Which equation is the inverse of [tex][tex]$5y + 4 = (x + 3)^2 + \frac{1}{2}$[/tex][/tex]?

A. [tex][tex]$y = \frac{1}{5}x^2 + \frac{6}{5}x + \frac{11}{10}$[/tex][/tex]
B. [tex][tex]$y = 3 \pm \sqrt{5x + \frac{7}{2}}$[/tex][/tex]
C. [tex][tex]$-5y - 4 = -(x + 3)^2 - \frac{1}{2}$[/tex][/tex]
D. [tex][tex]$y = -3 \pm \sqrt{5x + \frac{7}{2}}$[/tex][/tex]



Answer :

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]

Other Questions