The problem asks us to identify the coefficients $a$, $b$, and $c$ of quadratic functions and then calculate the x-coordinate of the vertex ($x_s$) for each function. The general form of a quadratic function is $f(x) = ax^2 + bx + c$. The x-coordinate of the vertex (or the axis of symmetry) is given by the formula $x_s = -\frac{b}{2a}$.

1. Problem Description

The problem asks us to identify the coefficients $a$, $b$, and $c$ of quadratic functions and then calculate the x-coordinate of the vertex ($x_s$) for each function. The general form of a quadratic function is $f(x) = ax^2 + bx + c$. The x-coordinate of the vertex (or the axis of symmetry) is given by the formula $x_s = -\frac{b}{2a}$.

2. Solution Steps

1) $f(x) = x^2 - 5x + 6$.

Here, $a = 1$, $b = -5$, and $c = 6$.

$x_s = -\frac{b}{2a} = -\frac{-5}{2(1)} = \frac{5}{2} = 2.5$.

2) $g(x) = -2x^2 + 12x - 18$.

Here, $a = -2$, $b = 12$, and $c = -18$.

$x_s = -\frac{b}{2a} = -\frac{12}{2(-2)} = -\frac{12}{-4} = 3$.

3) $h(x) = 45 + 5x^2 - 30x$. Rewriting in standard form: $h(x) = 5x^2 - 30x + 45$.

Here, $a = 5$, $b = -30$, and $c = 45$.

$x_s = -\frac{b}{2a} = -\frac{-30}{2(5)} = \frac{30}{10} = 3$.

4) $l(x) = -3x + 4 + 0.5x^2$. Rewriting in standard form: $l(x) = 0.5x^2 - 3x + 4$.

Here, $a = 0.5$, $b = -3$, and $c = 4$.

$x_s = -\frac{b}{2a} = -\frac{-3}{2(0.5)} = -\frac{-3}{1} = 3$.

3. Final Answer

1) $a = 1$, $b = -5$, $c = 6$, $x_s = 2.5$

2) $a = -2$, $b = 12$, $c = -18$, $x_s = 3$

3) $a = 5$, $b = -30$, $c = 45$, $x_s = 3$

4) $a = 0.5$, $b = -3$, $c = 4$, $x_s = 3$