vertex at $(3, -5)$.

Since $0 \le x \le 5$, we need to find the y-values at $x=0$ and $x=5$.

When $x=0$, $y = (0-3)^2 - 5 = 9 - 5 = 4$.

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

So the parabola passes through the points $(0,4)$ and $(5, -1)$.

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

Now we draw the parabola using the information we have. The parabola should pass through $(0, 4)$, $(3, -5)$, and $(5, -1)$. We need to limit the graph to the interval $0 \le x \le 5$.

The graph is a U-shaped curve. The graph should start at $(0, 4)$, go down to the vertex at $(3, -5)$, and then go up to $(5, -1)$.

3. Final Answer

The graph of the parabola $y = x^2 - 6x + 4$ for $0 \le x \le 5$ is a U-shaped curve starting at $(0, 4)$, reaching its minimum at $(3, -5)$, and ending at $(5, -1)$. The part of the parabola where x < 0 and x > 5 should not be drawn.

The graph from the original image already shows an accurate representation of the curve.