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Dvinal [7]
2 years ago
9

How can you tell from the equation of a rational function if the function has a hole in the graph ( a removable discontinuity) a

t x, rather than a vertical asymptote? Give an example​
Mathematics
1 answer:
Eva8 [605]2 years ago
5 0

Consider that,

x^2+4x+4 = (x+2)(x+2)

x^2+7x+10 = (x+2)(x+5)

Dividing those expressions leads to

(x^2+4x+4)/(x^2+7x+10) = (x+2)/(x+5)

The intermediate step that happened is that we have (x+2)(x+2) all over (x+2)(x+5), then we have a pair of (x+2) terms cancel as the diagram indicates (see below). This is where the removable discontinuity happens. Specifically when x = -2. Plugging x = -2 into (x+2)/(x+5) produces an output, but it doesn't do the same for the original ratio of quadratics. So we must remove x = -2 from the domain.

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Hello!

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\boxed{=-7y+14}

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