which equation is the inverse of $5y + 4=(x + 3)^2+\frac{1}{2}$?\n$y=\frac{1}{5}x^2+\frac{6}{5}x+\frac{11}{10…

Explanation:

Step1: Isolate the y - term

First, start with the equation $5y + 4=(x + 3)^2+\frac{1}{2}$. Subtract 4 from both sides: $5y=(x + 3)^2+\frac{1}{2}-4$. Simplify the right - hand side: $\frac{1}{2}-4=\frac{1 - 8}{2}=-\frac{7}{2}$, so $5y=(x + 3)^2-\frac{7}{2}$.

Step2: Solve for y

Divide both sides by 5: $y=\frac{1}{5}(x + 3)^2-\frac{7}{10}$. To find the inverse, swap x and y: $x=\frac{1}{5}(y + 3)^2-\frac{7}{10}$.

Step3: Isolate the squared term

Add $\frac{7}{10}$ to both sides: $x+\frac{7}{10}=\frac{1}{5}(y + 3)^2$. Multiply both sides by 5: $5x+\frac{7}{2}=(y + 3)^2$.

Step4: Solve for y

Take the square root of both sides: $y + 3=\pm\sqrt{5x+\frac{7}{2}}$. Subtract 3 from both sides: $y=-3\pm\sqrt{5x+\frac{7}{2}}$.

Answer:

$y=-3\pm\sqrt{5x+\frac{7}{2}}$