Question

Help with the last (q 3a and b)

Answer 1

$\displaystyle\bf\\a)\\\\(-2 \ ; \ 0) \ , \ (-1 \ ; \ 1) \ , \ (0 \ ; \ 2) \ , \ (1 \ ; \ 3) \ , \ (2 \ ; \ 4)\\\\y=kx+b\\\\(0 \ ; \ 2) \ \ \Rightarrow \ \ x=0 \ \ , \ \ y=2\\\\2=k\cdot 0+b\\\\b=2\\\\(-2 \ ; \ 0) \ \ \Rightarrow \ \ x=-2 \ \ , \ \ y=0\\\\0=k\cdot(-2)+2\\\\-2k+2=0\\\\-2k=-2\\\\k=1\\\\\\\boxed{\boxed{y=x+2}}$

$\displaystyle\bf\\b)\\\\(4 \ ; \ -1) \ , \ (3 \ ; \ 0) \ , \ (2 \ ; \ 1) \ , \ (1 \ ; \ 2) \ , \ (0 \ ; \ 3)\\\\y=kx+b\\\\(0 \ ; \ 3) \ \ \Rightarrow \ \ x=0 \ \ , \ \ y=3\\\\3=k\cdot 0+b\\\\b=3\\\\(3 \ ; \ 0) \ \ \Rightarrow \ \ x=3 \ \ , \ \ y=0\\\\0=k\cdot 3+2\\\\3k+2=0\\\\3k=-2\\\\k=-\frac{2}{3} \\\\\\\boxed{\boxed{y=-\frac{2}{3} x+3}}$