Answer:
The process below should work.
Step-by-step explanation:
Let's pretend we have these two points we are trying to find an exponential equation for: (-2,6) and (2,1).
Exponential equations are of the form [tex]y=a \cdot b^x[/tex] where we must find [tex]a[/tex] and [tex]b[/tex].
So you enter both points into that equation giving you:
[tex]6=a \cdot b^{-2}[/tex]
[tex]1=a \cdot b^{2}[/tex]
I'm going to divide equation 1 by 2 because if I do the a's will cancel and I could solve or b.
[tex]\frac{6}{1}=\frac{a \cdot b^{-2}}{a \cdot b^2}[/tex]
[tex]6=\frac{b^{-2}}{b^2}[/tex]
By law of exponent, I can rewrite the right hand side:
[tex]6=b^{-2-2}[/tex]
[tex]6=b^{-4}[/tex]
Now do ^(-1/4) on both sides to solve for b:
[tex]6^\frac{-1}{4}=b[/tex]
Now we use one of the equations along with our value for b to find a:
[tex]1=a \cdot b^2[/tex] with [tex]b=6^{\frac{-1}{4}}[/tex]
[tex]1=a \cdot (6^{\frac{-1}{4}})^2[/tex]
Simplify using law of exponents:
[tex]1=a \cdot 6^{-\frac{1}{2}}[/tex]
Multiply both sides by 6^(1/2) to solve for a:
[tex]6^{\frac{1}{2}}=a[/tex]
[tex]y=a \cdot b^x[/tex] with [tex]a=6^{\frac{1}{2}} \text{ and } b=6^{\frac{-1}{4}}[/tex] is:
[tex]y=6^\frac{1}{2} \cdot (6^{\frac{-1}{4})^x[/tex]
We can simplify a smidgen:
[tex]y=6^\frac{1}{2} \cdot (6)^\frac{-x}{4}[/tex]
