I'm pretty sure A, and C but I'm not very sure about B I hope that helped.
Let's do

Release k for some while
If
So
So vertical asymptote is at origin now
It mentioned that it's at x=-5 so we need to change x


- Vertical asymptote at x=-5
Now
- for k=0 horizontal asymptote at origin
But it's given
Same put y=12 in place of k

Graph attached for verification
Answer:
A. 1
B.
.
Step-by-step explanation:
We have been given an equation of a line
.
A. We know that slope of parallel lines is always equal,
We can see that slope of our given line is 1, therefore the slope of the line parallel to our given line would be 1.
B. We know that the product of slopes of two perpendicular lines is
.
Let m represent slope of perpendicular to our given line, then:


Therefore the slope of the line perpendicular to our given line would be
.
3 is correct , base is never equal to 1 cause all of solves of logarithm will be 0
Answer:
y=81x^4+648x^3+1944x^2+2592x+1289
Step-by-step explanation:
1. Interchange x and y.
- y=\frac{\sqrt[4]{x+7}}{3}-2
- x=\frac{\sqrt[4]{y+7}}{3}-2
2. Then solve x=\frac{\sqrt[4]{y+7}}{3}-2 for y.
3. y=81x^4+648x^3+1944x^2+2592x+1289
I hope this helps.