PART A: To determine k, you need to throw x + 3 into the original function:
[tex]f(x)= x^{2} -2x-24[/tex]
So, [tex]f(x+3)[/tex] would be:
[tex]f(x+3)= (x+3)^{2} -2(x+3)-24[/tex]
Doing some FOILing/distributing, we get:
[tex]f(x+3)= (x^{2}+6x+9) -2x-6-24[/tex]
Doing some clean up/combining like terms, we get:
[tex]f(x+3)= x^{2}+4x-21[/tex]
Now, it should be clear that k=4, since k is the coefficient in front of the x term (it's the "b" in [tex]ax^2+bx+c[/tex]).
PART B: There are several ways to find the zeros, but perhaps the easiest is to factor the result from Part A:
[tex]f(x+3)= x^{2}+4x-21=(x+7)(x-3)[/tex]
From those factors, we know that zeros occur at x = -7 and x = 3 (answers A and E).
By the way, you can check the logic of these zeros if you know about shifts/translations. [tex]f(x+3)[/tex] is a 3 unit left shift from the original function, which has zeros at x = -4 and x = 6. Well, x = -7 is 3 units to the left of x = -4. And x = 3 is 3 units to the left of x = 6. So everything seems to check out.