
What is the nature of the graph: \[y = \dfrac{4}{x}\]
A.Rectangular hyperbola in first and third quadrant
B.Rectangular hyperbola in first and second quadrant
C.Hyperbola but not rectangular hyperbola
D.Rectangular hyperbola in second and fourth quadrant
Answer
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Hint: This question involves the arithmetic operations like addition/ subtraction/ multiplication/ division. We need to know how to draw a graph for the given equation. Also, we need to know how to find \[x\& y\] from the given equation to make an easy calculation for the given question. Also, we need to know how to plot the points in the graph.
Complete step-by-step answer:
To find the nature of the graph we would draw the graph for the given equation.
For that, we would assume \[x = ...., - 2, - 1,0,1,2,...\]
We have \[y = \dfrac{4}{x} \to \left( 1 \right)\]
Let’s take \[x = - 2\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{{ - 2}} \\
y = - 2 \;
\]
Let’s take \[x = - 1\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{{ - 1}} \\
y = - 4 \;
\]
Let’s take \[x = 0\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{0} \\
y = \infty \;
\]
Let’s take \[x = 1\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{1} \\
y = 4 \;
\]
Let’s take \[x = 2\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{2} \\
y = 2 \;
\]
Let’s make a tabular column with various \[x\& y\] values as given below,
By using the above tabular column we can make the following graph,
The above graph defines the equation \[y = \dfrac{4}{x}\] . In the above graph, we can see that the rectangular hyperbola is present in the first quadrant and third quadrant.
So, the final answer is,
\[A)\] Rectangular hyperbola in first and third quadrant
So, the correct answer is “OPTION A ”.
Note: This question describes the arithmetic operations like addition/ subtraction/ multiplication/ division. Note that to solve these types of questions we would draw the graph for the given equation. Also, note that for finding the value \[y\] we have to assume the \[x\] values. By using \[x\& y\] values we can easily draw the graph.
Complete step-by-step answer:
To find the nature of the graph we would draw the graph for the given equation.
For that, we would assume \[x = ...., - 2, - 1,0,1,2,...\]
We have \[y = \dfrac{4}{x} \to \left( 1 \right)\]
Let’s take \[x = - 2\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{{ - 2}} \\
y = - 2 \;
\]
Let’s take \[x = - 1\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{{ - 1}} \\
y = - 4 \;
\]
Let’s take \[x = 0\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{0} \\
y = \infty \;
\]
Let’s take \[x = 1\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{1} \\
y = 4 \;
\]
Let’s take \[x = 2\] in the equation \[\left( 1 \right)\] , we get
\[\left( 1 \right) \to y = \dfrac{4}{x}\]
\[
y = \dfrac{4}{2} \\
y = 2 \;
\]
Let’s make a tabular column with various \[x\& y\] values as given below,
| \[x\] | \[ - 2\] | \[ - 1\] | \[0\] | \[1\] | \[2\] |
| \[y\] | \[ - 2\] | \[ - 4\] | \[\infty \] | \[4\] | \[2\] |
By using the above tabular column we can make the following graph,
The above graph defines the equation \[y = \dfrac{4}{x}\] . In the above graph, we can see that the rectangular hyperbola is present in the first quadrant and third quadrant.
So, the final answer is,
\[A)\] Rectangular hyperbola in first and third quadrant
So, the correct answer is “OPTION A ”.
Note: This question describes the arithmetic operations like addition/ subtraction/ multiplication/ division. Note that to solve these types of questions we would draw the graph for the given equation. Also, note that for finding the value \[y\] we have to assume the \[x\] values. By using \[x\& y\] values we can easily draw the graph.
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