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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.
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.
also x-6= y^2 is not a fn
butx/y-6/y=y is a fn in that form itself
thank you so much.
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