### Instruction: If we express $3x^2 + x - 4$ in the form $a(x - h)^2 + k$, then what is $k$?### Response: To express $3x^2 + x - 4$ in the form $a(x - h)^2 + k$, I need to complete the square for the quadratic term.
I can factor out 3 from the first two terms and get $3(x^2 + \frac{1}{3}x) - 4$.
Now I need to add and subtract a term inside the parentheses that will make it a perfect square.
The term I need is half of the coefficient of x, squared. That is, $(\frac{1}{6})^2 = \frac{1}{36}$.
So I add and subtract $\frac{1}{36}$ inside the parentheses and get $3(x^2 + \frac{1}{3}x + \frac{1}{36} - \frac{1}{36}) - 4$.
I can rewrite the expression inside the parentheses as a square of a binomial: $(x + \frac{1}{6})^2 - \frac{1}{36}$.
So the whole expression becomes $3((x + \frac{1}{6})^2 - \frac{1}{36}) - 4$.
Now I can distribute the 3 and simplify: $3(x + \frac{1}{6})^2 - \frac{1}{12} - 4$.
Finally, I can combine the constant terms and get $3(x + \frac{1}{6})^2 - \frac{49}{12}$.
This is in the form $a(x - h)^2 + k$, where $a = 3$, $h = -\frac{1}{6}$, and $k = -\frac{49}{12}$.