Linear rule from a function table (ACT). Game on.

The outputs decrease by $3$ each time $x$ increases by $1\textsf{,}$ so the constant rate is $-3\textsf{.}$ When $x = 0\textsf{,}$ $g(0) = 15\textsf{,}$ so $g(x) = -3x + 15\textsf{.}$

The height increases by $3$ centimeters each week, and the starting height is $8$ centimeters. So $k(x) = 3x + 8\textsf{.}$

The outputs increase by $3$ each time $x$ increases by $1\textsf{,}$ so the constant rate is $3\textsf{.}$ When $x = 0\textsf{,}$ $s(0) = 7\textsf{,}$ so $s(x) = 3x + 7\textsf{.}$

The temperature decreases by $3$ degrees each hour, and at noon the temperature is $72\textsf{.}$ So $w(x) = -3x + 72\textsf{.}$

The outputs increase by $3$ each time $x$ increases by $1\textsf{,}$ so the constant rate is $3\textsf{.}$ Using the point $(1, 7)$ gives $3(1) + b = 7\textsf{,}$ so $b = 4\textsf{.}$ Therefore $h(x) = 3x + 4\textsf{.}$

From $x = 0$ to $x = 2\textsf{,}$ $q(x)$ increases by $1\textsf{,}$ so the constant rate is $\dfrac{1}{2}\textsf{.}$ When $x = 0\textsf{,}$ $q(0) = 5\textsf{,}$ so $q(x) = \dfrac{1}{2}x + 5\textsf{.}$

The outputs decrease by $4$ each time $x$ increases by $1\textsf{,}$ so the constant rate is $-4\textsf{.}$ Using the point $(1, 9)$ gives $-4(1) + b = 9\textsf{,}$ so $b = 13\textsf{.}$ Therefore $r(x) = -4x + 13\textsf{.}$

From $x = 2$ to $x = 4\textsf{,}$ $n(x)$ increases by $4\textsf{,}$ so the constant rate is $\dfrac{4}{2} = 2\textsf{.}$ Using the point $(2, 11)$ gives $2(2) + b = 11\textsf{,}$ so $b = 7\textsf{.}$ Therefore $n(x) = 2x + 7\textsf{.}$