17) f(x) = 16/(13-x).
In order to find domain, we need to set denominator expression equal to 0 and solve for x.
And that would be excluded value of domain.
13-x =0
Adding x on both sides, we get
13-x +x = x.
13=x.
Therefore, domain is All real numbers except 13.
18).f(x) = (x-4)(x+9)/(x^2-1).
In order to find the vertical asymptote, set denominator equal to 0 and solve for x.
x^2 -1 = 0
x^2 -1^2 = 0.
Factoring out
(x-1)(x+1) =0.
x-1=0 and x+1 =0.
x=1 and x=-1.
Therefore, Vertical asymptote would be
x=1 and x=-1
19) f(x) = (7x^2-3x-9)/(2x^2-4x+5)
We have degrees of numberator and denominator are same.
Therefore, Horizontal asymptote is the fraction of leading coefficents.
That is 7/2.
20) f(x)=(x^2+3x-2)/(x-2).
The degree of numerator is 2 and degree of denominator is 1.
2>1.
Degree of numerator > degree of denominator .
Therefore, there would no any Horizontal asymptote.
Parenthases
Equation
Multiplication
Division
Addition
Subtraction
(-3y^2 − 8) − (-5y^2<span> + 1)
= </span>-3y^2 − 8 + 5y^2<span> - 1
= 2</span>y^2 <span>- 9
answer is
</span>2y^2 - 9<span>
</span>
I think its 8 im not sure but i have the same paper and she marked mine right
