Exponential function - with one point in fraction form Question: determine the exponential function when given 2 points: (-1,2) and (3/2,1/2). The examples I've found are with integers, not fractions, and I'm not sure how/if that effects the answer...
One way to do this problem is using the formula $y=ab^x$. Now if $y=2$ then $x=-1$ so that gives $2=ab^{-1}$. And also if $y=0.5$ then $x=1.5$ which gives $0.5=ab^{1.5}$. When you divide first equation by second, the $a$ cancels and you are left with $4=b^{-2.5}$ from which you can find $b$. This can be written as $\frac{1}{b^{2.5}}=4$ or $b^{2.5}=\frac{1}{4}$. Raising both sides to the power $2.5$ gives $b=\frac{1}{32}$. You can now backsub to find $a$
More generally,
if
$ab^x$
passes through
$(p, q)$
and
$(r, s)$,
then
$ab^p = q$
and
$ab^r = s$.

Dividing these,
$b^{p-r} = \frac{q}{s}$
or
$b
=(q/s)^{1/(p-r)}
$.

Then
$a
=qb^{-p}
=q(q/s)^{-p/(p-r)}
=q^{1-p/(p-r)}s^{p/(p-r)}
=q^{-r/(p-r)}s^{p/(p-r)}
$.

In your case,
$p, q, r, s
=-1, 2, 3/2, 1/2
$.

Then 
$1/(p-r)
=1/(-1-3/2)
=-1/(5/2)
=-2/5
$
so
$b
=(2/(1/2))^{-2/5}
=4^{-2/5}
=2^{-4/5}
$
and
$a
=qb^{-p}
=2(2^{-4/5})^{1}
=2(2^{-4/5})
=2^{1/5}
$.