Find axis of symmetry and vertex of x=5y^2 -20y +23x=5y2−20y+23?

1 Answer
mimi
Nov 14, 2017

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

Explanation:

STANDARD FORM: x=ay^2+by+cx=ay2+by+c
AXIS OF SYMMETRY: y=-b/(2a)y=−b2a
REFERENCE STANDARD FORM TO FIND a, b, ca,b,c, THEN FIND y=-b/(2a)y=−b2a.
y = (-(-20))/(2(5)) = 20/10 = 2y=−(−20)2(5)=2010=2
y=2y=2

VERTEX FORM: x=a(y-k)^2+hx=a(y−k)2+h
VERTEX: (h,k)(h,k)
PUT INTO VERTEX FORM, THEN FIND (h,k)(h,k).
x=5y^2-20y+23x=5y2−20y+23 [Subtract constant.]
x-23=5y^2-20yx−23=5y2−20y [Factor to isolate y^2y2.]
x-23=5(y^2-4y)x−23=5(y2−4y) [Complete the square.]
x-23+5(4)=5(y^2-4y+4)x−23+5(4)=5(y2−4y+4) [Factor & Simplify.]
x-3=5(y-2)^2x−3=5(y−2)2 [Isolate xx.]
x=5(y-2)^2+3x=5(y−2)2+3
(h,k)=(3,2)(h,k)=(3,2)
(3,2)(3,2)

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