44-20
-6-+20
-8a-8b
10b+40
-2p-14
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
Slope/gradient = 5
y= MX + C
(-6,-26)
-6 is x1 and -26 is y1
so therefore y= 5x -26