Given that the curve y=p-(x-q)^2 cuts the x-axis at (-1,0) & (3,0), find the value of p and of q.

This is what i get,

Since the curve y=p-(x-q)^2 cuts the x-axis at (-1,0) and (3,0)

y=(x+1)(x-3)
= x^2-2x-3
=(x-1)^2-4

but i do not know how to convert the answer into the needed form :)

Question

Given that the curve y=p-(x-q)^2 cuts the x-axis at (-1,0) & (3,0), find the value of p and of q. 

This is what i get,

Since the curve y=p-(x-q)^2 cuts the x-axis at (-1,0) and (3,0)

y=(x+1)(x-3) 
  = x^2-2x-3
   =(x-1)^2-4

but i do not know how to convert the answer into the needed form :)

anonymous · Answer

i think i see the problem

anonymous · Answer

you are making the assumption that the parabola faces up, but it could also face down
i.e. it could be 
$$y=-(x+1)(x-3)$$ still has the same zeros

anonymous · Answer

then you can write it as 
$$y=4-(x-1)^2$$ and it looks like what you want