Question #20549

Decide whether y is a function of x.

2a. x^2+y^2=1

2b. 2x+y=7

Expert's answer

1. Decide whether yy is a function of xx .

2a. x2+y2=1x^{2} + y^{2} = 1

2b. 2x+y=72x + y = 7

Solution:

To test to see if each of these equations represents a function, you would need to examine their graphs and do the vertical line on them. The vertical line test just consists of passing a vertical line over the graph from left to right (or vice versa) and checking that the line never crosses the graph at more than one point while doing so. This works because the definition of a function says that it must have only one yy for every xx value put into the function, only one output for every input.

First of all we have to check if yy is a function of xx , we need to solve for yy squared first and then check to see if there is only one output for every input.

x2+y2=1x^{2} + y^{2} = 1

y2=1x2y^{2} = 1 - x^{2}

y=±(1x2)y = \pm \sqrt{(1 - x^2)}

Solving the equation we get two values, because the quadratic equations have two solutions for yy for a valid value of xx . This means that at least one input value is associated with more than one output value, so by definition, yy is not a function of xx .

Graphing equations is as follows:



Reciprocally solve the problem for the second equation. Check if yy is a function of xx , we need to solve for yy first and then check to see if there is only one output for every input:

2x+y=72x + y = 7

y=72xy = 7 - 2x

y=2x+7y = - 2 x + 7


We get one value for yy if you plug in any value for xx . Obtain a linear function that can be expressed by the formula y=ax+by = ax + b , where xx - the argument, and a,xa, x given numbers. Linear function defined on the set of all real numbers.

For example, if we plugged in a 1 for x, then y would only equal one value 5. Since it is solved for y, y is our output value and x is our input value.



Graphing equations is as follows:

Since we get one value for yy if you plug in any value for xx , which means by definition, yy is a function of xx .

Answer:

2a. x2+y2=1yx^{2} + y^{2} = 1 - y is not a function of xx .

2b. 2x+y=7y2x + y = 7 - y is a function of xx .

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