# How do you find the exponential function that contains (0,-1)(-1,-3) and (-2,-9) and has an x-axis as an asyptote?

Jun 6, 2015

It's easy to see that $f \left(x\right) = - {3}^{-} x$ fits the bill, but how would you find the answer in general?

Suppose $f \left(x\right) = k {a}^{-} x$ where $k \in \mathbb{R}$ and $a \in \mathbb{R}$, $a > 0$. This will have the $x$ axis as an asymptote, regardless of the values of the constants $k$ and $a$.

Then if $f \left(x\right) > 0$:

$\log \left(f \left(x\right)\right) = \log \left(k {a}^{-} x\right) = \log \left(k\right) - x \log \left(a\right)$

or if $f \left(x\right) < 0$:

$\log \left(- f \left(x\right)\right) = \log \left(- k {a}^{-} x\right) = \log \left(- k\right) - x \log \left(a\right)$

In either case,

$\log \left(\left\mid f \left(x\right) \right\mid\right) = \log \left(\left\mid k \right\mid\right) - x \log \left(a\right)$

$= \left(- \log \left(a\right)\right) x + \log \left(\left\mid k \right\mid\right)$

This is in the form of the equation of a line, with slope $- \log \left(a\right)$ and intercept $\log \left(\left\mid k \right\mid\right)$.

Let's try applying this to our given points:

$\left({x}_{1} , {y}_{1}\right) = \left(0 , - 1\right)$
$\left({x}_{2} , {y}_{2}\right) = \left(- 1 , - 3\right)$
$\left({x}_{3} , {y}_{3}\right) = \left(- 2 , - 9\right)$

$\left({x}_{1} , \log \left(\left\mid {y}_{1} \right\mid\right)\right) = \left(0 , \log \left(1\right)\right) = \left(0 , 0\right)$
$\left({x}_{2} , \log \left(\left\mid {y}_{2} \right\mid\right)\right) = \left(- 1 , \log \left(3\right)\right)$
$\left({x}_{3} , \log \left(\left\mid {y}_{3} \right\mid\right)\right) = \left(- 2 , \log \left(9\right)\right) = \left(- 2 , 2 \log \left(3\right)\right)$

These three points, lie along a line of slope $- \log \left(3\right)$ allowing us to deduce that $a = 3$

The intercept, $\log \left(\left\mid k \right\mid\right) = 0$ so $k = \pm 1$

Since $f \left(x\right) < 0$, $k < 0$ so $k = - 1$ and $f \left(x\right) = - {3}^{-} x$