Answer
(a) domain $(-\infty,\infty)$, range $(-\infty,\infty)$.
(b) $f(x)=\frac{1}{4}x+\frac{3}{2}$, $f(-2)= 1$
Work Step by Step
(a) Given $x-4y=-6$, we can find its domain as $(-\infty,\infty)$, range as $(-\infty,\infty)$.
(b) We can rewrite the equation as $y=\frac{1}{4}x+\frac{3}{2}$ which represents a function $f(x)=\frac{1}{4}x+\frac{3}{2}$, and we can find $f(-2)=\frac{1}{4}(-2)+\frac{3}{2}=1$
