I am sure you meant
y= (3x^2+10x-8)/(x^2+7x+12 )
= (x+4)(3x-2)/(x+3)(x+4)
= (3x-2)/(x+3)
so you have a vertical asymptotes at x = -3
since we cancelled (x+4)/(x+4) and that would be 0/0 if x = -4
we have a hole at x = -4 , y = -14/-1 or 14, hole at (-4,14)
since the numerator is degree 3 and the denominator of degree 2, there is no vertical asymptote but the functions approaches y = 3x
confirmation by Wolfram:
http://www.wolframalpha.com/input/?i=plot+y%3D+(3x%5E2%2B10x-8)%2F(x%5E2%2B7x%2B12+)
Determine the holes, vertical asymptotes and horizontal asymptotes of the
rational function. y=3x^2+10x-8/x^2+7x+12
Can you also show your work. I need to do corrections. I have a 68 I just need to get to a 70. This would mean so much to me if anybody could help. I'm begging here~~~
1 answer
1 answer