Question

Imaginary zeros of y = 1/4(x-4)^2 +2

Answer 1

Answer:y=4+2√(2)i and y=4-2√(2)i

Explanation:

y= 1/4(x-4)^2 +2

y=1/4(x²-8x+16)+2

y=(1/4)x²-2x+4+2

y=(1/4)x²-2x+6

using the Quadratic Equation

y=(-b±√(b²-4ac))/(2a)

Where a=1/4, b=-2, c=6

y=(-(-2)±√((-2)²-4(1/4)(6)))/(2(1/4))

y=(2±√(4-6))/(1/2)

y=4±2√(-2)

y=4±2√(2)i

y=4+2√(2)i and y=4-2√(2)i