Final answer:
The function H(x)=(x-2)² +32 has no zeros.
Step-by-step explanation:
The zeros of the function H(x)=(x-2)² +32 can be determined by setting the function equal to zero and solving for x:
(x-2)² + 32 = 0
Expanding the squared term:
x² - 4x + 4 + 32 = 0
Combining like terms:
x² - 4x + 36 = 0
This is a quadratic equation that can be solved using the quadratic formula:
x = (-b ± √(b² - 4ac)) / (2a)
Plugging in the values a = 1, b = -4, and c = 36:
x = (-(-4) ± √((-4)² - 4(1)(36))) / (2(1))
Simplifying:
x = (4 ± √(16 - 144)) / 2
Further simplifying:
x = (4 ± √(-128)) / 2
The discriminant is negative, which means the quadratic equation has no real solutions. Therefore, the function H(x)=(x-2)² +32 has no zeros.