Let that be

Two vertical asymptotes at -1 and 0

If we simply

- Denominator has degree 2
- Numerator should have degree as 2 and coefficient as 3 inorder to get horizontal asymptote y=3 means the quadratic equation should contain 3x²
- But there should be a x intercept at -3 so one zeros should be -3
Find a equation
Find zeros
Horizontal asymptote
So our equation is

Graph attached
Answer:
Graph A
Step-by-step explanation:
-y=-2-3x
y=3x+2
Slope:3
y-intercept:2
Answer:
16.5
Step-by-step explanation:
Answer:

Step-by-step explanation:
Let the equation of the perpendicular line is,
y = mx + b
where m = slope of the line
b = y-intercept
From the graph, slope of the line passing through (0, -1) and (3, 1),
m' =
m' = 
m' = 
To get the slope (m) of this line we will use the property of perpendicular lines,
m × m' = (-1)
m ×
= -1
m = 
Equation of the perpendicular line will be,

x-intercept of the line is (-3) therefore, point on the line is (-3, 0)
0 = 
b = 
Equation of the line will be,
