Answer: [tex]y=-3+-\sqrt{5x+\frac{7}{2}}[/tex]
Step-by-step explanation: Given equation [tex]5y+4=(x+3)^2+\frac{1}{2}[/tex].
We need to find it's inverse.
In order to find the inverse, we need to switch x and y's and solve for y.
Step 1: Switching x and y's, we get
[tex]5x+4=(y+3)^2+\frac{1}{2}[/tex]
Step 2: Solving it for y.
Subtracting [tex]\frac{1}{2}[/tex] from both sides, we get
[tex]5x+4-\frac{1}{2}=(y+3)^2+\frac{1}{2}-\frac{1}{2}[/tex]
[tex]5x+4-\frac{1}{2}=(y+3)^2[/tex]
[tex]5x+\frac{8}{2}-\frac{1}{2}=(y+3)^2[/tex]
[tex]5x+\frac{7}{2}=(y+3)^2[/tex]
Taking square root on both sides, we get
[tex]\sqrt{5x+\frac{7}{2}}=\sqrt{(y+3)^2}[/tex]
[tex]\sqrt{5x+\frac{7}{2}}=y+3[/tex]
Subtracting 3 from both sides, we get
[tex]\sqrt{5x+\frac{7}{2}}-3=y[/tex]
Switching sides, we get
[tex]y=-3+-\sqrt{5x+\frac{7}{2}}[/tex]
Therefore, correct option is D option [tex]y=-3+-\sqrt{5x+\frac{7}{2}}[/tex].
