How do you find the domain for $f (q) = \frac{q + 1}{q^{2} + 6 q - 27}$ $f(q)=(q+1)/(q^2+6q-27)$ ?

Question

1402 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-05T11:57:22

The domain is $q \in (- \infty, - 9) \cup (- 9, 3) \cup (3, + \infty)$ $q in (-oo, -9) uu(-9,3)uu(3,+oo)$

The denominator is

$q^{2} + 6 q - 27 = (q - 3) (q + 9)$ $q^2+6q-27=(q-3)(q+9)$

As the denominator must be $\neq 0$ $!=0$

$(q - 3) (q + 9) \neq 0$ $(q-3)(q+9)!=0$

${(q-3!=0),(q+9!=0):}$

$=>$ , ${(q!=3),(q!=-9):}$

The domain is $q in (-oo, -9) uu(-9,3)uu(3,+oo)$

graph{(x+1)/(x^2+6x-27) [-20.28, 20.26, -10.14, 10.14]}

How do you find the domain for f(q)=(q+1)/(q^2+6q-27)f(q)=q+1q2+6q−27f(q)=(q+1)/(q^2+6q-27)?