To graph a parabola we need three points: the two #x#-intercepts and the vertex. To find the intercepts we need to set #y=0# and solve for #x#.
#0=-1/4x^2#
#0/(-1/4)=cancel(-1/4)/cancel(-1/4)x^2#
#0=x^2#
#+-sqrt0=+-sqrtx^2#
#x=+-0#, which is just #0#. So our two intercepts are at #0#.
Now let's find the vertex. The easiest way to do that is to convert the equation into vertex form. #y=-1/4x^2# becomes #y=-1/4(x+0)^0-0#. That means that the vertex is at #(0, 0)#, also known as the origin.
Now, we should know that the #-# if front of #1/4# means that the graph will be flipped, so that instead of looking like a #uu# it'll look like an #nn#. The #1/4# tells that there is a strech factor, and because it is less that #1# (a fraction), it is going to make the graph wider by a factor of #1/4#.
Our graph should have a vertex at #(0, 0)# and the intercepts will be in the same place. The graph should be inverted and wide. If we want some guiding points we just solve our equation for them, like so: #y=-1/4(2)^2# or #y=-1/4*4#, which becomes #y=-1#. That tells that one of our points is #(2, -1)#. I want another one, so let's solve for when #x=-2#. #y=-1/4(-2)^2# becomes #y=-1/4*4# which simplifies to #y=-1#. That means that we have another point: #(-2, -1)#. Let's graph the eqution and see if it fits our guesses.
graph{y=-1/4x^2}
And it does! Great work!!
