Question

Which of the following four equations defines a function?

A) `3x+2y-7=0`
B) `5x^2y=9`
C) `3x^2-4y^2=9`
D) `x=3y^2-1`

Answer 1

In A, B only

Explanation:

For y to be function of x, it should satisfy the condition that for any x, there is only one corresponding y. This condition is satisfied only in A and B. Observe the following:

A) y= `1/2 (7-3x)`

B) y=`9/(5x^2)`

C) y= `+-1/4 (3x^2-9)` , for any given x, y has two values.

D) y= `+- 1/3 (x+1)`, for any given x, y has two values.

Answer 2

A and B.

Explanation:

Example A) `3x+2y-7=0`

Subtract `3x-7` from both sides to get:

`2y=7-3x`

Divide both sides by `2` to get:

`y = (7-3x)/2`

So `y` is uniquely determined for any value of `x`.

In other words A passes the Vertical Line Test.

graph{3x+2y-7=0 [-10, 10, -5, 5]}

Example B) `5x^2y = 9`

Divide both sides by `5x^2` to get:

`y = 9/(5x^2)`

So `y` is uniquely determined for any value of `x` except `x=0`, where no value of `y` satisfies B

So B passes the vertical line test.

graph{5x^2y = 9 [-10.41, 9.59, -1.96, 8.04]}

Example C) `3x^2-4y^2 = 9`

Add `4y^2-9` to both sides to get:

`4y^2=3x^2-9`

Divide both sides by `4` to get:

`y^2=(3x^2-9)/4`

So

`y = +-sqrt((3x^2-9)/4)`

This fails the vertical line test. For example, when `x = 3`, we get:

`y = +-sqrt((27-9)/4) = +-sqrt(9/2)`

So `y` is not uniquely determined, failing the vertical line test.

Here's a graph of C with vertical line `x=3` :

graph{(3x^2-4y^2-9)(x-3+y*0.0001) = 0 [-9.365, 10.635, -4.56, 5.44]}

Example D) x = 3y^2 - 1

Add `1` to both sides and divide by `3` to get:

`y^2 = (x+1)/3`

So

`y = +-sqrt((x+1)/3)`

Again, `y` is not uniquely determined, failing the vertical line test.

For example, if `x=2` then `y = +-sqrt(3/3) = +-sqrt(1) = +-1`

graph{(x - 3y^2 - 1)(x-2+0.0001*y) = 0 [-3.405, 6.595, -2.22, 2.78]}

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