To find the inverse of the equation [tex]\( 5y + 4 = (x + 3)^2 + \frac{1}{2} \)[/tex], let us follow the steps below:
1. Isolate the [tex]\( y \)[/tex]-term in the given equation:
[tex]\[
5y + 4 = (x + 3)^2 + \frac{1}{2}
\][/tex]
Subtract 4 from both sides:
[tex]\[
5y = (x + 3)^2 + \frac{1}{2} - 4
\][/tex]
Combine the constants on the right side:
[tex]\[
5y = (x + 3)^2 - \frac{7}{2}
\][/tex]
2. Solve for [tex]\( y \)[/tex]:
Divide both sides by 5:
[tex]\[
y = \frac{1}{5} \left( (x + 3)^2 - \frac{7}{2} \right)
\][/tex]
Distribute the [tex]\(\frac{1}{5}\)[/tex]:
[tex]\[
y = \frac{1}{5} (x + 3)^2 - \frac{7}{10}
\][/tex]
3. Find the inverse function by swapping [tex]\( x \)[/tex] and [tex]\( y \)[/tex], and then solving for [tex]\( y \)[/tex]:
Start by swapping [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[
x = \frac{1}{5} (y + 3)^2 - \frac{7}{10}
\][/tex]
Multiply both sides by 5 to clear the fraction:
[tex]\[
5x = (y + 3)^2 - \frac{7}{2}
\][/tex]
Add [tex]\(\frac{7}{2}\)[/tex] to both sides:
[tex]\[
5x + \frac{7}{2} = (y + 3)^2
\][/tex]
Take the square root of both sides, remembering to include both the positive and negative roots:
[tex]\[
\sqrt{5x + \frac{7}{2}} = y + 3 \quad \text{or} \quad -\sqrt{5x + \frac{7}{2}} = y + 3
\][/tex]
Solving for [tex]\( y \)[/tex], we obtain two solutions:
[tex]\[
y = -3 + \sqrt{5x + \frac{7}{2}} \quad \text{or} \quad y = -3 - \sqrt{5x + \frac{7}{2}}
\][/tex]
We can combine these into one expression using the [tex]\( \pm \)[/tex] notation:
[tex]\[
y = -3 \pm \sqrt{5x + \frac{7}{2}}
\][/tex]
Thus, the equation which represents the inverse of the given equation is:
[tex]\[ \boxed{y = -3 \pm \sqrt{5x + \frac{7}{2}}} \][/tex]
The corresponding answer choice from the given options is:
[tex]\[ y = -3 \pm \sqrt{5x + \frac{7}{2}} \][/tex]
The correct answer is:
[tex]\[ \boxed{4} \][/tex]
