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

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In summary, if a vertex of a parabola is on the x-axis and its value is positive, there would be an infinite number of functions. The equation given, y = x^2 - 2x + 1, is in standard form and can be changed to vertex form as y = (x-1)^2. However, since the value is never negative, there is no need for a piecewise function. The question asking for a piecewise function may have been referring to the equation in absolute value brackets, but since it is never negative, the absolute value is not necessary.
  • #1
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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  • #2
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.
 
  • #3
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.
 
  • #4
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.
 
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  • #5
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.
 
  • #6
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
 
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  • #7
So the function is actually [itex]f(x)= |x^2- 2x+ 1|[/itex]? It would have helped to tell us that to begin with!

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

If it had been, say [itex]f(x)= |x^2- 3x+ 2|= |(x- 2)(x- 1)[/itex] then we would have f(x) equal to [itex]x^2- 3x+ 2[/itex] for x< 1 (since both x-2 and x-1 are negative, their product is positive), [itex]-x^2+ 3x- 2[/itex] for 1< x< 2 (now x-1 is positive but x-2 is still negative), and [itex]x^2- 3x+ 2[/itex] (since both x-2 and x-1 are positive).
 
  • #8
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
 

1. What is a piecewise function?

A piecewise function is a mathematical function that is defined by different equations or rules for different intervals or subintervals of the function's domain.

2. What does it mean for a parabola vertex to be on the x-axis in a piecewise function?

If a parabola vertex is on the x-axis in a piecewise function, it means that the function has a break or discontinuity at the vertex point. This can occur when the function switches from one equation to another at the vertex point.

3. How do I graph a piecewise function with a parabola vertex on the x-axis?

To graph a piecewise function with a parabola vertex on the x-axis, you can first graph each individual function separately. Then, plot the points where the functions meet and connect them to create a continuous graph. Be sure to label any breaks or discontinuities at the vertex point.

4. Can a piecewise function with a parabola vertex on the x-axis have more than two equations?

Yes, a piecewise function with a parabola vertex on the x-axis can have more than two equations. It can have as many equations as needed to define the function for different intervals or subintervals of the domain.

5. How can I determine the domain and range of a piecewise function with a parabola vertex on the x-axis?

To determine the domain and range of a piecewise function with a parabola vertex on the x-axis, you can look at the individual equations that make up the function and identify the intervals where they are defined. The domain will be the union of all these intervals. The range can be determined by looking at the minimum and maximum values of each equation, as well as any breaks or discontinuities at the vertex point.

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