Slope-intercept form
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now, the denominator has a higher degree, a degree of 1, than the numerator's, therefore the horizontal asymptote for this function is at y = 0.
now, the vertical asymptotes of it, are at the zeros of the denominator, namely x = 0, when is x = 0? at 0, so the vertical asymptote for this function is at x = 0, the y-axis.
now, if we want the horizontal asymptote moved over from its position at 0, over to x = 5, simply translate it by 5 units,
and if we want the vertical asymptote to move down from 0 to -3, simply translate the function vertically 3 units,