
How do you graph $y=\sec \left( \dfrac{x}{2} \right)$?
Answer
538.8k+ views
Hint:We explain the main function of the given equation $y=\sec \left( \dfrac{x}{2} \right)$. We take the general equation and explain the amplitude, period. Then we equate the given function $y=\sec\left( \dfrac{x}{2} \right)$ with the general one and find the graph.
Complete step by step solution:
We need to find the amplitude, period for $y=\sec \left( \dfrac{x}{2} \right)$.
The main function of the given equation is $\sec x$. The period of $\cos x$ is $2\pi $.
We define the general formula to explain the amplitude, period for $\sec x$.
The $\sec x$ changes to $A\sec \left[ B\left( x+C \right) \right]+D$, the amplitude and the period becomes $\left| A \right|$ and $\dfrac{2\pi }{\left| B \right|}$.
Now we explain the things for the given $y=\sec \left( \dfrac{x}{2} \right)$.
$y=\sec \left( \dfrac{x}{2} \right)=1\times \sec \left[ \dfrac{1}{2}\times x \right]$. We equate with $A\sec \left[ B\left( x+C \right) \right]+D$.
The values will be \[\left| A \right|=1;\left| B \right|=\dfrac{1}{2}\]. The period is $\dfrac{2\pi}{\dfrac{1}{2}}=2\pi \times 2=4\pi $.
Therefore, the amplitude and period for $y=\sec \left( \dfrac{x}{2} \right)$ is $1,4\pi $ respectively.
The usual common graph which is easier to plot on the graph is $y=\sec x$.
The graph is an oscillating graph with boundary being $\left( -\infty ,-1 \right)$ and $\left( 1,\infty\right)$.
The domain for the graph $y=\sec x$ is $\mathbb{R}$.
The range for the graph $y=\sec x$ is $\mathbb{R}\backslash \left( -1,1 \right)$.
Now depending on the above-mentioned graph, we are going to first find the graph of
$y=\sec \left( \dfrac{x}{2} \right)$.
The change between $y=\sec x$ and $y=\sec \left( \dfrac{x}{2} \right)$ is that the graph is now stretched for twice its previous area.
The domain for the graph $y=\sec \left( \dfrac{x}{2} \right)$ is $\mathbb{R}$.
The range for the graph $y=\sec \left( \dfrac{x}{2} \right)$ is $\mathbb{R}\backslash \left( -1,1 \right)$.
Note: Amplitude is the vertical distance from the X-axis to the highest (or lowest) point on a sin or cosine curve. Period of each generalized sine or cosine curve is the length of one complete cycle. The previous graphs were oscillating for an area of $\pi $ and now it’s for $2\pi $.
Complete step by step solution:
We need to find the amplitude, period for $y=\sec \left( \dfrac{x}{2} \right)$.
The main function of the given equation is $\sec x$. The period of $\cos x$ is $2\pi $.
We define the general formula to explain the amplitude, period for $\sec x$.
The $\sec x$ changes to $A\sec \left[ B\left( x+C \right) \right]+D$, the amplitude and the period becomes $\left| A \right|$ and $\dfrac{2\pi }{\left| B \right|}$.
Now we explain the things for the given $y=\sec \left( \dfrac{x}{2} \right)$.
$y=\sec \left( \dfrac{x}{2} \right)=1\times \sec \left[ \dfrac{1}{2}\times x \right]$. We equate with $A\sec \left[ B\left( x+C \right) \right]+D$.
The values will be \[\left| A \right|=1;\left| B \right|=\dfrac{1}{2}\]. The period is $\dfrac{2\pi}{\dfrac{1}{2}}=2\pi \times 2=4\pi $.
Therefore, the amplitude and period for $y=\sec \left( \dfrac{x}{2} \right)$ is $1,4\pi $ respectively.
The usual common graph which is easier to plot on the graph is $y=\sec x$.
The graph is an oscillating graph with boundary being $\left( -\infty ,-1 \right)$ and $\left( 1,\infty\right)$.
The domain for the graph $y=\sec x$ is $\mathbb{R}$.
The range for the graph $y=\sec x$ is $\mathbb{R}\backslash \left( -1,1 \right)$.
Now depending on the above-mentioned graph, we are going to first find the graph of
$y=\sec \left( \dfrac{x}{2} \right)$.
The change between $y=\sec x$ and $y=\sec \left( \dfrac{x}{2} \right)$ is that the graph is now stretched for twice its previous area.
The domain for the graph $y=\sec \left( \dfrac{x}{2} \right)$ is $\mathbb{R}$.
The range for the graph $y=\sec \left( \dfrac{x}{2} \right)$ is $\mathbb{R}\backslash \left( -1,1 \right)$.
Note: Amplitude is the vertical distance from the X-axis to the highest (or lowest) point on a sin or cosine curve. Period of each generalized sine or cosine curve is the length of one complete cycle. The previous graphs were oscillating for an area of $\pi $ and now it’s for $2\pi $.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

