Answer
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Hint: For answering this question we need to draw the graph of ${{x}^{3}}-4x+1$ . For drawing the graph of the given expression we will find the points lying on the given curve and plot them and join them and extend the curve.
Complete step-by-step solution:
Now considering from the question we have to draw the graph of \[{{x}^{3}}-4x+1\] .
From the basics of the concept we know that for drawing the graph of the curve we need to find the points lying on the given equation of the curve and plot them and join them and extend the curve formed.
When $x=0$ we will have
$\begin{align}
& \Rightarrow y={{\left( 0 \right)}^{3}}-4\left( 0 \right)+1 \\
& \Rightarrow y=1 \\
\end{align}$
When $x=1$ we will have
$\begin{align}
& \Rightarrow y={{\left( 1 \right)}^{3}}-4\left( 1 \right)+1 \\
& \Rightarrow y=-2 \\
\end{align}$
When $x=2$ we will have
$\begin{align}
& \Rightarrow y={{\left( 2 \right)}^{3}}-4\left( 2 \right)+1 \\
& \Rightarrow y=8-8+1 \\
& \Rightarrow y=1 \\
\end{align}$
When $x=-2$ we will have
$\begin{align}
& \Rightarrow y={{\left( -2 \right)}^{3}}-4\left( -2 \right)+1 \\
& \Rightarrow y=-8+8+1 \\
& \Rightarrow y=1 \\
\end{align}$
When $x=-1$ we will have
$\begin{align}
& \Rightarrow y={{\left( -1 \right)}^{3}}-4\left( -1 \right)+1 \\
& \Rightarrow y=-1+4+1 \\
& \Rightarrow y=4 \\
\end{align}$
So now we have four points lying on the curve are $\left( 0,1 \right)$, $\left( 1,-2 \right)$ , $\left( 2,1 \right)$ , $\left( -1,4 \right)$ and $\left( -2,1 \right)$
Now we will plot these points and join them and extend them to form a curve.
The graph of the curve is shown below.
Note: For answering this question we need to draw the graph of the given equation of the cubic polynomial so we need to be careful while plotting the points and extending the curve. We should be careful while performing the calculations. These graphs are similar to the graph for quadratic. For quadratic there will be two roots so there will be two shifts and for cubic there will be cubic it will have three shifts.
Complete step-by-step solution:
Now considering from the question we have to draw the graph of \[{{x}^{3}}-4x+1\] .
From the basics of the concept we know that for drawing the graph of the curve we need to find the points lying on the given equation of the curve and plot them and join them and extend the curve formed.
When $x=0$ we will have
$\begin{align}
& \Rightarrow y={{\left( 0 \right)}^{3}}-4\left( 0 \right)+1 \\
& \Rightarrow y=1 \\
\end{align}$
When $x=1$ we will have
$\begin{align}
& \Rightarrow y={{\left( 1 \right)}^{3}}-4\left( 1 \right)+1 \\
& \Rightarrow y=-2 \\
\end{align}$
When $x=2$ we will have
$\begin{align}
& \Rightarrow y={{\left( 2 \right)}^{3}}-4\left( 2 \right)+1 \\
& \Rightarrow y=8-8+1 \\
& \Rightarrow y=1 \\
\end{align}$
When $x=-2$ we will have
$\begin{align}
& \Rightarrow y={{\left( -2 \right)}^{3}}-4\left( -2 \right)+1 \\
& \Rightarrow y=-8+8+1 \\
& \Rightarrow y=1 \\
\end{align}$
When $x=-1$ we will have
$\begin{align}
& \Rightarrow y={{\left( -1 \right)}^{3}}-4\left( -1 \right)+1 \\
& \Rightarrow y=-1+4+1 \\
& \Rightarrow y=4 \\
\end{align}$
So now we have four points lying on the curve are $\left( 0,1 \right)$, $\left( 1,-2 \right)$ , $\left( 2,1 \right)$ , $\left( -1,4 \right)$ and $\left( -2,1 \right)$
Now we will plot these points and join them and extend them to form a curve.
The graph of the curve is shown below.
Note: For answering this question we need to draw the graph of the given equation of the cubic polynomial so we need to be careful while plotting the points and extending the curve. We should be careful while performing the calculations. These graphs are similar to the graph for quadratic. For quadratic there will be two roots so there will be two shifts and for cubic there will be cubic it will have three shifts.
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