The problem you provided reads as follows:

16. (a) The function $f(x) = ax^2 + bx + 3$, where $x \geq 2$, $x \in \mathbb{R}$.

Given that $f^{-1}(-3) = 4$ and $f(2) = 45$, show that $a = 2$ and $b = -8$.

For these values of $a$ and $b$, find an expression for $f^{-1}$.

Let's go step by step to solve it.

Step 1: Using $f(2) = 45$ to find a relation between $a$ and $b$

We are given that $f(x) = ax^2 + bx + 3$ and $f(2) = 45$, meaning when $x = 2$, the output is 45. Substituting $x = 2$ into the function:

$f(2) = a(2)^2 + b(2) + 3 = 45$

This simplifies to:

$4a + 2b + 3 = 45$

Subtract 3 from both sides:

$4a + 2b = 42$

Now divide through by 2:

$2a + b = 21 \quad \text{(Equation 1)}$

Step 2: Using $f^{-1}(-3) = 4$

This means when $f(4) = -3$. So substitute $x = 4$ into the original function $f(x)$ and set it equal to -3:

$f(4) = a(4)^2 + b(4) + 3 = -3$

This simplifies to:

$16a + 4b + 3 = -3$

Subtract 3 from both sides:

$16a + 4b = -6$

Now divide through by 2:

$8a + 2b = -3 \quad \text{(Equation 2)}$

Step 3: Solving the system of equations

We now have the following system of equations:

1. $2a + b = 21$
2. $8a + 2b = -3$

Solve the first equation for $b$:

$b = 21 - 2a$

Substitute this into the second equation:

$8a + 2(21 - 2a) = -3$

Simplify and solve for $a$:

$8a + 42 - 4a = -3$

$4a = -3 - 42$

$4a = -45$

[ a = -\frac{45}{4}