Question 135709

 



{{{4y^2 + x-xy^2-1 = 0}}} Start with the given equation



{{{4y^2-xy^2= 1-x}}} Subtract x from both sides



{{{y^2(4-x)= 1-x}}} Factor out the GCF {{{y^2}}}



{{{y^2 =  (1-x)/(4-x) }}} Divide both sides by {{{4-x}}}





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{{{y^2 =  (1-0)/(4-0) }}} To find the y-intercept, plug in {{{x=0}}}




{{{y^2 =  1/4 }}} Simplify


{{{y =  0+-sqrt(1/4) }}} Take the square root of both sides



{{{y=1/2}}} or {{{y=-1/2}}} Simplify


So the y-intercepts are (0,{{{1/2}}}) and (0,{{{-1/2}}})


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{{{y^2 =  (1-x)/(4-x) }}} Go back to the original equation



{{{0^2 =  (1-x)/(4-x) }}} To find the x-intercept, plug in {{{y=0}}}



{{{0 =  (1-x)/(4-x) }}} Square 0 to get 0



{{{0(4-x) =  1-x }}} Multiply both sides by 4-x



{{{0 =  1-x }}} Multiply




{{{x=  1 }}} Add x to both sides



So the x-intercept is (1,0)



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{{{4-x=0}}} To find the vertical asymptote(s), simply set the denominator equal to zero



{{{x=4}}} Solve for x


So the vertical asymptote is at {{{x=4}}} 



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{{{y^2 =  (1-x)/(4-x) }}} Go back to the original equation



{{{y^2 =  (-x+1)/(-x+4) }}} Rearrange the terms



Notice how the coefficient of "x" for the numerator and denominator is -1. So the horizontal asymptote is the ratio {{{-1/-1=1}}}. Since we are dealing with a square in {{{y^2}}}, this means that the final equation looks like {{{y =  0+-sqrt((1-x)/(4-x))}}}. So there are two final parts {{{y =  sqrt((1-x)/(4-x))}}} and {{{y =  -sqrt((1-x)/(4-x))}}}



This means that there is symmetry with respect with the x-axis and that there are two horizontal asymptotes {{{y=1}}} and {{{y=-1}}}





Notice if we graph both {{{y =  sqrt((1-x)/(4-x))}}} and {{{y =  -sqrt((1-x)/(4-x))}}}, we can visually verify our answer




{{{ graph( 500, 500, -10, 10, -10, 10, sqrt((1-x)/(4-x)) ,-sqrt((1-x)/(4-x)) ) }}} 




Notice how that if you pass a vertical line through the graph, the line will intersect with the graph more than once. So this tells us that this graph is <b>not</b> a function since it fails the vertical line test.