Okay, wow thanks for the explanation!
What I was doing was sketching the individual graphs (x+1) and x^2-x-6 on the same axis, and determiningwhere the graph would be positive or negative (eg the numerator is negative from neginifity to -1 while denominator is positive from neg infinity to -2 ) dividing the values means the graph would be negative from negative infinity to - 2. I had trouble using calculus bc my knowledge is limited to unit1/2 methods calc, but i'm going to try your method from now on as it seems much more reliable.
Thank youu
1/2 methods should cover quotient rule and differentiation of polynomials, which is all you need to differentiate these functions (unless I'm mistaken with whether quotient rule is in the course; if not, learn it yourself. It's not difficult, just be careful with the subtraction). Let's see how much we can do without calculus though.
Asymptotes are at x = 3, x = -2, y = 0. The third one is a little more complicated to see, as if we let x->inf, we get inf/inf. To see this, divide the top and bottom by the highest power of x, which is x
2:
and it is readily seen that this vanishes as x becomes larger; the numerator goes to 0 and the denominator goes to 1.
Next, we want to see the shape of this function. The x-intercept is at x=-1, so we have to check from -inf to -2, -2 to -1, -1 to 3, 3 to inf (consider x intercepts and asymptotes).
If x > 3, then all three brackets are positive, and f(x) > 0. So, f(x) is a large positive number to the right of the asymptote x = 3 and approaches 0 for large positive x.
If -1 < x < 3, then x + 1 > 0, x + 2 > 0, but x - 3 < 0, so f(x) < 0. f(x) is large but negative to the left of the asymptote x = 3.
If -2 < x < -1, then x + 1 < 0, x - 3 < 0, x + 2 > 0 so f(x) > 0. f(x) is large and positive to the right of the asymptote x = -2.
If x < -2, then x + 1 < 0, x - 3 < 0, x + 2 < 0 so f(x) < 0. f(x) is large and negative to the left of the asymptote x = -2 and approaches 0 for large negative x.
Hopefully you can draw a graph with just this knowledge. It doesn't tell you if there are turning points, however.