Piecewise Functions: Parabola Vertex on x-axis +ve, One Function?

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Discussion Overview

The discussion revolves around the nature of piecewise functions in relation to a parabola with its vertex on the x-axis and a positive value, specifically examining whether such a parabola can be represented as a single function or if it necessitates a piecewise definition. The conversation includes aspects of mathematical representation, function continuity, and the implications of absolute values.

Discussion Character

  • Debate/contested
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • Some participants propose that if a parabola opens upward with its vertex on the x-axis, it could be represented by a single function.
  • Others argue that there are infinitely many functions possible due to unspecified parameters, such as the value of 'a' in the parabola's equation.
  • There is confusion regarding the notation used, particularly the interpretation of "sqrt" and the representation of exponents.
  • One participant clarifies that the equation in question is in standard form and has been converted to vertex form, questioning the need for a piecewise function since it does not yield negative y-values.
  • Another participant seeks clarification on what constitutes a piecewise function, suggesting that it involves ensuring the function does not take negative values.
  • It is noted that if the function is expressed as f(x) = |x^2 - 2x + 1|, it simplifies to f(x) = |(x - 1)^2|, which is always non-negative, thus questioning the necessity of a piecewise representation.
  • One participant provides an example of a different function that would require a piecewise definition due to its behavior across specified intervals.

Areas of Agreement / Disagreement

Participants express differing views on whether the parabola can be represented as a single function or if it requires a piecewise definition. The discussion remains unresolved, with multiple competing perspectives on the nature of the function.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the parameters of the parabola and the definitions of piecewise functions. The conversation also highlights potential misunderstandings related to mathematical notation.

Coco12
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If a vertex of a parabola is on the x-axis and it's a value is positive so it opens upward, would there only be one function? For example x sqrt-2x +1 , the vertex form would be y= (x-1)sqrt
since it will never have neg y values , will u need to consider both functions like in other piecewise functions?
 
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Coco12 said:
If a vertex of a parabola is on the x-axis and it's a value is positive so it opens upward, would there only be one function?
No, there would be an infinite number of them. You didn't specify a, which would change the shape of the parabola, and you didn't specify where the vertex is on the x-axis.
Coco12 said:
For example x sqrt-2x +1 , the vertex form would be y= (x-1)sqrt
This is very confusing, but I think I understand what you're trying to say. "sqrt" does not mean "squared" - it's short for square root.

Your equation seems to be y = x2 - 2x + 1 = (x - 1)2. An easy way to indicate an exponent is using the ^ symbol.
Coco12 said:
since it will never have neg y values , will u need to consider both functions like in other piecewise functions?

Now I don't understand what you're asking. The parabola in your example is continuous for all values of x. There's nothing piecewise about it.
 
Mark44 said:
No, there would be an infinite number of them. You didn't specify a, which would change the shape of the parabola, and you didn't specify where the vertex is on the x-axis.
This is very confusing, but I think I understand what you're trying to say. "sqrt" does not mean "squared" - it's short for square root.

Your equation seems to be y = x2 - 2x + 1 = (x - 1)2. An easy way to indicate an exponent is using the ^ symbol.


Now I don't understand what you're asking. The parabola in your example is continuous for all values of x. There's nothing piecewise about it.

BTW, don't use "textspeak" like "u" for "you" here at PF. It's not allowed.
 
Sorry I mean squared for them all, the first equation is in standard form and I changed it to vertex form. The a value is 1 as can be seen in the vertex form. So for this equation , would it be correct to say that there is no piecewise function since it never goes into the negative y value?I am just asking because the question asked for a piecewise function of this equation but I don't think that it is possible.
 
Last edited:
Exactly what do you mean by "piecewise function"? It might help if you told us what the question that asks "for a piecewise function of this equation" is itself.
 
The equation is in absolute brackets so so a piecewise function is basically two functions that makes sure the equation doesn't go into the negative y value. The question simply gave u the equation in absolute brackets and then to write a piecewise function. I was the one who changed it to vertex form so I could see where it was on a graph
 
Last edited:
So the function is actually f(x)= |x^2- 2x+ 1|? It would have helped to tell us that to begin with!

Yes, that gives f(x)= |(x- 1)^2|. But (x- 1)^2 is never negative so the absolute value doesn't matter.

If it had been, say f(x)= |x^2- 3x+ 2|= |(x- 2)(x- 1) then we would have f(x) equal to x^2- 3x+ 2 for x< 1 (since both x-2 and x-1 are negative, their product is positive), -x^2+ 3x- 2 for 1< x< 2 (now x-1 is positive but x-2 is still negative), and x^2- 3x+ 2 (since both x-2 and x-1 are positive).
 
Sorry for the confusion. So the equation will not have a piecewise function? I didn't think that there was going to be according to the equation and put x> or equal to 1 and x < or equal to 1
 

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