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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

Verified Answer :

matthewdickin

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 :)




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