Register to reply 
Piecewise functions 
Share this thread: 
#1
Jan2213, 04:40 PM

P: 272

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 sqrt2x +1 , the vertex form would be y= (x1)sqrt
since it will never have neg y values , will u need to consider both functions like in other piecewise functions? 


#2
Jan2213, 09:06 PM

Mentor
P: 21,215

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


#3
Jan2213, 09:07 PM

Mentor
P: 21,215




#4
Jan2313, 05:14 AM

P: 272

Piecewise functions
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.



#5
Jan2313, 06:53 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,345

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
Jan2313, 01:56 PM

P: 272

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



#7
Jan2313, 02:42 PM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,345

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 x2 and x1 are negative, their product is positive), [itex]x^2+ 3x 2[/itex] for 1< x< 2 (now x1 is positive but x2 is still negative), and [itex]x^2 3x+ 2[/itex] (since both x2 and x1 are positive). 


#8
Jan2313, 05:37 PM

P: 272

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



Register to reply 
Related Discussions  
Piecewise functions  Calculus & Beyond Homework  7  
Piecewise functions  Calculus & Beyond Homework  9  
Piecewise functions  Calculus & Beyond Homework  5  
Piecewise functions  Calculus  1 