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Sorry I am not 100% sure where to begin with this. I think finding the slope would be helpful, yes?

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    @EuYu I believe it's an exponential funciton2012-12-10

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Without more information, this is only one possible solution. Let us suppose the curve is modelled by an exponential $f(x) = a\exp(bx)$ Then we have $f(2) = a\exp(2b) = 2600,\ \ \ f(7)=a\exp(7b)=5230$ Solving for the coefficients gives us $\exp(5b) = \frac{5230}{2600} \implies b = \frac{1}{5}\ln\left(\frac{523}{260}\right)\approx 0.1398$ Similarly, we find that $a$ is given by $a = \frac{2600}{\exp(2b)}=1965.9$ This gives $f(x) = 1965.9\exp(0.1398x)$