SOLUTION: Find the equation of a parabola that has a Minimum turning point at (2,-5). I am totally confused how to do this. .I looked up the answer; which is listed as being y=x^2-4x-1

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Find the equation of a parabola that has a Minimum turning point at (2,-5). I am totally confused how to do this. .I looked up the answer; which is listed as being y=x^2-4x-1      Log On


   



Question 932594: Find the equation of a parabola that has a Minimum turning point at (2,-5).
I am totally confused how to do this.
.I looked up the answer; which is listed as being y=x^2-4x-1
I tried to complete the square; so; (x-2)^2--2]-1
But that would give me a y value of 11
So x; seems to correct...but not sure if that is a fluke,


Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation of a parabola that has a Minimum turning point at (2,-5).
y = a(x-2)^2 - 5 for a = 1 (which is one such parabola)
y = (x-2)^2 - 5
y = x^2 -4x + 4 - 5
y = x^2 - 4x - 1