Find the inverse of the function:

[tex]y = x^2 + 4x + 4[/tex]

A. [tex]y = \sqrt{\frac{1}{2}x + 2}[/tex]

B. [tex]y = \pm \sqrt{\frac{1}{2}x + 2}[/tex]

C. [tex]y = \pm \sqrt{x} - 2[/tex]

D. [tex]y = \sqrt{x} - 2[/tex]



Answer :

To find the inverse of the function [tex]\( y = x^2 + 4x + 4 \)[/tex], we need to reverse the roles of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] and then solve for [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex]. Here are the detailed steps to find the inverse:

1. Express the given function:
[tex]\[ y = x^2 + 4x + 4 \][/tex]

2. Rewrite the equation by switching [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x = y^2 + 4y + 4 \][/tex]

3. Solve the quadratic equation for [tex]\( y \)[/tex]:
Rearrange the equation:
[tex]\[ y^2 + 4y + 4 = x \][/tex]
Subtract [tex]\( x \)[/tex] from both sides:
[tex]\[ y^2 + 4y + 4 - x = 0 \][/tex]

4. Use the quadratic formula to solve for [tex]\( y \)[/tex]:
The quadratic formula is given by:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
For our equation [tex]\( y^2 + 4y + (4 - x) = 0 \)[/tex], [tex]\( a = 1 \)[/tex], [tex]\( b = 4 \)[/tex], and [tex]\( c = 4 - x \)[/tex]. Plugging these values into the quadratic formula:
[tex]\[ y = \frac{-4 \pm \sqrt{16 - 4 \cdot 1 \cdot (4 - x)}}{2 \cdot 1} \][/tex]
[tex]\[ y = \frac{-4 \pm \sqrt{16 - 16 + 4x}}{2} \][/tex]
Simplify under the square root:
[tex]\[ y = \frac{-4 \pm \sqrt{4x}}{2} \][/tex]
Simplify further:
[tex]\[ y = \frac{-4 \pm 2\sqrt{x}}{2} \][/tex]
[tex]\[ y = -2 \pm \sqrt{x} \][/tex]

5. Express the inverse function:
Thus, we have two potential solutions for the inverse:
[tex]\[ y = -2 + \sqrt{x} \quad \text{and} \quad y = -2 - \sqrt{x} \][/tex]

Hence, the correct inverse functions of the given function [tex]\( y = x^2 + 4x + 4 \)[/tex] are:
[tex]\[ \boxed{-2 + \sqrt{x} \quad \text{and} \quad -2 - \sqrt{x}} \][/tex]

From the given answer choices, these match:
[tex]\[ y = \pm \sqrt{x} - 2 \][/tex]