Question

Hello please help me with my homework ! I would really appreciate it! :)))

Answer 1

At graph of Number 4: Let's first determine the equation of each line.

LINE 1: Let's use point A: (x1,y1) = (-5,-5) and point B: (x2,y2) = (-3,-4).

Getting the slope (m):

`\text{ m = }\frac{y_2-y_1_{}}{x_2-x_1}`
`\text{ m = }\frac{-4\text{ - (-5)}}{-3\text{ - (-5)}}\text{ = }\frac{-4\text{ + 5}}{-3\text{ + 5}}\text{ = }(1)/(2)`

Getting the y-intercept (b): Use m = 1/2 and x,y = -5,-5

`\text{ y = mx + b}`
`-5\text{ = (}(1)/(2))(-5)\text{ + b}`
`\text{ -5 = }(-5)/(2)\text{ + b }\rightarrow\text{ b = -5 + }(5)/(2)`
`\text{b = }\frac{\text{-10}}{2}\text{ + }(5)/(2)\text{ = }\frac{-10\text{ + 5}}{2}`
`\text{ b = -}(5)/(2)`

Let's complete the equation: Substitute m and b in y = mx + b.

`\text{ y = mx + b}`
`\text{ y = (}(1)/(2))x\text{ + (-}(5)/(2))`
`\text{ y = }(1)/(2)x\text{ - }(5)/(2)\text{ ; f(x) = }(1)/(2)x\text{ - }(5)/(2)`

LINE 2: Let's use point A: (x1,y1) = (-3,-2) and point B: (x2,y2) = (1,0).

Getting the slope (m):

`\text{ m = }(y_2-y_1)/(x_2-x_1)`
`\text{ m = }(0-(-2))/(1-(-3))=\frac{0\text{ + 2}}{1\text{ + 3}}\text{ = }(2)/(4)\text{ = }(1)/(2)`

Getting the y-intercept (b): Use m = 1/2 and x,y = 1,0

`\text{ y = mx + b}`
`\text{ 0 = (}(1)/(2))(1)\text{ + b}`
`\text{ 0 = }(1)/(2)\text{ + b}`
`\text{ b = -}(1)/(2)`

Let's complete the equation: Substitute m and b in y = mx + b.

`\text{ y = mx + b}`
`\text{ y = (}(1)/(2))x\text{ + }(-(1)/(2))`
`\text{ y = }(1)/(2)x\text{ - }(1)/(2)\text{ ; f(x) = }(1)/(2)x\text{ - }(1)/(2)`

LINE 3: Let's use point A: (x1,y1) = (1,2) and point B: (x2,y2) = (3,3).

Getting the slope (m):