14. The monthly cost of driving a car depends on the number of miles driven. Lynn found that in May it cost her $380 to drive 480 mi and in June it cost her $460 to drive 800 mi.
a) Express the monthly cost C as a function of the distance driven d, assuming that a linear relationship gives a suitable model.
b) Use part (a) to predict the cost of driving 1500 miles per month.
c) Draw the graph of the linear function. What does the slope represent?
d) What does the y-intercept represent?
e) Why does a linear function give a suitable model in this situation?
I assume you know how to handle equations of a straight line of the form y = mx + b if you are given two ordered pairs of the form (x,y)
So interpret your data as two ordered pairs of the form (m,c), m for miles c for cost.
so you would have (480,380) and (800,460)
so proceed by finding the slope etc
just like you would for the more familiar points.
the rest of the questions are of a type very common for the y = mx + b notation.
a) To find a linear relationship between the monthly cost C and the distance driven d, we need to determine the equation of a line in slope-intercept form (y = mx + b), where y represents the dependent variable (monthly cost) and x represents the independent variable (distance driven).
Using the given data, we can find the slope (m) and the y-intercept (b).
First, we calculate the slope:
m = (change in y) / (change in x)
= (460 - 380) / (800 - 480)
= 80 / 320
= 0.25
Next, we substitute the slope and one point (480 mi and $380) into the slope-intercept form to find the y-intercept:
380 = 0.25 * 480 + b
b = 380 - 0.25 * 480
b = 380 - 120
b = 260
Therefore, the monthly cost C can be expressed as the linear function:
C = 0.25d + 260
b) To predict the cost of driving 1500 miles per month, we substitute d = 1500 into the linear function we found in part (a):
C = 0.25 * 1500 + 260
C = 375 + 260
C = $635
The predicted cost of driving 1500 miles per month is $635.
c) To draw the graph of the linear function C = 0.25d + 260, we plot a graph on a coordinate plane. The x-axis represents the distance driven (d), and the y-axis represents the monthly cost (C). The graph will be a straight line with a positive slope of 0.25 and a y-intercept of 260.
d) In the context of this problem, the y-intercept (b = 260) represents the fixed monthly cost of driving a car, regardless of the distance driven. It includes costs such as insurance, registration fees, or car payments that do not depend on the number of miles driven.
e) A linear function gives a suitable model in this situation because it assumes a constant rate of change, meaning the monthly cost increases at a constant rate for every additional mile driven. The given data points lie on a straight line, indicating a linear relationship between the monthly cost and the distance driven.
a) To express the monthly cost C as a function of the distance driven d, assuming a linear relationship, we can use the formula for a straight line: y = mx + b.
Let's assign C as the dependent variable (y) representing the cost, and d as the independent variable (x) representing the distance driven. Using the given data points (480 mi, $380) and (800 mi, $460), we can calculate the slope (m) and y-intercept (b) of the linear function.
First, we calculate the slope (m):
m = (y2 - y1) / (x2 - x1)
= ($460 - $380) / (800 mi - 480 mi)
= $80 / 320 mi
= $0.25/mi
Next, we can substitute one of the data points and the slope into the equation y = mx + b to solve for the y-intercept (b). Let's use the point (480 mi, $380):
$380 = ($0.25/mi)(480 mi) + b
$380 = $120 + b
b = $380 - $120
b = $260
Thus, the equation representing the monthly cost C as a function of the distance driven d is:
C = $0.25d + $260
b) To predict the cost of driving 1500 miles per month, we can substitute the value of d = 1500 into the equation we derived in part (a):
C = $0.25(1500) + $260
C = $375 + $260
C = $635
Therefore, the predicted cost of driving 1500 miles per month is $635.
c) The graph of the linear function C = $0.25d + $260 will show a straight line. The x-axis represents the distance driven (d), and the y-axis represents the monthly cost (C). The slope of the line is 0.25, which means that for each additional mile driven, the monthly cost increases by $0.25.
d) The y-intercept of the graph is at the point (0, $260). In this context, the y-intercept represents the fixed monthly cost of driving a car, regardless of the number of miles driven. It includes expenses like insurance, maintenance, registration, etc. In this case, the y-intercept of $260 represents Lynn's fixed monthly cost.
e) A linear function gives a suitable model in this situation because the relationship between the monthly cost and the distance driven is approximately linear. It fits the data points provided (480 mi, $380) and (800 mi, $460), which suggests a linear trend. Additionally, it is reasonable to assume that other factors affecting the monthly cost, such as fuel consumption and wear and tear, increase proportionally with the distance driven, supporting a linear relationship.