Question

A telephone company charges a fixed monthly rate plus a rate per minute of usage. The company charges $120 for 100 minutes of usage and $95 for 50 minutes of usage. An equation can be written to show the relationship between the total minutes used (x) and the total monthly charges (y). Which of the following best describes the steps to draw the graph of y against x?

Draw a graph which joins the points (100, 120) and (50, 95) and has a slope = 0.50

Draw a graph which joins the points (100, 120) and (50, 95) and has a slope = 2

Draw a graph which joins the points (120, 100) and (95, 50) and has a slope = 2

Draw a graph which joins the points (120, 100) and (95, 50) and has a slope = 0.50

Answer 1

Draw a graph which joins the points (100, 120) and (50, 95) and has a slope = 0.50

Explanation:

Answer 2

Draw a graph which joins the points (100, 120) and (50, 95) and has a slope = 0.50

Explanation

A graph of y against x means that x is the independent variable and y is the dependent variable. Notice that the charge (y) depends on the number of minutes (x). Therefore, the points that we are going to join to make our graph will have coordinates (x, y)

We know that the company charges $120 for 100 minutes, so x=100 and y=120. Therefore, our first point will be (100, 120)

We also know that the company charges $95 for 50 minutes, so x=95 and y=50. Therefore, our second point will be (50, 95)

Now, to find the slope between our tow points, we are going to use the slope formula:
`m=(y_(2)-y_(1))/(x_(2)-x_(1))`

where

`(x_(1),y_(1))` are the coordinates of the first point

`(x_(2),y_(2))` are the coordinates of the second point

We know from our points that
`x_(1)=100`,
`y_(1)=120`,
`x_(2)=50`, and
`y_(2)=95`. So let's replace those values in our formula to find
`m`

`m=(y_(2)-y_(1))/(x_(2)-x_(1))`

`m=(95-120)/(50-100)`

`m=(-25)/(-50)`

`m=(1)/(2)`

`m=0.50`

Now that we have our slope, we can use the point-slope formula to complete the equation of the line joining our two points

`y-y_(1)=m(x-x_(1))`

`y-120=0.50(x-100)`

`y-120=0.50x-50`

`y=0.50x+70`