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.