
Plot a graph for the equation \[y = {x^2} - 4x\].
Answer
562.8k+ views
Hint:
Here we need to plot the given equation. For that, we will first simplify the equation and after simplification, we will find that the given equation is the equation of an upward parabola and its vertex is not at the center. We will write the equation in standard form and from there, we will find the vertex of the parabola and then we will find the point of the curve with both the coordinates. After obtaining the points, we will plot the graph accordingly.
Complete step by step solution:
Here we need to plot the graph of the given equation \[y = {x^2} - 4x\].
We will first simplify the equation and we will write it in the standard form.
Now we will add 4 on both sides of the equation. Therefore, we get
\[ \Rightarrow y + 4 = {x^2} - 4x + 4\]
Rewriting the equation, we get
\[ \Rightarrow y + 4 = {x^2} - 2 \cdot 2 \cdot x + {2^2}\]
Using the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2 \cdot a \cdot b\], we get
\[ \Rightarrow y + 4 = {\left( {x - 2} \right)^2}\]
Rewriting the equation, we get
\[ \Rightarrow {\left( {x - 2} \right)^2} = y - \left( { - 4} \right)\] ………. \[\left( 1 \right)\]
We can see that the given equation is the equation of the upward parabola and it is a shifted parabola i.e. vertex is not at the center.
We know that the equation of shifted upward parabola is given by \[{\left( {x - h} \right)^2} = y - k\].
Here, vertex of this shifted upward parabola is \[\left( {h,k} \right)\].
Comparing this general equation with standard equation, we get the vertex of the shifted upward parabola as \[\left( {2, - 4} \right)\].
Now, we will first find the intersection of the parabola with \[x\] axis by the putting the value \[y = 0\] in the equation of the parabola i.e. in equation \[\left( 1 \right)\].
\[ \Rightarrow {\left( {x - 2} \right)^2} = 0 - \left( { - 4} \right)\]
On further simplification, we get
\[ \Rightarrow {\left( {x - 2} \right)^2} = 4\]
Now, taking square roots on both sides, we get
\[\begin{array}{l} \Rightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = \sqrt 4 \\ \Rightarrow x - 2 = 2\end{array}\]
On adding 2 on both sides, we get
\[\begin{array}{l} \Rightarrow x - 2 + 2 = 2 + 2\\ \Rightarrow x = 4\end{array}\]
Therefore, point of intersection of the parabola with the \[x\] axis is equal to \[\left( {0,4} \right)\].
Similarly, we will first find the intersection of the parabola with \[y\] axis by the putting the value \[x = 0\] in the equation of the parabola i.e. in equation \[\left( 1 \right)\].
\[ \Rightarrow {\left( {0 - 2} \right)^2} = y - \left( { - 4} \right)\]
On further simplification, we get
\[ \Rightarrow 4 = y + 4\]
On subtracting 4 from both sides, we get
\[\begin{array}{l} \Rightarrow 4 - 4 = y + 4 - 4\\ \Rightarrow 0 = y\\ \Rightarrow y = 0\end{array}\]
Therefore, point of intersection of the parabola with the \[y\] axis is equal to \[\left( {0,0} \right)\]
Now, we will plot the graph using all these points.
This is the required plot of the given equation.
Note:
The given equation is an equation of upward parabola. A parabola is a type of conic shape or a U-shaped plane curve in which any point is at an equal distance from a fixed point and from a fixed straight line known as the directrix. In order to plot any graph we always find the point of intersection of the graph with the coordinate axes.
Here we need to plot the given equation. For that, we will first simplify the equation and after simplification, we will find that the given equation is the equation of an upward parabola and its vertex is not at the center. We will write the equation in standard form and from there, we will find the vertex of the parabola and then we will find the point of the curve with both the coordinates. After obtaining the points, we will plot the graph accordingly.
Complete step by step solution:
Here we need to plot the graph of the given equation \[y = {x^2} - 4x\].
We will first simplify the equation and we will write it in the standard form.
Now we will add 4 on both sides of the equation. Therefore, we get
\[ \Rightarrow y + 4 = {x^2} - 4x + 4\]
Rewriting the equation, we get
\[ \Rightarrow y + 4 = {x^2} - 2 \cdot 2 \cdot x + {2^2}\]
Using the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2 \cdot a \cdot b\], we get
\[ \Rightarrow y + 4 = {\left( {x - 2} \right)^2}\]
Rewriting the equation, we get
\[ \Rightarrow {\left( {x - 2} \right)^2} = y - \left( { - 4} \right)\] ………. \[\left( 1 \right)\]
We can see that the given equation is the equation of the upward parabola and it is a shifted parabola i.e. vertex is not at the center.
We know that the equation of shifted upward parabola is given by \[{\left( {x - h} \right)^2} = y - k\].
Here, vertex of this shifted upward parabola is \[\left( {h,k} \right)\].
Comparing this general equation with standard equation, we get the vertex of the shifted upward parabola as \[\left( {2, - 4} \right)\].
Now, we will first find the intersection of the parabola with \[x\] axis by the putting the value \[y = 0\] in the equation of the parabola i.e. in equation \[\left( 1 \right)\].
\[ \Rightarrow {\left( {x - 2} \right)^2} = 0 - \left( { - 4} \right)\]
On further simplification, we get
\[ \Rightarrow {\left( {x - 2} \right)^2} = 4\]
Now, taking square roots on both sides, we get
\[\begin{array}{l} \Rightarrow \sqrt {{{\left( {x - 2} \right)}^2}} = \sqrt 4 \\ \Rightarrow x - 2 = 2\end{array}\]
On adding 2 on both sides, we get
\[\begin{array}{l} \Rightarrow x - 2 + 2 = 2 + 2\\ \Rightarrow x = 4\end{array}\]
Therefore, point of intersection of the parabola with the \[x\] axis is equal to \[\left( {0,4} \right)\].
Similarly, we will first find the intersection of the parabola with \[y\] axis by the putting the value \[x = 0\] in the equation of the parabola i.e. in equation \[\left( 1 \right)\].
\[ \Rightarrow {\left( {0 - 2} \right)^2} = y - \left( { - 4} \right)\]
On further simplification, we get
\[ \Rightarrow 4 = y + 4\]
On subtracting 4 from both sides, we get
\[\begin{array}{l} \Rightarrow 4 - 4 = y + 4 - 4\\ \Rightarrow 0 = y\\ \Rightarrow y = 0\end{array}\]
Therefore, point of intersection of the parabola with the \[y\] axis is equal to \[\left( {0,0} \right)\]
Now, we will plot the graph using all these points.
This is the required plot of the given equation.
Note:
The given equation is an equation of upward parabola. A parabola is a type of conic shape or a U-shaped plane curve in which any point is at an equal distance from a fixed point and from a fixed straight line known as the directrix. In order to plot any graph we always find the point of intersection of the graph with the coordinate axes.
Recently Updated Pages
Master Class 12 Economics: Engaging Questions & Answers for Success

Master Class 12 Physics: Engaging Questions & Answers for Success

Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Maths: Engaging Questions & Answers for Success

Master Class 12 Business Studies: Engaging Questions & Answers for Success

Trending doubts
Which are the Top 10 Largest Countries of the World?

What are the major means of transport Explain each class 12 social science CBSE

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

Draw a neat and well labeled diagram of TS of ovary class 12 biology CBSE

Give 10 examples of unisexual and bisexual flowers

