Question

question 1-11
determine the inverse equation of ( y = (x - 4)^2 + 2 ), ( x > 4 ).
\\( \circ\\ y = (x - 2)^2 + 4 \\)
\\( \circ\\ y = \sqrt{x - 2} + 4 \\)
\\( \circ\\ y = (x^2 - 8x + 18)^{-1} \\)
\\( \circ\\ y = \sqrt{x - 4} \\)

Explanation:

Step1: Swap x and y

Given the equation \( y=(x - 4)^2+2 \) with \( x>4 \), we first swap \( x \) and \( y \) to get \( x=(y - 4)^2+2 \).

Step2: Solve for y

Subtract 2 from both sides: \( x - 2=(y - 4)^2 \).
Since \( x>4 \), the original function \( y=(x - 4)^2+2 \) is increasing for \( x>4 \), so we take the square root of both sides. We get \( \sqrt{x - 2}=y - 4 \) (we take the positive square root because the original function has \( x>4 \), so the inverse should be a function, and the positive root will give the correct inverse).
Then add 4 to both sides: \( y=\sqrt{x - 2}+4 \).

Answer:

\( y=\sqrt{x - 2}+4 \) (the second option)