Question

Use the information below to sketch the function.
There is a horizontal asymptote at f (x)=1
The denominator of the function is x^2 + x - 6
There is a y intercept at 0.
f(-2)=f(-4)=0
f(4)=f(1)=f(2.5)=1
(all of the < > are less then and greater then no less then/greater the or equal to)
f(x) < 0 when - 4 < x < -3, - 2 < x < 0
f(x) > 0 when x < -4, -3 < x < -2, 0 < x < 2, x>2

Answer 1

There is a horizontal asymptote at f(x)=1
means that the degree of the numerator and denominator are the same, and the ratio of their leading coefficients is 1. That is, both are of the form
(x-a)(x-b)... with all the coefficients just 1

The denominator of the function is x^2 + x - 6 = (x-2)(x+3)

There is a y intercept at 0.
means that f(x) is of the form x^n/((x-2)(x+3))
so now were at the spot where it looks like
y = x^2/((x-2)(x+3))

Now we start getting into murky waters.
0 < x < 2, x>2
means that f(x) has a double root at x=2, so we need
y = x(x-2)^2/((x-2)(x+3))
But now we need a cubic polynomial in the bottom, but you said it was just x^2+x-6.

And the rest of the constraints make even more changes necessary. I suspect you have mangled the question somehow.

Answer 2

Nope I just checked. I made no typos.