What is the domain of #(-3x^2)/(x^2+4x-45)#?

1 Answer
Jun 19, 2015

Answer:

The domain is all the real #x# except:
#x=-9# and #x=5#

Explanation:

In this division you must ensure to avoid a division by zero, i.e., to have a zero in the denominator.
The denominator is equal to zero when:
#x^2+4x-45=0#
This is a quadratic equation that you can solve, say, using the Quadratic Formula.
So:
#x_(1,2)=(-4+-sqrt(16+180))/2=(-4+-14)/2=#
so you have two values of #x# that makes the denominator equal to zero:
#x_1=(-4+14)/2=5#
#x_2(-4-14)/2=-9#
These two values cannot be used by your function. All the other values of #x# are allowed: