DSC1520 – Quantitative Modelling
Assignment 01, Semester 02
Unique Number: 837043
Due Date: 10 August 2022
Question 1
Find the equation of the straight line passing through the points (4; 2) and (2; 4).
𝑦 = 𝑚𝑥 + 𝑐
𝑠𝑙𝑜𝑝𝑒, 𝑚 = 𝑦 2−𝑦1=42−−24=−1
𝑥2−𝑥1
𝑦 = −1𝑥 + 𝑐
Use the point (4; 2) to find the 𝑦-intercept, 𝑐
2 = −1(4) + 𝑐
2 = −4 + 𝑐
2+4=𝑐=6
Note: Using the other point (2; 4) will yield the same result.
The equation of the line is therefore
𝑦 = −1𝑥 + 6 which simplifies to
𝑦 = −𝑥 + 6 Option [1]
Graph of 𝑦=−𝑥+1 passes through the two given points
, Question 2
A R200 000 car depreciates linearly to R40 000 in 8 years’ time. Derive a linear equation for the
value of the car after 𝑥 years with .
Two points can be drawn from the given information: the initial value of R200 000 at time 𝑥 = 0
years and the final value of R40 000 after 𝑥 = 8 years. These two points are therefore (0; 200
000) and (8; 40 000).
𝑦 = 𝑚𝑥 + 𝑐
𝑠𝑙𝑜𝑝𝑒, 𝑚 = =−20 000
𝑦 = −20 000𝑥 + 𝑐
Use (8; 40 000) to find 𝑐, the 𝑦-intercept
40 000 = −20 000(8) + 𝑐
40 000 = −160 000 + 𝑐
40 000 + 160 000 = 𝑐 = 200 000
Note: Using the other point (0; 200 000) to find 𝑐 will yield the same result.
The equation is therefore 𝑦 = −20 000𝑥 + 200 000 Option [1]
Assignment 01, Semester 02
Unique Number: 837043
Due Date: 10 August 2022
Question 1
Find the equation of the straight line passing through the points (4; 2) and (2; 4).
𝑦 = 𝑚𝑥 + 𝑐
𝑠𝑙𝑜𝑝𝑒, 𝑚 = 𝑦 2−𝑦1=42−−24=−1
𝑥2−𝑥1
𝑦 = −1𝑥 + 𝑐
Use the point (4; 2) to find the 𝑦-intercept, 𝑐
2 = −1(4) + 𝑐
2 = −4 + 𝑐
2+4=𝑐=6
Note: Using the other point (2; 4) will yield the same result.
The equation of the line is therefore
𝑦 = −1𝑥 + 6 which simplifies to
𝑦 = −𝑥 + 6 Option [1]
Graph of 𝑦=−𝑥+1 passes through the two given points
, Question 2
A R200 000 car depreciates linearly to R40 000 in 8 years’ time. Derive a linear equation for the
value of the car after 𝑥 years with .
Two points can be drawn from the given information: the initial value of R200 000 at time 𝑥 = 0
years and the final value of R40 000 after 𝑥 = 8 years. These two points are therefore (0; 200
000) and (8; 40 000).
𝑦 = 𝑚𝑥 + 𝑐
𝑠𝑙𝑜𝑝𝑒, 𝑚 = =−20 000
𝑦 = −20 000𝑥 + 𝑐
Use (8; 40 000) to find 𝑐, the 𝑦-intercept
40 000 = −20 000(8) + 𝑐
40 000 = −160 000 + 𝑐
40 000 + 160 000 = 𝑐 = 200 000
Note: Using the other point (0; 200 000) to find 𝑐 will yield the same result.
The equation is therefore 𝑦 = −20 000𝑥 + 200 000 Option [1]
