Let's solve the given question step by step.

The equation is:

$x^2 - 6x + 1 = (x - a)^2 - b$

Part (i) Find the values of $a$ and $b$:

We need to express $x^2 - 6x + 1$ in the form $(x - a)^2 - b$.

First, expand $(x - a)^2$: $(x - a)^2 = x^2 - 2ax + a^2$ So, the equation becomes: $x^2 - 6x + 1 = x^2 - 2ax + a^2 - b$

Now, compare the coefficients of like terms from both sides of the equation.

• Coefficient of $x^2$ is the same on both sides.

• For the $x$-term: On the left-hand side, the coefficient of $x$ is $-6$, and on the right-hand side, it is $-2a$. Therefore: $-2a = -6 \quad \Rightarrow \quad a = 3$

• For the constant term: On the left-hand side, the constant is $1$, and on the right-hand side, it is $a^2 - b$. Thus: $a^2 - b = 1 \quad \Rightarrow \quad 9 - b = 1 \quad \Rightarrow \quad b = 8$

So, the values are: $a = 3, \quad b = 8$

Part (ii) Coordinates of the turning point:

The equation $y = x^2 - 6x + 1$ can be rewritten as:

$y = (x - a)^2 - b = (x - 3)^2 - 8$

The turning point (vertex) of a quadratic function $y = (x - a)^2 - b$ occurs at the point $(a, -b)$. Therefore, the coordinates of the turning point are:

$(3, -8)$

Final Answers:

1. $a = 3$
2. $b = 8$
3. The coordinates of the turning point are $(3, -8)$.

Would you like any further details or explanations? Here are some related questions to explore:

1. What is the general method to complete the square for a quadratic equation?
2. How does the form $(x - a)^2 - b$ help in identifying the turning point?
3. What is the significance of the discriminant in quadratic equations?
4. How can you find the axis of symmetry for a quadratic function?
5. How does the graph of $y = x^2 - 6x + 1$ compare to the graph of $y = x^2$?

Tip: Completing the square is a useful technique for rewriting quadratics to easily identify key features like the vertex (turning point).