How do you identify the important parts of #y= 5x^2-x# to graph it?

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Answer 1 · 2018-04-13T05:29:01

See below....

The #x#-intercepts can be found out by putting #y=0# in the equation.

#" "5x^2-x=0#
#rArr" "x(5x-1=0)#

So #x=0# and #x=1/5=0.2#

Also the minimum of the function is at the point where #dy/dx=0#

or,#" "10x-1=0#
or,#" "x=0.1#

The function is a parabola of the form #x^2=4ay# and #y=+oo# when #x->+-oo#.

graph{5x^2-x [-1.108, 1.292, -0.239, 0.96]}