\(\frac{2x+3}{log_4x}>0\\ log_4x\qquad \text{Is positive when x>1}\\ if\;\;x>1\quad then\\ 2x+3>0\\ 2x>-3\\ x>-3/2\\ \text {So positive when x>1 }\)
\(\frac{2x+3}{log_4x}>0\\ log_4x\qquad \text{Is negative when 0<x<1}\\ if\;\; 0<x<1\quad then\\ 2x+3<0\\ 2x<-3\\ x<-3/2\\ \text {No solution here. }\)
So the domain of the solution is \(x>1\)
Here is the graph
https://www.desmos.com/calculator/fgp5ahuson
Here is one you tube, I have not watched it but it will probably help.
There are many others if you don't like this one :)
https://www.youtube.com/watch?v=1Lwi8cOdqCU
Parabola
\(\boxed{(x-h)^2=4a(y-k)}\)
Vertex is (h,k)
a not equal 0
concave up if a>0 positive
concave down if a<0 negative
The focal length is |a|
The focal point will be (h,k+a)
The length of the latus rerctum is 4a The equation of it will be y= k+a
so
The points at the end of the latus r****m will be (h-2a,k+a) and (h+2a,k+a)
What else do you want to know?
Here is an excellent graphing calc so you can play around with some different equations and see what happens.