Answer:
d and e are 1 and 5
b and c are -1 and 2
a is -1
Explanation:
To find b and c, we have to notice that if (x+b) or (x+c) is equal to zero, then f(x) is equal to 0. We see two xs where it happens. 1 and -2.
Let's assume that f(1)=0 corresponds to (x+b)=0 and f(-2)=0 corresponds to (x+c)=0 (could also be the other way around).
(1+b) = 0
b = -1
(-2+c) = 0
c = 2
To find d and e, we have to notice that if (x+d) or (x+e) is equal to zero, then the denominator is equal to zero, and as such, the function has no value. We clearly see two xs where it happens, -1 and -5.
As before (also interchangably between d and e)
(-1+d)=0
d = 1
(-5+e)=0
e = 5
To find a, the easiest way is to notice that the limit of the function is a. It happens because for very big xs, the x^2 component of both the numerator and denominator dominates other components, see
f(x) = a(x+b)(x+c)/((x+d)(x+e)) = a(x^2 + bx + cx + bc)/(x^2 + dx + ex + de)
because x is the only variable, we can set it so that x^2 is arbitrarily bigger than bx+cx+bc or dx+ex+de. So the value at x -> infinity is approaching a.
We were given the asymptote at y = -1, so a = -1
Another thing you COULD do is approximate using e.g. x = -3, where f(-3) appears to be around 1 (looking at the graph).
f(-3) = a * (-3-1)(-3+2) / ((-3+1)(-3+5)) = a * -4 * -1 / (-2 * 2) = a * -1
so if f(-3) is about 1, then a is about 1/-1 so a is about -1. I'm saying "about" because it's not proper mathematics to just read the graph like this. But it gives us the right answer in this case.