How do you graph the parabola #y=-1/2x^2+3# using vertex, intercepts and additional points?

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Answer 1 · 2018-02-21T16:42:56

Take any point as #x#, find #y#, and find the point of #(x,y)# on the graph.

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

The minus sign before #x^2# means it's upside down, so it is a reflection of #y =3#.

#"vertex"= -b/(2a)#

So

#v=-0/(2*-1/2)=0#

Substitute #x=0#

#"vertex"(0,3)#

To get any point, you take any #x# point, like #x=2#

#y=-1/2(2)^2+3 #

#y=1#

So #(2,1)# is a point. You can use any point like

#"vertex" -2 -> 0-2=-2 #

#"vertex"-1 -> 0-1=-1 #

#"vertex" -> 0 #

#"vertex"+1 -> 0+1=1 #

#"vertex"+2 -> 0+2=2#

You can use the #x#-intercept, which means #y=0#, and the #y#-intercept, which means #x=0#.

#y=-1/2(x)^2+3#

#0=-1/2(x)^2+3#

#1/2(x)^2=3#

#x^2=3/(1/2)#

#x^2=6#

#x=+-sqrt( 6)#

Then the points are #(sqrt 6 , 0)# and #(-sqrt 6 , 0)#.

#y=-1/2(x)^2+3#

#y=-1/2(0)^2+3#

#y=3#

Then the point is #(0,3)#.