Question

Which equation is the inverse of 5y+4 = (x+3)^2 + 1/2?

Answer 1

we have

$5y+4=(x+3)^{2}+\frac{1}{2}$

Step 1

Swap x and y

$5x+4=(y+3)^{2}+\frac{1}{2}$

Step 2

Solve for y

$5x+4-\frac{1}{2}=(y+3)^{2}$

$(y+3)^{2}=5x+\frac{7}{2}$

Square root both sides

$(y+3)=(+/-)\sqrt{5x+\frac{7}{2}}$

$y=(+/-)\sqrt{5x+\frac{7}{2}}-3$

Step 3

Let

$f(x)^{-1}=y$

$f(x)^{-1}=(+/-)\sqrt{5x+\frac{7}{2}}-3$

therefore

the answer is the option D

$y=-3(+/-)\sqrt{5x+\frac{7}{2}}$

Answer 2

5y+4=(x+3)^2+1/2

(x+3)^2=5y+3.5

x+3=√(5y+3.5)

x=√(5y+3.5)-3

y=-3±√(5x+7/2)