1. **Stating the problem:** Given pairs of function values, find the linear function $f(x) = mx + b$ that satisfies these values. 2. **Formula and rules:** A linear function is $f(x) = mx + b$ where $m$ is the slope and $b$ is the y-intercept. To find $m$ and $b$, use the points $(x_1, y_1)$ and $(x_2, y_2)$: $$m = \frac{y_2 - y_1}{x_2 - x_1}$$ Then solve for $b$ using $y_1 = mx_1 + b$. 3. **Solve each function:** **i)** Given $f(1) = 2$, $f(-1) = -6$: $$m = \frac{2 - (-6)}{1 - (-1)} = \frac{8}{2} = 4$$ $$2 = 4 \times 1 + b \Rightarrow b = 2 - 4 = -2$$ So, $$f(x) = 4x - 2$$ **ii)** Given $f(1) = 1$, $f(3) = 0$: $$m = \frac{0 - 1}{3 - 1} = \frac{-1}{2} = -\frac{1}{2}$$ $$1 = -\frac{1}{2} \times 1 + b \Rightarrow b = 1 + \frac{1}{2} = \frac{3}{2}$$ So, $$f(x) = -\frac{1}{2}x + \frac{3}{2}$$ **iii)** Given $f(2) = 15$, $f(3) = 35$: $$m = \frac{35 - 15}{3 - 2} = \frac{20}{1} = 20$$ $$15 = 20 \times 2 + b \Rightarrow b = 15 - 40 = -25$$ So, $$f(x) = 20x - 25$$ **iv)** Given $f(3) = 2$, $f(-6) = -1$: $$m = \frac{2 - (-1)}{3 - (-6)} = \frac{3}{9} = \frac{1}{3}$$ $$2 = \frac{1}{3} \times 3 + b \Rightarrow b = 2 - 1 = 1$$ So, $$f(x) = \frac{1}{3}x + 1$$ **v)** Given $f(-2) = 5$, $f(4) = -1$: $$m = \frac{-1 - 5}{4 - (-2)} = \frac{-6}{6} = -1$$ $$5 = -1 \times (-2) + b \Rightarrow b = 5 - 2 = 3$$ So, $$f(x) = -x + 3$$ **vi)** Given $f(-1) = 3$, $f(2) = -6$: $$m = \frac{-6 - 3}{2 - (-1)} = \frac{-9}{3} = -3$$ $$3 = -3 \times (-1) + b \Rightarrow b = 3 - 3 = 0$$ So, $$f(x) = -3x$$