Question 1105987
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For a vertical asymptote at x=-4, you need a factor of (x+4) in the denominator, with no like factor in the numerator.<br>
For the removable discontinuity at x=4, you need factors of (x-4) in both numerator and denominator.<br>
To have no x-intercept, there can be no other linear factors in the numerator.<br>
Using those constraints, we know parts of the equation are<br>
{{{(a(x-4))/((x-4)(x+4))}}}<br>
where a is a constant.<br>
With the equation as it is, it will have a horizontal asymptote of y=0, because the degree of the denominator is greater than the degree of the numerator.<br>
We want the y-intercept to be (0,1/4); so the equation evaluated at 0 should be 1/4:<br>
{{{(a(x-4))/((x-4)(x+4)) = (-4a)/-16 = a/4 - 1/4}}}  -->  a = 1<br>
So an equation that has the required features is<br>
{{{y=((x-4))/((x-4)(x+4))}}}