How do you graph $y = 4 tan (\frac{π}{2} t)$ $y=4tan(pi/2t)$ over the interval $[- 2, 2]$ $[-2,2]$ ?

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How do you graph $y = 4 tan (\frac{π}{2} t)$ $y=4tan(pi/2t)$ over the interval $[- 2, 2]$ $[-2,2]$ ?

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Answer 1 · 2017-02-18T14:04:54

(see below)

Explanation:

If we let $θ = \frac{π}{2} t$ $theta=pi/2t$
then $t \in [- 2, 2]$ $tin[-2,2]$
is equivalent to $θ \in [- π, + π]$ $thetain[-pi,+pi]$
and we know $y = tan (θ), θ \in [- π, + π]$ $y=tan(theta), thetain[-pi,+pi]$ looks like:
enter image source here
Replacing the horizontal axis with the $t$ $t$ variable based scale:

Replacing $y = tan (\frac{π}{2} t)$ $y=tan(pi/2t)$ with $y = 4 tan (\frac{π}{2} t)$ $y=4tan(pi/2t)$
simply pushes every $y$ $y$ coordinate value $4$ $4$ further away from the horizontal axis;
the easiest way to show this is to modify the scale along the vertical axis to $4$ $4$ times their previous values:
enter image source here

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How do you graph $y = 4 tan (\frac{π}{2} t)$ $y=4tan(pi/2t)$ over the interval $[- 2, 2]$ $[-2,2]$ ?

1 Answer

Explanation:

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How do you graph $y = 4 tan (\frac{π}{2} t)$ $y=4tan(pi/2t)$ over the interval $[- 2, 2]$ $[-2,2]$ ?