Piecewise functions

Coco12

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?

Mark44

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.

Mark44

BTW, don't use "textspeak" like "u" for "you" here at PF. It's not allowed.

Coco12

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.

HallsofIvy

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.

Coco12

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

HallsofIvy

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).

Coco12

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