Answer:
is valid for all values of x ≥ 0
Step-by-step explanation:
Select all of the statements that are true about the following expression: -4x^3+3x^2-1
1 answer:
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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.
