Let's go through each part of the question step by step.

Part (a)

You are asked to express $f(x) = x^2 - 4x + 5$ in the form $f(x) = (x - a)^2 + b$.

This requires completing the square:

1. Start with the quadratic $f(x) = x^2 - 4x + 5$.

2. To complete the square, take half of the coefficient of $x$ (which is -4), square it, and add/subtract that inside the expression.

   $f(x) = (x^2 - 4x + 4) + 5 - 4$

3. This simplifies to:

   $f(x) = (x - 2)^2 + 1$

So, $f(x) = (x - 2)^2 + 1$, where $a = 2$ and $b = 1$.

Part (b)

You need to sketch two graphs on the same diagram:

1. The graph of $y = f(x)$, which is $y = (x - 2)^2 + 1$, a parabola shifted 2 units to the right and 1 unit upwards.

2. The graph of $y = 10 - f(x)$, which becomes:

   $y = 10 - \left( (x - 2)^2 + 1 \right) = 9 - (x - 2)^2$

   This is an upside-down parabola shifted 2 units to the right and 9 units upwards.

Part (c)

You need to find the range of values of $x$ for which $10 - f(x)$ is positive. That is, solve:

$10 - f(x) > 0 \quad \text{or} \quad 10 > (x - 2)^2 + 1$

This simplifies to:

$9 > (x - 2)^2$

Taking the square root of both sides:

$-3 < x - 2 < 3$

So:

$-1 < x < 5$

Thus, the range of values for which $10 - f(x)$ is positive is $x \in (-1, 5)$.

Would you like more details or have any questions?

5 Related Questions:

1. How do you complete the square for any quadratic expression?
2. What transformations occur when a parabola is shifted horizontally and vertically?
3. How can you determine the vertex of a quadratic function from its completed square form?
4. What happens graphically when you subtract a quadratic function from a constant?
5. How do you solve inequalities involving quadratic functions?

Tip:

When completing the square, remember that the goal is to rewrite a quadratic function in a form that makes it easier to identify shifts and the vertex of the parabola.