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I found this extremely confusing because there is so called "rose pedal function"..and circle standard form... Does it mean that if a relation failed either vertical or horizontal test, then we have to use its shape to identify it? Is there really a rose pedal function or my prof was incorrect?

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Your confusion is stemming from a misunderstanding of the definition of a function. Formally, a function $f$ from a set $A$ to a set $B$ is a subset of $A\times B = \{(a,b):a\in A\,\text{and}\, b\in B\}$, such that for each $a\in A$ there is a unique $b\in B$ with $(a,b)\in f$. In other words, every input gets one and only one output.

In the case of the rose petal function you are describing, we are forming ordered pairs by first picking an angle $\theta$ and then specifying a radius $r$ for that particular $\theta$. Hence, the rose petal function is a collection of ordered pairs $(\theta, r)$, such that to each angle corresponds a unique radius. When we plot this function in polar coordinates, it looks as though it is violating the "vertical line" test (we are specifying more than one radius for each angle), but this is not actually the case. If we plotted the rose petal function in the $\theta,r$-plane, we would see that this function did indeed pass the vertical line test, and is a well-defined function.

See here for a glimpse of what these "petal" functions look like in the $\theta,r$-plane so you can see how these functions actually do pass the vertical line test.

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  • $\begingroup$ Excellent answer. $\endgroup$ – RJM Oct 14 '16 at 1:47
  • $\begingroup$ Got it. A function of whom is important for defining whether it is a function. $\endgroup$ – Newbornalive Oct 14 '16 at 1:51
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There can be mappings into $\mathbb{R^2}$ that do not pass the vertical line test for the graph of a function. Ex. g := {(x,y) $\in$ $\mathbb{R}$ x $\mathbb{R}$| $x^2 + y^2 = 1$}

Your professor may have been referring to graphs in polar coordinates. For example r = $\sin 2\theta$

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  • $\begingroup$ I see. So a function failed vertical test in R can still be a function in other space. $\endgroup$ – Newbornalive Oct 14 '16 at 1:48
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    $\begingroup$ Yes, unfortunately at high school and early college level the definition of function is too strongly tied to $f: \mathbb{R} \rightarrow \mathbb{R}$ $\endgroup$ – RJM Oct 14 '16 at 1:52

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