What is the axis of symmetry and vertex for the graph #y=1/20x^2#?

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Answer 1 · 2017-05-23T13:47:44

vertex: #(0, 0)#; axis of symmetry: #x = 0#

Given: #y = 1/20 x^2#

Find the vertex : When #y= Ax^2 + Bx + C = 0# the vertex is #(h, k)#, where #h = (-B)/(2A)#:

#h = -0/(2*1/20) = 0#

#k = f(h) = 1/20 (0)^2 = 0#

#"vertex":(0, 0)#

Find the axis of symmetry , #x = h#:

axis of symmetry, #x = 0#