Based on the information you've provided, the equation for your exponential decay function appears to be in the form:
\[ y = a \cdot b^x \]
where \( a \) is the initial value, \( b \) is the decay factor, and \( x \) is the independent variable.
From the point (0, 3), we see that when \( x = 0 \), \( y = 3 \). This indicates that the initial value \( a \) is 3.
The equation you've written seems to be a bit incomplete, but it can be inferred to perhaps be:
\[ y = 3(0.75)^x \]
In this equation:
- The horizontal asymptote represents the value that \( y \) approaches as \( x \) goes to infinity. For exponential decay functions of this form (where \( 0 < b < 1 \)), the horizontal asymptote is usually at \( y = 0 \).
Therefore, for the function modeled by \( y = 3(0.75)^x \), the horizontal asymptote is:
\[ y = 0 \]
So, in summary, the equation for the horizontal asymptote of the sequence modeled by your graph is:
Horizontal Asymptote: \( y = 0 \)