Question

In 2000, Canada produced 3.0 million new motor vehicles. In 2006, it produced 2.5 million. Write the linear equation (in slope-intercept form)that models the number of vehicles
a.) -.071 x+9
b.) -.083 x+3
c.) -.074 x+4
d.) -.065 x+0
e.) -.068 x+2
f.) -.058 x+12
g.) -.062 x+10
h.) -.079 x+5

Answer 1

To begin, you are given two point that this function would intercept: (0,3) where the 0 is the original year (2000) and 3 is the amount of million cars sold. The next point would be (6,2.5) Where the 6 is the 6 years later and the 2.5 is the 2.5 million cars sold. Since we have 2 points, we can work out the gradient by finding the rate of change in y and dividing by the rate of change in x:
$\frac{y_{1}- y_{2} }{x_{1}- x_{2} }$

We then substitute in the points:
$x_{1} =3$
$y_{1} =0$
$x_{2} =2.5$
$y_{2} =6$

$\frac{3-2.5}{0-6}$
$=\frac{0.5}{-6}$
$=-.083$

Because we started on the y-intercept of (0,3), we already have our intercept value of 3. If we did not start with the y-intercept given, we could substitute in a point to solve for it:

Using point (2.5,6):

$2.5=-0.083*6+c$
$2.5=-0.5+c$
$2.5+0.5=c$
$c=3$

Hope this helps :)