Consider the standard form of:#" "y=ax^2+bx+c#
Write this as:#" "y=a(x^2+b/ax)+c#
Note that #c# is the y-intercept
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Considering the parts of the general equation")#
#color(blue)("in relation to "y=1/4x^2)#
#color(brown)("The shape of the graph")#
If #a>=0 # then the graph is of general shape #uu#
if #a<0# then the graph is of general shape #nn#
#color(brown)(a=1/4 >0 " so the graph is of shape "uu)#
'....................................................................................
#color(brown)("Determine " x_("vertex"))#
#x_("vertex") = (-1/2)xxb/a#
Your equation is #y=1/4x^2+0x+0->y=1/4[x^2+(4xx0)x]+0#
So #b/a=0#
#color(brown)(=> x_("vertex")=(-1/2)xx0=0)#
So the graph is symmetrical about the y-axis
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(brown)("Determine " y_("vertex"))#
Substitute #x=0# giving
#color(blue)(y=1/4x^2+0x+0)color(brown)(" "->" "y_("vertex")=1/4xx0^2 = 0)#
,.............................................................................
#color(brown)(" Vertex "->(x,y)=(0,0))#
#color(green)(=>y_("intercept")=0)#
#color(green)(=>x_("intercept") = 0" the x-axis is a tangent to the")# #color(green)("curve at the vertex")#
'.................................................................................
#color(brown)(" Determine any other points")#
Substitute values for #x# into the equation and determine the corresponding value of #y#
Suppose #x=2# then #y=1/4(2)^2 = 4/4=1#
This is also true for #x=-2#
So at #y=1" ; " x= +-2" as it is symmetrical about the y-axis"#
