For each of the following, determine whether the equation defines y as a function of x.

Question

1=|y|+x^2

4+y^2=x^2

5x=y^3

y=2|x|-4

Answer 1

Hi,

1=|y|+x^2 ==> |y| = 1 - x²

Base of definition of function You choose a number such as 1/2 = x then find y

|y| = 1 - (1/2)²====> |y| = 1 - 1/4= 3/4 ====> y = ±3/4 it means:

( 1/2, 3/4) & ( 1/2,-3/4) are on the graph |y| = 1 - x² so it is not function

4+y^2=x^2====== x²- y² = 4 choose x = 3 =====> 9 - y² = 4 ===> y² = 5 ====> y = ±√5

it means: ( 3, √5 )& ( 3, -√5 ) are on the graph 4+y^2=x^2, so it is not function

5x=y^3=======> y = (5x)^1/3 base of definition of function it is function because:

x₁ = x₂ ===> 5x₁ = 5x₂ =====> (5x₁ )^1/3= (5x₂)^1/3 =====> y₁=y₂

y=2|x|-4 you can draw this graph and show it is function

I hope it is helpful, I tried to deliver three methods for saying it is function or not.

Minoo

Answer 2

the definition of a function of x for variable y

is that for any x there is only one value of y.

Any specific y value can be duplicated for other x

but at any x, only one value of y..

1=|y|+x^2

4+y^2=x^2

5x=y^3

y=2|x|-4

edited 1st equation, |y|=1-x² ,,,,,a good way to approach is to say |y|=√y² so will have

y²=(1-x²)²,,,,y=+/-(1-x²).so any x has two y values and the expression is not a function..

The values of x are constricted -1≤x≤1.

2nd equation, y=+/-√(x²-4) does not meet function requirement two values of y for any x

3rd equation y=(5x)^(1/3) is a function

4th equation, y=2x-4 when x>0 y=-2x-4 when x<0 is always one x generating just one y

Answer 3