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I'm trying to figure out a future value on a graph from an average rate of change.

So this is what I have.

Avg rate of change $= \frac{y_2 - y_1}{x_2 - x_1}$

$= \frac{5.9 - 7.2}{2006 - 1979}$

$= -\frac{1.3}{27}$

$= -0.048$

I've then turned this to scientific notation of $-4.8 \times 10^6 \text{ km}^2 \text{yr}^{-1}$

So I think I have got that portion correct, would like to know if not of course. But my question is to find out what the value of $y$ would be if $x=2015$. How do I calculate further points from what I have already figured out?

Thank you for any advice.

2 Answers 2

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This is a linear projection. It assumes a straight line to project future values of $y$ depending upon the future value of $x$.

You in addition to the slope $m=-0.048$ of the projection line you need to know at least one data point $(x^1,y^1)$. Then to compute the projected value of $y$ from some other value of $x$ you would use the equation

$$ y-y^1=-0.048(x-x^1)$$

which can be written in the form

$$ y=y^1-0.048(x-x^1)$$

Note: When dealing with years it is customary to let the earliest year be year $0$ and count later years as the number of years since the earliest year. For example, if the first year in the data is $1979$ then $x=0$ for $1979$ and for $2006$, $x=2006-1979=27$.

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We have for our $x$ values: $x_1=1979$ and $x_2=2006$.

We have for our $y$ values: $y_1=7.2$ and $y_2=5.9$.

You've evaluated your average rate of change correctly as $m=-\frac{1.3}{27}$.

Now you must find a linear function to extrapolate your data.

Since the equation of a line is of the form $y=mx+c$, you can use any of $(x_1, y_1)$ or $(x_2, y_2)$ to evaluate the value of the constant $c$. We will use the values of $x_1$ and $y_1$.

We use $y_1=mx_1+c$ and obtain $7.2=-\frac{1.3}{27} \times 1979 + c$. Rearrange this to obtain $c \approx 102.485$.

Hence you now have the equation: $$y=-\frac{1.3}{27} x + 102.485 \tag{1}$$ You can now evaluate your value for $y$ by substituting $x=2015$ to obtain $y \approx 5.466$ (3.d.p)

To evaluate further values of $y$ given a value for $x$, just simply substitute your value for $x$ on equation $(1)$.