Instructions: Given the quadratic function, state whether the parabola opens up or down, and whether it has a maximum or minimum.y=x2−6x+1𝑦=𝑥2−6𝑥+1The parabola opens and has a vertex of (( ,, )), which is a value.

The given quadratic function is y = x^2 - 6x + 1.

Step 1: Identify the coefficient of x^2. In this case, it is 1, which is positive.

Step 2: If the coefficient of x^2 is positive, the parabola opens upwards. If it is negative, the parabola opens downwards. So, in this case, the parabola opens upwards.

Step 3: If the parabola opens upwards, it has a minimum value. If it opens downwards, it has a maximum value. So, in this case, the parabola has a minimum value.

Step 4: The vertex of a parabola given in the form y = ax^2 + bx + c is at the point (-b/2a , f(-b/2a)).

Step 5: Here, a = 1 and b = -6. So, the x-coordinate of the vertex is -(-6)/(2*1) = 3.

Step 6: Substitute x = 3 into the equation to find the y-coordinate of the vertex. y = (3)^2 - 6*(3) + 1 = 9 - 18 + 1 = -8.

So, the vertex of the parabola is at (3, -8), and it is a minimum value.