151k views
1 vote
If y = 2^2 - 12x - p > 0, where p is a constant, find the range of values of p.

(a) p < -18
(b) p > -18
(c) p < 18
(d) p > 18

User Uri Lukach
by
7.4k points

1 Answer

0 votes

Final answer:

The range of values of p in the equation y = 2^2 - 12x - p > 0 is p ≤ -12x, so the correct option is p < -18.

Step-by-step explanation:

To find the range of values of p, we need to solve the equation y = 2^2 - 12x - p > 0 for p. Let's analyze the expression 2^2 - 12x - p. Since the exponent is 2 and positive, the term 2^2 is always positive. The term -12x can be positive or negative depending on the value of x. To make sure the entire expression is greater than 0, we want the sum of the terms -12x and -p to be less than or equal to 0. So, the range of values for p can be expressed as p ≤ -12x. Therefore, the correct option is (a) p < -18.

User Daja
by
7.5k points