Go to Desmos.com to graph your functions
If these of your functions
s(x) = (x - 1)1/2 + 3
x must be 1 or greater because the square root for a negative number does not exist in the set of real numbers
y will always be 3 or greater
your plot starts at x = 1
your plot starts at y = 3
your plot starts at (1,3) and keeps increasing based your positive input of x values
t(x) = -1/(x +3)2
Negative 3 is a vertical asymptote since it yields division by zero and the y value of your function will always be less than zero, because of the negative sign, and anything squared in the denominator yields a postive number, but a negative/positive is always a negative check it at x = -4
