Question 644528
{{{y^2=x^3-4x}}}

set {{{x=0}}} and solve for {{{y}}} to find {{{y-intercept}}}:

{{{y^2=0^3-4*0=0}}}.........{{{y=0}}}

{{{y-intercept}}} is at origin ({{{0}}},{{{0}}})

set {{{y=0}}} and solve for {{{y}}} to find {{{y-intercept}}}:

{{{0=x^3-4x}}}...=>..{{{0=x(x^2-4)}}}..=>.{{{0=x(x-2)(x+2)}}}.....so, {{{0=x}}} or {{{x=2}}} or {{{x=-2}}}

{{{ graph( 600, 600, -10, 10, -10, 10, sqrt(x^3-4x),-sqrt(x^3-4x)) }}}