Question 350939
{{{y=2x^2-8}}} Start with the given equation.



{{{0=2x^2-8}}} Plug in {{{y=0}}}



{{{8=2x^2}}} Add 8 to both sides.



{{{8/2=x^2}}} Divide both sides by 2.



{{{4=x^2}}} Reduce.



{{{x^2=4}}} Flip the equation.



{{{x=""+-sqrt(4)}}} Take the square root of both sides.



{{{x=sqrt(4)}}} or {{{x=-sqrt(4)}}} Break up the plus/minus.



{{{x=2}}} or {{{x=-2}}} Take the square root of 4 to get 2.



So when {{{y=0}}}, {{{x=2}}} or {{{x=-2}}}



Put another way, we have the points (2,0) and (-2,0)



So the x-intercepts are (2,0) and (-2,0)



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Jim