

Tag, Which of the following rules represents a function? Welche der folgenden Regeln stellt eine Funktion dar? can you explain why these are called rules?. $i\phantom{\rule{0.12em}{0ex}}$ know about functions but confused here about they call these rules? please, any hinto with the subject/topic at hand. can you refer a tutorial or begin a solution. however way you think it is best, now you know $i\phantom{\rule{0.12em}{0ex}}$ am here to get you to do my homework but to explain the way I should go and the materials available. you can refer german sites too. $i\phantom{\rule{0.12em}{0ex}}$ dont speak much but can read German. Danke im Voraus. Which of the following rules represents a function? Choose an answer $\{(x\phantom{\rule{0.12em}{0ex}},y\phantom{\rule{0.12em}{0ex}}):y\phantom{\rule{0.12em}{0ex}}2=x\phantom{\rule{0.12em}{0ex}}2\}$ $\{(x\phantom{\rule{0.12em}{0ex}},y\phantom{\rule{0.12em}{0ex}}):y\phantom{\rule{0.12em}{0ex}}=x\phantom{\rule{0.12em}{0ex}}2\}$ $\{(x\phantom{\rule{0.12em}{0ex}},y\phantom{\rule{0.12em}{0ex}}):\lefty\phantom{\rule{0.12em}{0ex}}\right=x\phantom{\rule{0.12em}{0ex}}2\}$ $\{(x\phantom{\rule{0.12em}{0ex}},y\phantom{\rule{0.12em}{0ex}}):y\phantom{\rule{0.12em}{0ex}}2=x\phantom{\rule{0.12em}{0ex}}\}$ 

123mathe.de/relationenundfunktionen 

thank you. Ich melde mich noch!. 

A "function rule" simply is an equation which defines a function  it describes a functional relationship. $y\phantom{\rule{0.12em}{0ex}}={x\phantom{\rule{0.12em}{0ex}}}^{2}+1$ is an example for a function rule. $y\phantom{\rule{0.12em}{0ex}}}^{4}=x\phantom{\rule{0.12em}{0ex}$ is a rule, but its not a function rule, because a function has to be unique (every $x\phantom{\rule{0.12em}{0ex}}$ is assigned just one $y\phantom{\rule{0.12em}{0ex}})$. Here for $x\phantom{\rule{0.12em}{0ex}}=1$ we have two possible values for $y\phantom{\rule{0.12em}{0ex}},$ either $+1$ or $1$. So this rules defines a relation which is not unique and therefore is not a function. Now check your given rules if they define a function (unique relation) or not. 

I do not understand well, can you provide an example?. Graphing the equations given in Desmos I could see that $b\phantom{\rule{0.12em}{0ex}})$ is the right choice becasue it passes the vertical test, but, they won't let me do it with the graph, how can $i\phantom{\rule{0.12em}{0ex}}$ do it assigning values to what variable, $x\phantom{\rule{0.12em}{0ex}}$ or y? don't really understand your explanation. thanks! 

$>$ can you provide an example?. I did. It was $y\phantom{\rule{0.12em}{0ex}}}^{4}=x\phantom{\rule{0.12em}{0ex}$. And, yes, only $b\phantom{\rule{0.12em}{0ex}})$ is a function and what you describe with "vertical test" is exactly what I meant when I wrote that the relation has to be unique to be called a function. At every $x\phantom{\rule{0.12em}{0ex}}$ position there must be no more than just one assigned $y\phantom{\rule{0.12em}{0ex}}$ value. To "see" if the relation is unique without plotting the relation, you would do the plotting in your mind. If you find just one xvalue with two or more assigned yvalues, than its not a function. For example $\lefty\phantom{\rule{0.12em}{0ex}}\right={x\phantom{\rule{0.12em}{0ex}}}^{2}$. You know that $e\phantom{\rule{0.12em}{0ex}}.g\phantom{\rule{0.12em}{0ex}}$. both $+4$ and $4$ simplify to $+4$. This means that for $x\phantom{\rule{0.12em}{0ex}}=2(\to {x\phantom{\rule{0.12em}{0ex}}}^{2}=4)$ there are two possible values for $y\phantom{\rule{0.12em}{0ex}}(\to +4$ and $4)$ and this contradicts the characterizing property of a function. 

I got clearly! I gave a value to $x\phantom{\rule{0.12em}{0ex}},$  if I get 2 values of $y\phantom{\rule{0.12em}{0ex}}$ for the same $x\phantom{\rule{0.12em}{0ex}}$ then it is not a function. So for the equation $A\phantom{\rule{0.12em}{0ex}}>{x\phantom{\rule{0.12em}{0ex}}}^{2}={y\phantom{\rule{0.12em}{0ex}}}^{2}$. For $x\phantom{\rule{0.12em}{0ex}}=1,$ you can get $y\phantom{\rule{0.12em}{0ex}}=1$ or 1. So it is not a function. Equation $B\phantom{\rule{0.12em}{0ex}}$ gave me only one value so it's a function. Because there is one yvalue for the entered xvalue, and only one. $x\phantom{\rule{0.12em}{0ex}}=2,$ then, $y\phantom{\rule{0.12em}{0ex}}=4$ when $x\phantom{\rule{0.12em}{0ex}}=2$ $y\phantom{\rule{0.12em}{0ex}}={x\phantom{\rule{0.12em}{0ex}}}^{2}$ $y\phantom{\rule{0.12em}{0ex}}=4$ Vielen Danke. 