Question

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

Answer 1

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)`