Analyze the function `R(x)=x^2+x-30/(x+3)` :
By inspection this is a rational function.
(1) The y-intercept is found by letting x be 0: `x=0==>y=-10`
Thus the y-intercept is y=-10
The x-intercept(s) are found by solving `R(x)=0` :
`x^2+x-30/(x+3)=0==>x^2+x=30/(x+3)` Multiply both sides by x+3:
`x-2=0==>x=2` but `x^2+6x+15=0` has no real zeros.
Thus the x-intercept is x=2
(2) The domain of the function is all real numbers except -3 (you cannot divide by zero). There is a vertical asymptote at x=-3.
For large values of x (positive or negative), the expression `30/(x+3)` goes to zero. (If you are in a calculus class take the limits). Thus `R(x)` looks like `R^*(x)=x^2+x` for large x, so there are no horizontal asymptotes. (The function `R(x)` approaches `y=x^2+x` asymptotically)
(3) Investigating near x=-3 we find on the right side the function goes to negative infinity. (The term `30/(x+3)` grows without bound as x approches -3 from the right -- since we are subtracting this term, the function decreases without bound)
As x approaches -3 from the left, the term `30/(x+3)` decreases without bound -- subtracting a largenegative makes a large positive, so the function increases without bound.
The graph; the first graph shows details near 0, while the second graph shows the "big picture":