Question 1181009
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Duplicate -- answer repeated below<br>
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f(x) = a(1/2)^b(x+2)+k<br>
NOTE: To make the form of the function absolutely clear, the whole exponent "b(x+2)" should be in parentheses: f(x) = a(1/2)^(b(x+2))+k<br>
{{{f(x) = a(1/2)^(b(x+2))+k}}}<br>
Use the two given points and the horizontal asymptote to get three equations that can be solved to determine the constants a, b, and k.<br>
(1) horizontal asymptote:
For large values of x, the decaying exponential goes to zero, making the function close to f(x)=k.  Since the asymptote is y=4, we have k=4.<br>
(2) f(-2) = 1:
{{{f(-2) = a(1/2)^b(-2+2)+4}}}
{{{1 = a(1/2)^0+4}}}
{{{1 = a+4}}}
{{{a=-3}}}<br>
(3) f(-3) = -8:
{{{f(-3) = -3(1/2)^b(-3+2)+4}}}
{{{-8 = -3(1/2)^b(-1)+4}}}
{{{-12 = -3(2^b)}}}
{{{2^b=4}}}
{{{b=2}}}<br>
a=-3; b=2; k=4<br>
The function is<br>
{{{f(x) = -3(1/2)^(2(x+2))+4}}}<br>
A graph:<br>
{{{graph(400,400,-4,4,-6,6,-3(1/2)^(2(x+2))+4)}}}<br>