How to graph $\eta(t)=(t,|t|)$?

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We are studying lines (I am not quite sure about the correct term in English and could not find any other term for this from googling so please correct me if I am wrong) in our calculus class right now and this might be a dumb question but I am having a hard time understanding how to draw graphs.

As an example, we were given $\gamma:[0,2\pi]\to\mathbb{R}^2$, $\gamma(t)=(\cos{t},\sin{t})$ and said that it graphs a unit circle. However, it is not quite clear to me how does this happen. Does $\gamma$ take values from $[0,2\pi]$ and I plug them in such as $\gamma(0)=(\cos{0},\sin{0})=(1,0)$, which is the starting point, and $\gamma(2\pi)=(\cos{2\pi},\sin{2\pi})=(1,0)$, which is the ending point? I mean $t\in [0,2\pi]$ right? And when I connect the starting and ending points I get an unit circle.

Another example which actually confused me was we were given $\eta: I\to\mathbb{R}^2$, $\eta(t)=(t,|t|)$ where $I=[-1,1]$. This was graphed as

enter image description here

but I don't know how we got that. If I plug in the values as I did on the first example, $\eta(-1)=(-1,|-1|)=(-1,1)$ and $\eta(1)=(1,|1|)=(1,1)$ but if it is now true that $(-1,1)$ is the starting point and $(1,1)$ the ending point, why the graph is not just a straight line between these points? How did we suddenly connected the points to the origin $(0,0)$?

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For the unit circle, if you know that $\cos^2 t + \sin^2 t = 1$, then it follows by deparameterizing the curve. Since $x = \cos t$, we have that $y^2 = \sin^2 t = 1-\cos^2 t = 1-x^2$, so the curve is given by $x^2+y^2 = 1$, or $y = \pm\sqrt{1-x^2}$. Since $\cos t$ and $\sin t$ are periodic with period $2\pi$, the graph for $t\in [0,2\pi]$ should be the entire circle, with the point $(1,0)$ being the starting and ending point.

For the absolute value, we do the same thing. We have $x = t$, so $y = |x|$ with $x\in [-1,1]$ Thus, we just graph the absolute value function only over the values $[-1,1]$ on the $x$-axis.

If you want to graph some parameterized curve like this without solving it algebraically (and sometimes you can't), you can simply plug in a bunch of values for $t$ in order and play connect-the-dots. The more $t$ values you use, the closer your approximation will be to the actual curve.