Find axis of symmetry and vertex of x=5y^2 -20y +23?

1 Answer
Nov 14, 2017

AXIS OF SYMMETRY: y=2
VERTEX: (3,2)

Explanation:

STANDARD FORM: x=ay^2+by+c
AXIS OF SYMMETRY: y=-b/(2a)
REFERENCE STANDARD FORM TO FIND a, b, c, THEN FIND y=-b/(2a).
y = (-(-20))/(2(5)) = 20/10 = 2
y=2

VERTEX FORM: x=a(y-k)^2+h
VERTEX: (h,k)
PUT INTO VERTEX FORM, THEN FIND (h,k).
x=5y^2-20y+23 [Subtract constant.]
x-23=5y^2-20y [Factor to isolate y^2.]
x-23=5(y^2-4y) [Complete the square.]
x-23+5(4)=5(y^2-4y+4) [Factor & Simplify.]
x-3=5(y-2)^2 [Isolate x.]
x=5(y-2)^2+3
(h,k)=(3,2)
(3,2)