How do you graph the parabola #y=3/4x^2# using vertex, intercepts and additional points?

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Answer 1 · 2018-06-07T14:57:56

See Graph

Graph:

#y=3/4x^2#

y-intercept, set #x=0# and solve for #y#

#y=3/4(0)^2#

#y=0#

x-intercept(s) if they exist, set #y=0# and solve for #x#

#0=3/4x^2#

#x=0#

Axis of Symmetry (aos) in the form #ax^2+bx+c#

#aos = (-b)/(2a)#

#y=3/4x^2 + 0x +0#

#aos = (-0)/(2(3/4))#

#aos=0#

Finally find the vertex it is a minimum if #a>0# a maximum if #a<0#

#vertex = (aos, f(aso))#

#vertex = (0, f(0))#

#vertex = (0, 3/4(0)^2)#

#vertex = (0, 0)#

graph{(3/4)x^2 [-10.125, 9.875, -0.48, 9.52]}