How do you find the vertex and intercepts for #x^2 = -8y#?

Multiply both sides by (-1). Makes #-8y# positive

#8y=-x^2#

Divide both sides by 8

#y=-x^2/8#
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Compare to the standard form of #y=ax^2+bx+c#

All of these #a, b's and c's# change the basis of #y=x^2# in particular ways. Lets consider them one at a time.

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#color(blue)("Point 1 - general shape is " nn)#

Compare #-x^2/8# to #ax^2#

#a->-1/8#

As this is negative the graph is of general shape #nn# so it has a maximum.

#color(purple)("As "a" is a fraction it widens the graph of "y=-x^2)#
#color(purple)("If "a>1" then it make the graph of "y=-x^2" narrower.")#
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#color(blue)("Point 2 - the y-axis is the axis of symmetry")#

The part of the equation #+bx# moves the graph left or write.

Given that #y=ax^2+bx+c# write as #y=a(x^2+b/ax) +c#

#x_("vertex")=(-1/2)xxb/a#

But our equation is #y=-1/8x^2" "->" "y=-1/8x^2+0x+0#

So #b=0" "->" "x_("vertex")=(-1/2)xx0/(" "-1/8" ") = 0#

As #x_("vertex")=0# the graph is symmetrical about the y-axis

So at this point we can sate that Vertex#->(x,y)=(0,y_("vertex"))#
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#color(blue)("Determine the y-intercept")#

The constant #c# in #y=ax^2+bc+c# is the y-intercept.

And we have #y=-1/8x+0x+0# so #c=0#

As the graph has the y-axis is the axis of symmetry the y-intercept can only be at the vertex.

Vertex#->(x,y)=(0,0)#

Graph of #y=-x^2# in red. Notice the way that #y=-1/8x^2# in blue is wider