When b=0
f(x)=ax^2+c
test each
A. with x^2-1, A is false
B. with -x^2-1, B is false
C. cannot find contradiction
D. the axis is actually x=0, 0 is nithere positive nor negative, false
answer is C
It would be biased because one would assume that students can get to school by means other than those that put them 8n the parking lor. (Parking lot = car drivers who would be the only students sampled.). Others may ride to school by either parent or school bus who could drop them off somewhere else or possibly walk to school. The freshmen/sophomores would be excluded generally as they would be too young to drive altogether
General Idea:
Domain of a function means the values of x which will give a DEFINED output for the function.
Applying the concept:
Given that the x represent the time in seconds, f(x) represent the height of food packet.
Time cannot be a negative value, so

The height of the food packet cannot be a negative value, so

We need to replace
for f(x) in the above inequality to find the domain.
![-15x^2+6000\geq 0 \; \; [Divide \; by\; -15\; on\; both\; sides]\\ \\ \frac{-15x^2}{-15} +\frac{6000}{-15} \leq \frac{0}{-15} \\ \\ x^2-400\leq 0\;[Factoring\;on\;left\;side]\\ \\ (x+200)(x-200)\leq 0](https://tex.z-dn.net/?f=%20-15x%5E2%2B6000%5Cgeq%200%20%5C%3B%20%5C%3B%20%20%5BDivide%20%5C%3B%20by%5C%3B%20-15%5C%3B%20on%5C%3B%20both%5C%3B%20sides%5D%5C%5C%20%5C%5C%20%5Cfrac%7B-15x%5E2%7D%7B-15%7D%20%2B%5Cfrac%7B6000%7D%7B-15%7D%20%5Cleq%20%5Cfrac%7B0%7D%7B-15%7D%20%5C%5C%20%5C%5C%20x%5E2-400%5Cleq%200%5C%3B%5BFactoring%5C%3Bon%5C%3Bleft%5C%3Bside%5D%5C%5C%20%5C%5C%20%28x%2B200%29%28x-200%29%5Cleq%200%20)
The possible solutions of the above inequality are given by the intervals
. We need to pick test point from each possible solution interval and check whether that test point make the inequality
true. Only the test point from the solution interval [-200, 200] make the inequality true.
The values of x which will make the above inequality TRUE is 
But we already know x should be positive, because time cannot be negative.
Conclusion:
Domain of the given function is 