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

Sorry I am not 100% sure where to begin with this. I think finding the slope would be helpful, yes?
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 $$f(x) = 1965.9\exp(0.1398x)$$