Question 1012572
{{{f(x)=2/(x^2-2x-3)=2/((x+1)(x-3))}}}
There are only two restrictions on x that make the denominator go to zero. 
So {{{x=-1}}} and {{{x=3}}} are excluded from the domain.
Domain : ({{{-infinity}}},{{{-1}}})U({{{-1}}}{{{3}}})U({{{3}}}{{{infinity}}})
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We can plot to look for the range,
*[illustration vc2.JPG].
So from the graph, you see that as you move from left to right you go from {{{0}}} to {{{infinity}}}, {{{-infinity}}} to {{{-1/2}}}, and then from {{{infinity}}} back to {{{0}}}.
So the range is,
Range: ({{{-infinity}}},{{{-1/2}}}]U({{{0}}}{{{infinity}}})
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There is no x-intercept, the graph never crosses the x-axis.
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When {{{x=0}}},
{{{y=2/(-3)=-(2/3)}}}
So the y-intercept is 
({{{0}}},{{{-2/3}}})