How can you determine if this equation: x-6=y2 is a function or not?



Asked on

4 Answers | Add Yours

embizze's profile pic

Posted on (Answer #1)

Determine if `x-6=y^2` is a function.

(1) If a relation is a function, each input corresponds to exactly one output. If x is the independent variable, and y the dependent variable, then this relation is not a function. The following ordered pairs are in the relation: (6,0),(7,1),(7,-1),(10,2),(10,-2). Notice that the input 7 has two outputs; i.e. 1 and -1. Likewise the input 10 has two outputs. If any input has more than one output then the relation is not a function.

(2) Draw the graph. If you pass a vertical line across the graph, then the relation is a function if the line crosses the graph at no  more than 1 point. If the vertical line cuts the graph twice or more times, the relation is not a function.

Notice that the vertical line x=7 crosses the graph at two points, so the relation is not a function.

(3) Try to write the relation in "function" form; i.e. y=f(x). Here solving for y yields `y=+-sqrt(x-6)` . Notice that there are really two functions here, the positive root and the negative root. But if this relation was a function, you would just get one function when solving for y.

sweetskittles24's profile pic

Posted on (Answer #2)

when you draw the graph, each x-value has to correspond to only one y-value to be a function. In this graph, each x-value corresponds to more than one y-value. therfore, not a function.

alyssab223's profile pic

Posted on (Answer #3)

thank you so much.

jishnudeepkar's profile pic

Posted on (Answer #4)

also x-6= y^2 is not a fn

butx/y-6/y=y is a fn in that form itself

We’ve answered 396,832 questions. We can answer yours, too.

Ask a question