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.
You might be interested in
Step-by-step explanation:
If a variables varies jointly, we can just divide it by the other variables in relation to it.
For example, since p variables jointly as q and square of r, then
where k is a constant
First, let find k. Substitute p= 200
q= 2, and r=3.
Now, since we know our constant, let find p.
Q is 5, and r is 2.