<PRE><B>Question 1138844</B>
&lt;br&gt;
The equation you show for the temperature is not correct:&lt;br&gt;
{{{T(t)=80(1/2)^t/30+18}}}&lt;br&gt;
Assuming you know something about how a hot liquid cools, you know this is not the right equation.  You should know that parentheses are required:&lt;br&gt;
{{{T(t)=80(1/2)^(t/30)+18}}}&lt;br&gt;
As with any function, the y-intercept is the value when x=0.  For this equation,&lt;br&gt;
{{{T(0) = 80(1/2)^(0/30)+18 = 80(1)+18 = 98}}}&lt;br&gt;
The horizontal asymptote is the value of the function when t gets very large.  For this equation, since 1/2 to a very large power approaches 0,&lt;br&gt;
{{{T(infinity) = 0+18 = 18}}}&lt;br&gt;
In this problem, it should be clear that the y-intercept 98 is the initial temperature of the liquid (in °C).&lt;br&gt;
The horizontal asymptote is 18, which is the temperature in °C of the room.  That makes sense; a hot liquid left in a room for a very long time will eventually have a temperature that is the same as the temperature of the room.&lt;br&gt;
Note that, while the problem doesn&#39;t ask anything about it, the exponent &quot;t/30&quot; means that the difference between the temperature of the liquid and the room temperature gets cut in half every 30 units of time (presumably minutes).</PRE>