If f(x) = 2x + 3, then find the value of f f(-1). 12. Find the vertex of the parabola formed from the quadratic equation y = x2 − 5x + 6.

Substitute $x = -1$ into the function $f(x) = 2x + 3$.

$$f(-1) = 2(-1) + 3$$ $$f(-1) = -2 + 3$$ $$f(-1) = 1$$

From Step 1, we know $f(-1) = 1$. So, we need to find $f(1)$.
Substitute $x = 1$ into the function $f(x) = 2x + 3$.

$$f(1) = 2(1) + 3$$ $$f(1) = 2 + 3$$ $$f(1) = 5$$

The value of $f \circ f(-1)$ is $\boxed{5}$.

12. Find the vertex of the parabola formed from the quadratic equation $y = x^2 - 5x + 6$.

The given quadratic equation is $y = x^2 - 5x + 6$.
This is in the standard form $y = ax^2 + bx + c$, where $a=1$, $b=-5$, and $c=6$.

The x-coordinate of the vertex is given by the formula $x = -\frac{b}{2a}$.

$$x = -\frac{-5}{2(1)}$$ $$x = \frac{5}{2}$$

Substitute the x-coordinate ($x = \frac{5}{2}$) back into the quadratic equation $y = x^2 - 5x + 6$.

$$y = \left(\frac{5}{2}\right)^2 - 5\left(\frac{5}{2}\right) + 6$$ $$y = \frac{25}{4} - \frac{25}{2} + 6$$

To combine these terms, find a common denominator, which is 4.

$$y = \frac{25}{4} - \frac{25 \times 2}{2 \times 2} + \frac{6 \times 4}{1 \times 4}$$ $$y = \frac{25}{4} - \frac{50}{4} + \frac{24}{4}$$ $$y = \frac{25 - 50 + 24}{4}$$ $$y = \frac{-25 + 24}{4}$$ $$y = \frac{-1}{4}$$

The vertex of the parabola is $(x, y) = \left(\frac{5}{2}, -\frac{1}{4}\right)$.

The vertex of the parabola is $\boxed{\left(\frac{5}{2}, -\frac{1}{4}\right)}$.