Question 943039


{{{x^2 +3y=0}}}


{{{3y=-x^2 }}}


{{{y=-x^2/3 }}}

{{{y=-(1/3)x^2 }}}

since the vertex form of parabola is {{{y=a(x-h)^2+k}}} where {{{h}}} and {{{k}}} coordinates of the vertex, if we compare it to {{{y=-(1/3)x^2 }}} we see that {{{a=-1/3}}}, {{{h=0}}} and {{{k=0}}}; so, the vertex is at origin ({{{0}}},{{{0}}})

make table:

{{{x}}},{{{y}}}

{{{0}}},{{{0}}}

{{{-3}}},{{{-3}}} .......{{{y=-(1/3)(-3)^2= -(1/3)(9)}}}

{{{3}}},{{{-3}}} .......{{{y=-(1/3)(3)^2= -(1/3)(9)}}}

{{{-2}}},{{{-4/3}}} .......{{{y=-(1/3)(-2)^2= -(1/3)(4)}}}

{{{2}}},{{{-4/3}}} .......{{{y=-(1/3)(-2)^2= -(1/3)(4)}}}

plot points and draw a line through:


{{{drawing( 600, 600, -10, 10, -10, 10,circle(0,0,.2),
circle(-3,-3,.2),circle(3,-3,.2),circle(-2,-4/3,.2),circle(2,-4/3,.2),
 graph( 600, 600, -10, 10, -10, 10, -(1/3)(x)^2)) }}}