Write an equation of a rational function with these conditions: No Vertical Asymptote Horizontal Asymptote at y=5 Y-intercept at (0,3)

The function has no vertical Asymptote means the denominator does not  equate to 0 at any value of x. In other words the polynomial should not have real roots. A simple form of this type function of function would be `ax^2+1` .

The function has a horizontal Asymptote at y=5. So the polynomial of the numerator would have a type like `5x^2+bx+c` . 

So from these data we can say the function is;

`f(x) = (5x^2+bx+c)/(ax^2+1)`

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The function has no vertical Asymptote means the denominator does not  equate to 0 at any value of x. In other words the polynomial should not have real roots. A simple form of this type function of function would be `ax^2+1` .

The function has a horizontal Asymptote at y=5. So the polynomial of the numerator would have a type like `5x^2+bx+c` . 

So from these data we can say the function is;

`f(x) = (5x^2+bx+c)/(ax^2+1)`

It is given that at x = 0 then y = 3.

`3 = c/1`

`c = 3`

`f(x) = (5x^2+bx+3)/(ax^2+1)`

So a and b can be any rational value where `a!=0` .

A simple form of the answer would be at a = 1 and b = 0;

`f(x) =(5x^2+3)/(x^2+1)`

So the answer can be given as;

`f(x) =(5x^2+3)/(x^2+1)`