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

Hint:
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 \[\tan \left( {\dfrac{x}{2}} \right) + 1\], 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 \[\tan \left( x \right)\] has a period of \[\pi \], so that \[\tan \left( {\dfrac{x}{2}} \right)\] has a period of \[2\pi \], since the angle is divided by 2 in the function given and\[\tan \left( x \right)\] is undefined when \[\cos \left( x \right) = 0\] and is zero when \[\sin \left( x \right) = 0\] because
\[\tan \left( x \right) = \dfrac{{\sin \left( x \right)}}{{\cos \left( x \right)}}\].
In the graph we can see that the function has vertical asymptotes at π intervals so the period is π and when \[x = 0,y = 0\].
So, if you have \[\tan x + 1\] it shifts all the y values up by one \[\tan \left( {x2} \right)\]is a vertical shift and it doubles the period to \[2\pi \].


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.