Question

Please solve both and tell me how!

Answer 1

PART A:
To determine k, you need to throw x + 3 into the original function:

$f(x)= x^{2} -2x-24$

So, $f(x+3)$ would be:

$f(x+3)= (x+3)^{2} -2(x+3)-24$

Doing some FOILing/distributing, we get:

$f(x+3)= (x^{2}+6x+9) -2x-6-24$

Doing some clean up/combining like terms, we get:

$f(x+3)= x^{2}+4x-21$

Now, it should be clear that k=4, since k is the coefficient in front of the x term (it's the "b" in $ax^2+bx+c$).

PART B:
There are several ways to find the zeros, but perhaps the easiest is to factor the result from Part A:

$f(x+3)= x^{2}+4x-21=(x+7)(x-3)$

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. $f(x+3)$ 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.