How to find vertex of a horizontal parabola in standard form

for example: x=1+y-y^2
please explain, Thank you

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x = 1+y-y²
x = -y²+y+1            ← standard form ... a= -1, b= +1, c= +1

y = -b/(2a) = -1/(2(-1)) = ½            ← y-coordinate of the vertex

x = (4ac-b²)/(4a) = ?

Note ... Consider this:
4ac-b²       4ac          b²                  b²
——— = ——– − ———  =  c − —— = c − (ᵇ∕₂ₐ)(ᵇ∕₂)  OR  -(ᵇ∕₂ₐ)(ᵇ∕₂)+c
   4a           4a           4a                 4a


So, the second coordinate of the vertex (the x-coordinate for this parabola)
can be determined by taking the y-coordinate (i.e.  -(ᵇ∕₂ₐ)) and multiplying
it by (ᵇ∕₂) and then adding c.

So,
x-coordinate of this vertex = -(ᵇ∕₂ₐ)(ᵇ∕₂)+c  =  (½)(½)+1  =  ⁵∕₄


                                 ANSWER
                                   V(⁵∕₄,½)


Hope you found this useful.
Have a good one!
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It's similar, except that the y-coordinate can be found via -b/(2a) and the x-coordinate can be found by (4ac - b^2)/(4a).

So, with x = 1 + y - y^2 = (-1)y^2 + y + 1, we see that:
a = -1, b = 1, and c = 1.

Hence, the required vertex is:
((4ac - b^2)/(4a), -b/(2a)) = ([4(1)(-1) - 1^2]/[4(-1)], -1/[2(-1)])
= ((-4 - 1)/(-4), 1/2)
= (5/4, 1/2).

I hope this helps!

x=1+y-y^2
x-1=-(y²-y)
x-1-.25=-(y²-y+.25)
x-1.25=-(y-.5)² vertex (1.25, 0.5)