What's the equation for this?

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Answer 1 · 2018-01-07T19:32:07

$y = \frac{x - 4}{x^{2} - 16}$ $y = (x-4)/(x^2-16)$

There is a vertical asymptote at x = -4 and a horizontal asymptote at y = 0 means:

$y = \frac{1}{x + 4}$ $y = 1/(x+4)$

There is a removable discontinuity at $x = 4$ $x = 4$ means there is a common factor in the numerator and denominator of $x - 4$ $x-4$ :

$y = \frac{1}{x + 4} \cdot \frac{x - 4}{x - 4}$ $y = 1/(x+4)*(x-4)/(x-4)$

$y = \frac{x - 4}{x^{2} - 16}$ $y = (x-4)/(x^2-16)$

Check that the y intercept is $(0, \frac{1}{4})$ $(0,1/4)$

$y = \frac{0 - 4}{0^{2} - 16}$ $y = (0-4)/(0^2-16)$

$y = \frac{1}{4}$ $y = 1/4$

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