Independent and Dependent Variables

[Jason76]

What is going on here? Any beginning hints? I do know about functions, dependent and independent variables.

Determine whether each equation defines \(\displaystyle y\) (dependent variable) as a function of \(\displaystyle x\) (independent).

a. \(\displaystyle x + 2y = 7\)

b. \(\displaystyle x^{2} + y = 16\)

c. \(\displaystyle y^{2} + x^{2} = 8\)

I think they are wanting you to plot a small graph for each equation, and then see if any "two y values for one x value" situations may arise. In that case, the equation would NOT represent a function.

 

[mmm4444bot]

I think they are wanting you to plot a small graph for each equation, and then see if any "two y values for one x value" situations may arise.

Yes, applying the "vertical-line test" on a graph works.

In this exercise, a shortcut comes from knowing that IF y is defined by a polynomial in x THEN y is a function of x.

I think that you should graph (c).

 

[Jason76]

Yes, applying the "vertical-line test" on a graph works.

In this exercise, a shortcut comes from knowing that IF y is defined by a polynomial in x THEN y is a function of x.

I think that you should graph (c).

So you can look for relations in which x maps to more than one y value (NOT functions), or just do the vertical line test.

 

[HallsofIvy]

The simplest way to determine if y is a function of x is to try solving for x. If you can do that "unambiguously" (for example to solve \(\displaystyle y^2= x\), you would take the square root of both sides giving you the "ambiguous" \(\displaystyle y= \pm\sqrt{x}\)) then y is a function of x.

 

[soroban]

Hello, Jason76!

Solve for \(\displaystyle y.\)
See if the equation produces a single value of \(\displaystyle y\) for each value of \(\displaystyle x.\)

Determine whether each equation defines \(\displaystyle y\) as a function of \(\displaystyle x.\)

\(\displaystyle (a)\;x + 2y \:=\:7\)

\(\displaystyle 2y \:=\:7-x \quad\Rightarrow\quad y \:=\:\dfrac{7-x}{2}\)

Yes, it is a function.

\(\displaystyle (b)\;x^2+y+16\)

\(\displaystyle y \:=\:16-x^2\)

Yes, it is a function.

\(\displaystyle (c)\;y^2+x^2 \:=\:8\)

\(\displaystyle y^2 \:=\:8-x^2 \quad\Rightarrow\quad y \:=\:\pm\sqrt{8-x^2}\)

No, it is not a function.