Answer:
define f(2) = 11/8
Step-by-step explanation:
To make the graph have a horizontal asymptote at y=3, the coefficient of x^2 in the numerator must be 3 times the coefficient of x^2 in the denominator. That is, a = 3·2 = 6.
In order for there to be a removable discontinuity at x=2, we must have (x-2) a factor of both numerator and denominator. We already know the denominator factors as 2(x-2)(x+2), so we need to find the value of b that makes the numerator have a factor of (x-2).
We can find the required value of "b" using synthetic division. The work shown in the attachment tells us that b=-13 will make division of the numerator by (x-2) have a remainder of zero. So the function is ...
[tex]f(x)=\dfrac{6x^2-13x+2}{2(x^2-4)}=\dfrac{(6x-1)(x-2)}{2(x+2)(x-2)}[/tex]
Canceling the factors of (x-2), the simplified function is ...
[tex]f_1(x)=\dfrac{6x-1}{2(x+2)}\\\\f_1(2)=\dfrac{6(2)-1}{2(2+2)}=\dfrac{11}{8}[/tex]
In order to remove the discontinuity at x=2, the function value must be defined as 11/8.
