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)#