# Part A: Marguerite rented a truck at $125 for 2 days. If she rents the same truck for 5 days, she has to pay a total rent

###### Question:

Part B: Write the equation obtained in Part A using function notation. (2 points)

Part C: Describe the steps to graph the equation obtained above on the coordinate axes. Mention the labels on the axes and the intervals. (4 points)

20 points if right you will get and p. s name the parts so i know what to right

## Answers

(2,125)(5,275)

slope = (275 - 125) / (5 - 2) = 150/3 = 50

y = mx + b

slope(m) = 50

use either of ur points...(2,125)...x = 2 and y = 125

now we sub and find b, the y int

125 = 50(2) + b

125 = 100 + b

125 - 100 = b

25 = b

so ur equation is : y = 50x + 25...but we need it in standard form

y = 50x + 25

-50x + y = 25

50x - y = -25 <== ur equation

62.50y × 5x because she spends $62.50 a day to rent the truck.

PART A:

To obtain an equation in the standard form to represent the total rent (y) that Marguerite has to pay for renting the truck for x days, we use the fomula for the equation of a straight line.

Recall that the equation of a straight line passing through points [tex](x_1,\ y_1)[/tex] and points [tex](x_2,\ y_2)[/tex] is given by

[tex]\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}[/tex]

Given that Marguerite rented a truck at $125 for 2 days. If she rents the same truck for 5 days, she has to pay a total rent of $275.

This means that the lne modelling this situation crosses points (2, 125) and (5, 275).

The equation modelling the total rent (y) that Marguerite has to pay for renting the truck for x days is given by,

[tex]\frac{y-125}{x-2} = \frac{275-125}{5-2} = \frac{150}{3} =50 \\ \\ y-125=50(x-2)=50x-100 \\ \\ y=50x+25[/tex]

Writing the equation in standard form, we have

50x - y = -25

PART B:

Writing the function using function notation means making 'y' the subject of the formula and then replacing the 'y' with 'f(x)'.

Recall that from part A, we have that the equation for the total rent (y) that Marguerite has to pay for renting the truck for x days is given by y = 50x + 25.

Writing the equation using the function notation, we have

f(x) = 50x + 25

PART C:

To graph the function, we label the x-axis 'number of days' and label the y-axis 'total rent'. the x-axis is numbered using the intervals of 1 while the y-axis is labelled using the intervals of 50.

The points (2, 125) and (5, 275) are marked on the coordinate axis and a straight line is drawn to pass through the two points.

Part A: To get an equation into standard form to represent the total amount rented (y) that Marguerite has to pay for renting the truck for x amount of days, we use the formula for the equation of a straight line.

Remember that the equation of a straight line passing through points is ( x_{1} , y_{1} ) and the points ( x_{2} , y_{2} ) is given by

y - y_{1} / x - x_{1} = y - y_{2} / x - x_{2}

Knowing that Marguerite rented a truck at $125 for 2 days, we know if she rents the exact same truck for 5 days, she has to pay a total of $275 for the rent.

This means that the line modeling this situation crosses points at (2, 125) and (5, 275).

The equation modeling the total rent (y) that Marguerite has to pay for renting the truck for x days is given by

y - 125 / x - 2 = 275 - 125 / 5 - 2 = 150 / 3 = 50

But if you are writing the equation in standard form it would be

50x - y = -25

Part B:

When writing the function using function notation it means you are making y the subject of the formula and then replacing the y with f(x).

If you remember that from part A, we have that the equation for the total rent which is y that Marguerite has to pay for renting the truck for x amount of days is given by

y = 50x + 25.

Writing the equation using the function notation would give us this

f(x) = 50x + 25

Part C:

To graph the function, we name the x-axis the number of days and name the y-axis total rent. The x-axis is numbered using the intervals of 1 while the y-axis is numbered using the intervals of 50.

The points of (2,125) and of (5,275) are marked on the coordinate axis and a straight line is drawn to pass through these two points.

Think in terms of a linear function whose graph (a straight line) goes thru the 2 points given:

(2 days, $125) and (5 days, $275 ).

Find the equation of this line.

$275-$125 $150

The slope is m = = = $50/day

5 days - 2 days 3 days

Use the slope-intercept form of the equation of a str. line next:

y=mx + b becomes $275 = ($50/day)(5 days) + b.

Solving for b: $275 - $250 = b = $25

Thus, the equation of this line is y = ($50/day)x + $25, where that $25 is the upfront charge (y-intercept).

We must change this into standard form. To do this, group all the terms on one side. I choose to do that on the right side, so that the x term will be + .

($50/day)x - 1y + $25 = 0 is the equation of this line in std. form.

Part B: writing this equation in function notation:

y = ($50/day)x + $25 becomes f(x) = ($50/day)x + $25

Part C: To graph this equation, plot the two points on the graph:

(2, $125) and (5, $275) ... and draw a straight line thru them.

Alternatively, let x = 0 in f(x) = ($50/day)x + $25 to find the y-intercept; it is (0, $25).

Next, let f(x) = y = 0 = ($50/day)x + $25 to find the x-intercept:

-$25

50x + 25 = 0 => x = = (-1/2) day

$50/day

The x-intercept is ([-1/2] day, $0).

Of course, x must be zero or greater, as a negative number of days would be meaningless.