Calculus by michael spivak (3rd edition) chapter 12 (apendix) question 2 goes as follows
Let $$f(t)= \begin{cases} t^2&,x\geq0\\ -t^2&,x\leq0 \end{cases}$$ Show that $c(t)=\left(f(t),t^2\right)$ lies along the graph of $h(x)=|x|$.
The answer book states
We have $$f(t)= \begin{cases} t^2&,x\geq0\\ -t^2&,x\leq0 \end{cases}$$ and these points are all on the graph of $h(x)=|x|$, since $|-t^2|=t^2$.
to me, these seems to be a pointles answer.
What is the actual reason $c$ lies along the graph of $h$? In other words, how could one know that it does?
Well, the graph of $h$ is the set $G = \{(x,|x|) :\, x \in \mathbb R\}$. A point in $G$ is of the form $(x,x)$ or $(x,-x)$, right? The points, $$(-2,2),(0,0),(-1,1),(\pi,\pi),(-\sin 1,\sin 1)$$ are point that lies in $G$ to be clear. Now, since $c(t)$ is either $(t^2,t^2)$ or $(-t^2,t^2)$, clearly lies in $G$. Are you more convinced?
To be formal, a point $(x,y)$ in $\mathbb R^2$ lies in $G$ if and only if $y=|x|$, that is, an ordered pair lies in $G$ if and only if their second entry is the absolute value of the first. So, in the case that $c(t) = (t^2,t^2)$ it is obvoius that $c(t) \in G$, and in the other case, the case that $c(t) = (-t^2,t^2)$, $c(t) \in G$ since $|-t^2| = t^2$.