let f(y) = (y^2 + 11y + 30)/(y+5) <---- those brackets are needed or else we would have a term of 30/y sitting there all by itself
= (y+5)(y+6)/(y+5)
= y+6
That is, f(y) behaves just like a linear function of
f(y) = y+6, EXCEPT when y = -5
in the original function we get
f(-5) = 0/0
in the reduced function we get f(-5) = 1
conclusion:
domain is any real number of y , except y = -5
in the 2nd:
f(a) = (5a - 3)/(a^2 - 49(
= (5a - 3)/((a+7)(a-7))
here nothing divides out, so we have to leave it like that.
notice when a = 7 , we get f(7) = 32/0 <--- undefined
when a = -7 , we get f(-7) = -38/0 <--- undefined
domain: all real values of a, except ±7
The undefined at ±7 results in vertical asymptotes at those two values, BUT in your first expression we had 0/0, which is called indeterminate instead of undefined.
With a good calculator, pick a value of y close to -5, say a = -4.9999, on mine f(-4.9999) = 1.0001 , close to 1
try a = -5.000001 and see what you get
so in our first expression the value of 0/0 gets closer to 1, the closer we get to y = -5
The expression has a "hole" at (-5,1)
I will graph both, changing the independent variable to x
http://www.wolframalpha.com/input/?i=plot+y+%3D+(x%5E2+%2B+11x+%2B+30)%2F(x%2B5)
http://www.wolframalpha.com/input/?i=plot+y+%3D+(5x-3)%2F(x%5E2-49)+for+-10+%E2%89%A4+x+%E2%89%A4+10
I am having a hard time finding the domain, set notation,excluded value.
Rational expression 1.
Y^2+11y+30/y+5
Rational expression 2.
5a-3/a^2-49
1 answer
2 answers