Asap 50 points] 17, 18, 19, 20

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.

if the eqn of the parabola is f(x) = (x-4)(x+2), we need to mult. this out to find the coordinates of the vertex.

then f(x) = x^2 + 2x - 4x - 8, and f(x) = x^2 - 2x - 8.

one way in which you can find the vertex is to calculate x = -b / (2a).   here, that comes to

        )

x = = 1.   f(1) = 1^2 - 2(1) - 8 = - 9.   vertex is at (1, -9).

        2(1) 

find another point on the curve:   let x=0, for which y=-8.   then (0, -8) is on the curve.   plot the vertex and this point and draw a smooth parabola thru both.

hope this

the answer is d on edge

step-by-step explanation: