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Question

Question: How do you graph \[\tan \left( {\dfrac{x}{2}} \right) + 1\]?...

How do you graph tan(x2)+1\tan \left( {\dfrac{x}{2}} \right) + 1?

Explanation

Solution

To graph the tangent function, we mark the angle along the horizontal x axis, and for each angle, we put the tangent of that angle on the vertical y-axis and the curve obtained is positive infinity in one direction and negative infinity in the other.

Complete step by step solution:
To graph the function tan(x2)+1\tan \left( {\dfrac{x}{2}} \right) + 1, as we know the shape of the tangent curve is the same for each full rotation of the angle and so the function is called 'periodic'.
And we must know a few points before sketching the graph.
The constant +1 in the function represents how much the graph is to be raised. We can find the period which are the lengths at which the function repeats itself, in which tan(x)\tan \left( x \right) has a period of π\pi , so that tan(x2)\tan \left( {\dfrac{x}{2}} \right) has a period of 2π2\pi , since the angle is divided by 2 in the function given andtan(x)\tan \left( x \right) is undefined when cos(x)=0\cos \left( x \right) = 0 and is zero when sin(x)=0\sin \left( x \right) = 0 because
tan(x)=sin(x)cos(x)\tan \left( x \right) = \dfrac{{\sin \left( x \right)}}{{\cos \left( x \right)}}.
In the graph we can see that the function has vertical asymptotes at π intervals so the period is π and when x=0,y=0x = 0,y = 0.
So, if you have tanx+1\tan x + 1 it shifts all the y values up by one tan(x2)\tan \left( {x2} \right)is a vertical shift and it doubles the period to 2π2\pi .

Note:
To sketch the trigonometry graphs of the functions – Sine, Cosine and Tangent, we need to know the period, phase, amplitude, maximum and minimum turning points. Hence based on this we can graph the derivative of a function. And the tangent function has a range that goes from positive infinity to negative infinity.