#color(green)("Consider each part of the standard equation "y=ax^2+bx+c" and what effect they have.")#

#color(blue)("Type 1" -> y=x^2#

The coefficient of #x# is (+1). Because this is positive the graph is of general shape #uu# and symmetric about the y-axis. The minimum is at #(x,y)->(0,0)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Type 2" -> y=-x^2#

The coefficient of #x# is (-1). This is negative so the graph is of general shape #nn# and symmetric about the y-axis. The maximum is at #(x,y)->(0,0)#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Type 3" -> y=ax^2#

If#" "-1 < a < +1 " "#then the graph is wider (Shallower gradient )

if#" "-1 > a > +1" "#then the graph narrows (Steeper gradient )

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Type 4" -> y=x^2+c#

If the value of #c# is positive it 'lifts' each point of the graph upwards by the value of #c#. Conversely if #c# is negative it lowers the graph.

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Type 5" -> y=x^2+bx+c#

The coefficient of #b# moves the plot sideways.

First write the equation as#" "y=a(x^2+b/ax)+c#

Now apply: #" "(-1/2)xxb/a -> x_("vertex")#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Answering your question")#

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

The coefficient of #x^2# is positive so of general shape #uu# thus the graph has a#color(blue)(" minimum.")#

There is #underline("no")# constant value c #underline("nor")# is there any #bx#

So it is a mixture of Type 1 and Type 3. Consequently

#color(blue)("Vertex"->(x,y)->(0,0)#

#color(blue)("Axis of symmetry"-> x=0#)

#color(blue)("It is a minimum")#

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(green)("This graph shows what happens when you change the coefficient of "x^2" ( changing the value of "a" in "y=ax^2"))#