#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"))#
