How do you find the vertex and axis of symmetry of #y=5/4(x+2)^2 -1#?

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Answer 1 · 2017-06-28T14:02:01

Vertex is #(-2,-1)# and axis of symmetry is #x+2=0#.

This is the equation of a horizontal parabola in vertex form i.e.

#y=a(x-h)^2+k#

where vertex is #(h,k)# and axis of symmetry is #x=h#.

Hence vertex and axis of symmetry of #y=1/4(x+2)^2-1#

or #y=1/4(x-(-2))^2-1#

are #(-2,-1)# and #x=-2# or #x+2=0# respectively

graph{((x+2)^2-4-4y)(x+2)((x+2)^2+(y+1)^2-0.02)=0 [-13, 7, -4.28, 5.72]}