Question

How do you graph \[y = {\left( {\dfrac{1}{3}} \right)^x} + 1?\]

Answer 1

Hint: First, understand the given equation. Apply the \[x\]value in the given equation and find the corresponding \[y\] value. We get a coordinate point\[\left( {a,b} \right)\] on the given equation. Now plot the point in the graph and connect it. The equation to find the graph is a strictly decreasing and unbounded graph. The graph lies above the \[x\] axis.

Complete answer:
Given equation is \[y = {\left( {\dfrac{1}{3}} \right)^x} + 1\]
This equation is exponential. In this equation to we draw a graph.
Assume that,
\[x = - 3\]apply the equation \[y\]
\[y = {\left( {\dfrac{1}{3}} \right)^{ - 3}} + 1\]
The first\[\dfrac{1}{3}\] is a rational number. That number power value is a negative number generally \[{\left( {\dfrac{1}{a}} \right)^{ - b}}\]
So that we will give power separately numerator and denominator. It means
\[
= {\left( {\dfrac{1}{3}} \right)^{ - 3}} + 1 \\
= \dfrac{{{{\left( 1 \right)}^{ - 3}}}}{{{{\left( 3 \right)}^{ - 3}}}} + 1 \\
\]
In \[1\] power any non-zero number the value is
\[
{\left( 1 \right)^a},\,a \ne 0 \\
{\left( 1 \right)^a} = 1 \\
\]
So that \[{\left( 1 \right)^{ - 1}} = 1\] and denominator \[{\left( 3 \right)^{ - 3}},\]
A non-zero number has a power value that is a negative number. We change to that nonzero in the denominator.
It means \[{\left( a \right)^{ - b}}\] is a change to \[\dfrac{1}{{{a^b}}}\]
So that \[{\left( 3 \right)^{ - 3}} = \dfrac{1}{{{3^3}}}\]
Now the negative power value is converted to a positive power value. Now easily get the three power three is \[27\]
So that,
\[
= \dfrac{1}{{{3^3}}} \\
= \dfrac{1}{{27}} \\
\]
and \[x = - 3\]we get \[y = 27 + 1\]
because apply \[1\] over \[\dfrac{1}{{27}}\] is \[27\]
\[
= 1 + 27 \\
y = 28 \\
\]
So the point is \[\left( { - 3,28} \right)\]
\[x = - 2\] apply the equation \[y\]
\[
y = {\left( {\dfrac{1}{3}} \right)^{ - 2}} + 1 \\
= {\left( {\dfrac{1}{3}} \right)^{ - 2}} + 1 \\
= \dfrac{{{{\left( 1 \right)}^{ - 2}}}}{{{{\left( 3 \right)}^{ - 2}}}} + 1 \\
\]
So that \[{\left( 3 \right)^{ - 2}} = \dfrac{1}{{{3^2}}}\]
Now the negative power value is converted to a positive power value. Now easily get the three power two is \[9\]
So that,
\[
= \dfrac{1}{{{3^2}}} \\
= \dfrac{1}{9} \\
\]
and \[x = - 2\]apply to \[y = 9 + 1\]
\[y = 10\]
So the point is \[\left( { - 2,10} \right)\]
\[x = - 1\] apply equation \[y\]
\[
y = {\left( {\dfrac{1}{3}} \right)^{ - 1}} + 1 \\
= {\left( {\dfrac{1}{3}} \right)^{ - 1}} + 1 \\
= \dfrac{{{{\left( 1 \right)}^{ - 1}}}}{{{{\left( 3 \right)}^{ - 1}}}} + 1 \\
\]
So that \[{\left( 3 \right)^{ - 1}} = \dfrac{1}{{{3^1}}}\]
Now the negative power value is converted to a positive power value. Now easily get the three power one is \[3\]
So that,
\[
= \dfrac{1}{{{3^1}}} \\
= \dfrac{1}{3} \\
\]
and \[x = - 1\]we get \[y = 3 + 1\]
\[y = 4\]
So the point is \[\left( { - 1,4} \right)\]
\[x = 0\] apply equation \[y\]
\[
y = {\left( {\dfrac{1}{3}} \right)^0} + 1 \\
= {\left( {\dfrac{1}{3}} \right)^0} + 1 \\
= \dfrac{{{{\left( 1 \right)}^0}}}{{{{\left( 3 \right)}^0}}} + 1 \\
\]
So that \[{\left( 3 \right)^0} = \dfrac{1}{{{3^0}}}\]
Now easily get any constant number power zero is equal to one.
So that,
\[
\dfrac{{{{\left( 1 \right)}^0}}}{{{{\left( 3 \right)}^0}}} = \dfrac{1}{1} \\
= 1 \\
\]
and \[x = 0\]we get \[y = 1 + 1\]
\[
= 1 + 1 \\
y = 2 \\
\]
So the point is \[\left( {0,2} \right)\]
\[x = 1\] apply equation \[y\]
\[
y = {\left( {\dfrac{1}{3}} \right)^1} + 1 \\
= {\left( {\dfrac{1}{3}} \right)^1} + 1 \\
= \dfrac{{{{\left( 1 \right)}^1}}}{{{{\left( 3 \right)}^1}}} + 1 \\
\]
\[
= \dfrac{{1 + 3}}{3} \\
y = \dfrac{4}{3} \\
\]
So the point is \[\left( {1,\dfrac{4}{3}} \right)\]
\[x = 2\] apply the equation \[y\]
\[
y = {\left( {\dfrac{1}{3}} \right)^2} + 1 \\
= \dfrac{{{{\left( 1 \right)}^2}}}{{{{\left( 3 \right)}^2}}} + 1 \\
\]
\[
= \dfrac{{1 + 9}}{9} \\
y = \dfrac{{10}}{9} \\
\]
So the point is \[\left( {2,\dfrac{{10}}{9}} \right)\]
\[x = 3\] apply equation \[y\]
\[
y = {\left( {\dfrac{1}{3}} \right)^3} + 1 \\
= \dfrac{{{{\left( 1 \right)}^3}}}{{{{\left( 3 \right)}^3}}} + 1 \\
\]
\[
= \dfrac{{1 + 27}}{{27}} \\
y = \dfrac{{28}}{{27}} \\
\]
So the point is \[\left( {3,\dfrac{{28}}{{27}}} \right)\]
Now we mention the values in tabular format,

\[x\]	\[ - 3\]	\[ - 2\]	\[ - 1\]	\[0\]	\[1\]	\[2\]	\[3\]
\[y\]	\[28\]	\[10\]	\[4\]	\[2\]	\[\dfrac{4}{3}\]	\[\dfrac{{10}}{9}\]	\[\dfrac{{28}}{{27}}\]

The Points are, \[( - 3,28),( - 2,10),( - 1,4),(0,2),(1,\dfrac{4}{3}),(2,\dfrac{{10}}{9}),(3,\dfrac{{28}}{{27}})\]
Then we get a graph of \[y = {\left( {\dfrac{1}{3}} \right)^x} + 1\]
The graph is given below

Note: Given problem is an exponential equation. The property of the exponential equation is unbounded. In \[{a^x}\] and \[{a^{ - x}}\] are the two forms of the exponent series. In \[{a^x}\] and \[{a^{ - x}}\] are strictly inequality forms. And that is strictly decreasing also. In the given equation the value \[y\]is always positive. Carefully labeled the point in the graph.