To write the quadratic function $g(x) = 6x^2 - x + 5$ in standard form, we need to rewrite it in the form:

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

where $(h, k)$ represents the vertex of the parabola.

We can achieve this by completing the square.

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Step 1: Factor out the coefficient of $x^2$ (which is 6) from the first two terms.

$g(x) = 6\left(x^2 - \frac{1}{6}x\right) + 5$

Step 2: Complete the square for the expression inside the parentheses.

To complete the square, take half of the coefficient of $x$, square it, and add it inside the parentheses.

The coefficient of $x$ inside the parentheses is $-\frac{1}{6}$. Half of that is $-\frac{1}{12}$, and squaring it gives $\left(-\frac{1}{12}\right)^2 = \frac{1}{144}$.

Now add and subtract $\frac{1}{144}$ inside the parentheses:

$g(x) = 6\left(x^2 - \frac{1}{6}x + \frac{1}{144} - \frac{1}{144}\right) + 5$

$g(x) = 6\left(\left(x - \frac{1}{12}\right)^2 - \frac{1}{144}\right) + 5$

Step 3: Simplify the expression.

Distribute the 6 across the terms inside the parentheses:

$g(x) = 6\left(x - \frac{1}{12}\right)^2 - 6 \times \frac{1}{144} + 5$

$g(x) = 6\left(x - \frac{1}{12}\right)^2 - \frac{1}{24} + 5$

Combine the constant terms:

$g(x) = 6\left(x - \frac{1}{12}\right)^2 + \left(5 - \frac{1}{24}\right)$

$g(x) = 6\left(x - \frac{1}{12}\right)^2 + \frac{119}{24}$

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Final Answer:

The quadratic function in standard form is:

$g(x) = 6\left(x - \frac{1}{12}\right)^2 + \frac{119}{24}$

The vertex of the parabola is:

$\left( \frac{1}{12}, \frac{119}{24} \right)$

Let me know if you'd like further explanation!