How can you tell from the equation of a rational function if the function has a hole in the graph ( a removable discontinuity) a
1 answer:
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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Answer:
We just need to evaluate and get f(2i)=0, f(-2i)=0.
Step-by-step explanation:
Since , then
, and we can apply this when we evaluate
for 2i and -2i.
First we have:
Which shows that 2i is a zero of f(x).
Then we have:
Which shows that -2i is a zero of f(x).