How do you graph # f(x) = 2/3x^2 - 5#?

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Answer 1 · 2017-08-03T18:38:18

See a solution process below:

Setting #x# to #0# gives us the point:

#f(x) = (2/3 * 0^2) - 5#

#f(x) = -5#

Or

#0, -5#

We can find the #0#s by solving the function for #0#:

#2/3x^2 - 5 = 0#

#2/3x^2 - 5 + 5 = 0 + 5#

#2/3x^2 - 0 = 5#

#3/2 xx 2/3x^2 = 3/2 xx 5#

#x^2 = 15/2#

#x^2 = +-sqrt(15/2)#

Or

#(-sqrt(15/2), 0)# and #(-sqrt(15/2), 0)#

graph{(y-((2x^2)/3)+5)(x^2+(y+5)^2-0.125)((x+(15/2)^(1/2))^2+(y)^2-0.125)((x-(15/2)^(1/2))^2+(y)^2-0.125)=0 [-20, 20, -10, 10]}