## Functions Modeling Change: A Preparation for Calculus, 5th Edition

(a) $g(0)=6$ (b) $x=2,3$
(a) To get $g(0)$, use $x=0$ in the expression to get $g(0)=0^2-5(0)+6=6$. (b) Using $g(x)=0$, the following equation is generated: $$x^2-5x+6=0$$Using the quadratic formula $$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ with $a=1$, $b=-5$, and $c=6$, the solutions are $$x=\frac{5 \pm \sqrt{(-5)^2-4(1)(6)}}{2}=\frac{5 \pm \sqrt{25-24}}{2}=\frac{5\pm1}{2}$$ $$x={2,3}$$