$\"\"$ \

The function is $\"h(x)=(0.2)^{x+2}\"$ .

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Make the table of values to find ordered pairs that satisfy the function.

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Choose values for $\"x\"$ and find the corresponding values for $\"y\"$ .

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$\"x\"$	$\"h(x)=(0.2)^{x+2}\"$	$\"(x,$
$\"-3.5\"$	$\"f(-3.5)=(0.2)^{-3.5+2}=11.18\"$	$\"(-3.5,$
$\"-3\"$	$\"f(-3)=(0.2)^{-3+2}=5\"$	$\"(-3,$
$\"-2\"$	$\"f(-2)=(0.2)^{-2+2}=1\"$	$\"(-2,$
$\"-1\"$	$\"f(-1)=(0.2)^{-1+2}=0.2\"$	$\"(-1,$
$\"0\"$	$\"f(0)=(0.2)^{0+2}=0.04\"$	$\"(0,$
$\"1\"$	$\"f(1)=(0.2)^{1+2}=0.008\"$	$\"(1,$
$\"2\"$	$\"f(2)=(0.2)^{2+2}=0.001\"$	$\"(2,$
$\"3\"$	$\"f(3)=(0.2)^{3+2}=0.0003\"$	$\"(3,$

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$\"\"$

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Graph:

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1. Draw a coordinate plane.

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2. Plot the coordinate points.

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3. Then sketch the graph, connecting the points with a smooth curve.

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$\"\"$

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Observe the above graph :

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Domain of the function is all real numbers.

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Range of the function is $\"(0,$ .

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The function does not have any $\"x\"$ - intercept.

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$\"y\"$ - intercepts is $\"0.04\"$ .

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The line $\"y=L.\"$ is the horizontal asymptote of the curve $\"y=f(x)\"$ .

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if either $\"\\lim_{x$ or $\"\\lim_{x$ .

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$\"\\lim_{x$

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$\"\\\\=(0.2)^{-\\infty+\\2}$

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$\"=0\"$ .

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$\"y=0\"$ is the horizontal asymptote of the function.

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Observe the above graph vertical asymptote does not exist.

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End behavior : $\"\\lim_{x\\rightarrow$ and $\"\\lim_{x\\rightarrow$ .

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The function is continuous for all real numbers.

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Decreasing on the interval : $\"\\left$ .

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$\"\"$

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The graph of the function $\"h(x)=(0.2)^{x+2}\"$ is :

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$\"\"$

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Domain of the function is all real numbers.

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Range of the function is $\"(0,$ .

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$\"y\"$ - intercepts is $\"0.04\"$ .

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Horizontal asymptote of the function is $\"y=0\"$ .

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End behavior : $\"\\lim_{x\\rightarrow$ and $\"\\lim_{x\\rightarrow$ ..

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The function is continuous for all real numbers.

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The function $\"h(x)\"$ decreasing over the interval : $\"\\left$ .