The given quadratic function is:

$y = 4x^2 - 5x + 13$

To rewrite this in vertex form, we will complete the square. The vertex form of a quadratic equation is:

$y = a(x - h)^2 + k$

where $(h, k)$ is the vertex of the parabola. Let's work through the process:

1. Start with the quadratic: $y = 4x^2 - 5x + 13$

2. Factor out the coefficient of $x^2$ (which is 4) from the first two terms: $y = 4(x^2 - \frac{5}{4}x) + 13$

3. To complete the square inside the parentheses, take half of the coefficient of $x$ (which is $-\frac{5}{4}$), square it, and add it inside the parentheses: $\left(\frac{-5}{4} \cdot \frac{1}{2} = \frac{-5}{8}\right) \quad \text{then} \quad \left(\frac{-5}{8}\right)^2 = \frac{25}{64}$

4. Add and subtract $\frac{25}{64}$ inside the parentheses: $y = 4\left(x^2 - \frac{5}{4}x + \frac{25}{64} - \frac{25}{64}\right) + 13$

5. Now, simplify by factoring the perfect square trinomial and adjusting the constants: $y = 4\left(\left(x - \frac{5}{8}\right)^2 - \frac{25}{64}\right) + 13$ $y = 4\left(x - \frac{5}{8}\right)^2 - 4 \times \frac{25}{64} + 13$ $y = 4\left(x - \frac{5}{8}\right)^2 - \frac{100}{64} + 13$ $y = 4\left(x - \frac{5}{8}\right)^2 - \frac{25}{16} + 13$

6. Finally, combine the constants: $y = 4\left(x - \frac{5}{8}\right)^2 + \frac{183}{16}$

So, the vertex form of the function is:

$y = 4\left(x - \frac{5}{8}\right)^2 + \frac{183}{16}$

Would you like more details or have any questions?

Here are five related questions for further exploration:

1. How do you find the vertex of a parabola from the standard form of a quadratic equation?
2. What is the significance of the vertex in the graph of a quadratic function?
3. How does completing the square help in converting to vertex form?
4. Can you derive the vertex form directly from the quadratic formula?
5. What are the advantages of using vertex form over standard form in graphing quadratics?

Tip: Completing the square is a valuable technique not only for rewriting quadratics but also for solving quadratic equations when factoring is difficult.