Y as a Function of X: Deciding the Equation

  • Thread starter Thread starter AznBoi
  • Start date Start date
  • Tags Tags
    Function
AI Thread Summary
The equation y=sqrt(x+3) represents y as a function of x because it passes the vertical line test, indicating that each x-value corresponds to only one y-value. Square roots do not automatically include a plus/minus sign; the principal square root is defined as the positive value. The term "sleeping parabola" refers to a parabola with a horizontal axis, but in this case, only the top half is relevant for the function. Therefore, the graph of y=sqrt(x+3) is indeed a function, representing the upper portion of a horizontal parabola. Understanding these concepts clarifies the nature of square root functions in relation to their graphical representations.
AznBoi
Messages
470
Reaction score
0
Hmm I'm not sure if this is correct:

Problem: Decide whether the equation represents y as a function of x:

y=sq.rt.(x+3)

Ok, so do square roots always have the plus/minus sign automatically? Radicals are the equivalent of sleep parabolas right?? Do I add the bottom half also??

My answer was No because it is a radical/sleeping parabola.
 
Physics news on Phys.org
AznBoi said:
Hmm I'm not sure if this is correct:

Problem: Decide whether the equation represents y as a function of x:

y=sq.rt.(x+3)

Ok, so do square roots always have the plus/minus sign automatically?
No, \sqrt{x} is defined as the positive number whose square is equal to x. Square root, like any real valued function, is single valued.

Radicals are the equivalent of sleep parabolas right?? Do I add the bottom half also??

My answer was No because it is a radical/sleeping parabola.
"Sleeping" parabola? You mean a parabola with its axis horizontal?
Unfortunately, you are wrong. \sqrt{x+3} is a function. It's graph is the top half of the horizontal parabola.
 
it is a function because it passes the vertical line test. usually the square root means the principal square root.
 
ok thanks.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top