Let's break down and solve each of the questions shown in the image one by one.

(a) Express $4x^2 - 12x + 13$ in the form $(2x + a)^2 + b$, where $a$ and $b$ are constants.

We start by completing the square on the quadratic expression $4x^2 - 12x + 13$.

1. Factor out 4 from the quadratic terms: $4(x^2 - 3x) + 13$
2. Complete the square on $x^2 - 3x$. The coefficient of $x$ is -3, and half of it squared is $\left(\frac{-3}{2}\right)^2 = \frac{9}{4}$. $4\left( (x - \frac{3}{2})^2 - \frac{9}{4} \right) + 13$
3. Distribute the 4: $4(x - \frac{3}{2})^2 - 9 + 13$ Simplifying: $4(x - \frac{3}{2})^2 + 4$
4. Now, rewrite the expression as: $(2x - 3)^2 + 4$ So, the constants $a$ and $b$ are: $a = -3, \quad b = 4$

(b) Given that it is possible to form the composite function $gf$, find the least possible value of $p$ and the greatest possible value of $q$.

The function $f(x) = 4x^2 - 12x + 13$ is defined for $p < x < q$, and the function $g(x) = 3x + 1$ is defined for $x < 8$.

For the composite function $gf(x)$ to exist, the output of $f(x)$ (i.e., $f(x) = 4x^2 - 12x + 13$) must be within the domain of $g(x)$, which is $(-\infty, 8)$.

We need to solve: $4x^2 - 12x + 13 < 8$ Simplify: $4x^2 - 12x + 5 < 0$ Divide by 4: $x^2 - 3x + \frac{5}{4} < 0$ Find the roots of the corresponding quadratic equation: $x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(\frac{5}{4})}}{2(1)} = \frac{3 \pm \sqrt{9 - 5}}{2} = \frac{3 \pm \sqrt{4}}{2} = \frac{3 \pm 2}{2}$ The solutions are: $x = \frac{3 + 2}{2} = 2.5 \quad \text{and} \quad x = \frac{3 - 2}{2} = 0.5$ Therefore, $f(x)$ is less than 8 between $x = 0.5$ and $x = 2.5$. So, the least possible value of $p$ is 0.5, and the greatest possible value of $q$ is 2.5.

(c) Find an expression for $gf(x)$.

We already know: $f(x) = 4x^2 - 12x + 13 \quad \text{and} \quad g(x) = 3x + 1$ The composite function $gf(x)$ is: $gf(x) = g(f(x)) = g(4x^2 - 12x + 13) = 3(4x^2 - 12x + 13) + 1$ Simplify: $gf(x) = 12x^2 - 36x + 39 + 1 = 12x^2 - 36x + 40$

(d) Find an expression for $h^{-1}(x)$, where $h(x) = 4x^2 - 12x + 13$ for $x < 0$.

We are tasked with finding the inverse function of $h(x)$. First, express $h(x)$ as a completed square: $h(x) = (2x - 3)^2 + 4$ Let $y = h(x)$, so: $y = (2x - 3)^2 + 4$ Solve for $x$: $y - 4 = (2x - 3)^2$ Take the square root of both sides: $\sqrt{y - 4} = 2x - 3$ Since $x < 0$, we take the negative square root: $-\sqrt{y - 4} = 2x - 3$ Solve for $x$: $x = \frac{-\sqrt{y - 4} + 3}{2}$ Thus, the inverse function is: $h^{-1}(x) = \frac{-\sqrt{x - 4} + 3}{2}$

Would you like more details on any part of the solution or explanations? Here are five related questions you might consider:

1. What are some other methods for completing the square?
2. How does the domain of a composite function depend on the original functions?
3. Could we solve $gf(x)$ by applying a different function first?
4. How can we confirm if the calculated inverse function is correct?
5. What happens to the range of $f(x)$ and $g(x)$ if their domains are extended?

Tip: Always check the domain restrictions when forming composite functions to avoid invalid outputs.