How do you use the important points to sketch the graph of #y=-2x^2 #?

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Answer 1 · 2017-09-23T12:18:04

See below.

#y=-2x^2#

If we consider a quadratic in standard for; #ax^2+bx+c#

Here: #a=-2, b=0, c=0#

Since #a<0# we know that #y# will have an absolute maximum value on its axis of symmetry, where #x=(-b)/(2a)# which is #x=0# in this case.

Hence, #y_max = y(0) = 0#

This gives us our "important point" #(0,0)# as the absolute maximum of #y#.

As there are no other intercepts we are left with plotting points to produce the graph of #y# below.

E.g. #(x=+-1, y=-2), (x=+-2, y=-8)# etc.

graph{-2x^2 [-8.96, 8.82, -6.54, 2.35]}