Answer:
1.
Vert. asymptote: x = {-3, 2}
Horiz. asymptote: y = 0
x-int: None
Question 3.
a. There is no hole
b. Vert. asymptote: x = {-2, 2}
c. f(x) = 0: x = {0, -1/2}
d. The graph has no hole at (-2, 4)
Question 4.
a. Vert. asymptote: x = {-2, 2}
b. f(x) = 0: x = {0, -1/2}
c. Horizontal asymptote: y = 2
d. The graph has no hole
I'm a bit confused. Some of the things stated in the question aren't true like how there are holes in places where there aren't.
You haven't shared the given line, so all I can do here is to invent a line and then show you how to write the equation of a new line which is parallel to mine and which has an x-intercept of 4.
My invented line: y = (2/3)x + 3
The new line MUST have the same slope: m = 2/3.
Then y = mx + b becomes y = (2/3)x + b. Find the x-intercept by setting y = 0 and solving for x: (2/3)x = 0 - b. Now replace x with 4 and find b:
-b = (2/3)(4) = 8/3. Then b = -8/3, and the new line is
y = (2/3)x - 8/3.
You’re answer is y=-1/1.5x+2.5
Explanation
2.5 is you’re y intersect and you go over one < and up 1.5
Answer:
1. 6x + 25
2. 60 + 28k
3. 5 + 19y
4. 3d + 1
5. 5v
6. w - 14.04
Step-by-step explanation:
These are what I got. I'm very sorry if I'm wrong.
