QUESTION IMAGE
Question
which equation is the inverse of $5y + 4 = (x + 3)^2 + \frac{1}{2}$?
\\(\circ\\) $y = \frac{1}{5}x^2 + \frac{6}{5}x + \frac{11}{10}$
\\(\circ\\) $y = 3 \pm \sqrt{5x + \frac{7}{2}}$
\\(\circ\\) $-5y - 4 = -(x + 3)^2 - \frac{1}{2}$
\\(\circ\\) $y = -3 \pm \sqrt{5x + \frac{7}{2}}$
Step1: Swap x and y
To find the inverse of a function, we first swap the variables \( x \) and \( y \) in the original equation. The original equation is \( 5y + 4=(x + 3)^{2}+\frac{1}{2} \). After swapping \( x \) and \( y \), we get \( 5x+4=(y + 3)^{2}+\frac{1}{2} \).
Step2: Solve for y
First, we isolate the squared term. Subtract \( \frac{1}{2} \) from both sides:
\( 5x + 4-\frac{1}{2}=(y + 3)^{2} \)
Simplify the left - hand side: \( 4-\frac{1}{2}=\frac{8 - 1}{2}=\frac{7}{2} \), so the equation becomes \( 5x+\frac{7}{2}=(y + 3)^{2} \)
Then, take the square root of both sides. Remember that when we take the square root of a number, we have both a positive and a negative solution. So we get \( y + 3=\pm\sqrt{5x+\frac{7}{2}} \)
Finally, solve for \( y \) by subtracting 3 from both sides: \( y=-3\pm\sqrt{5x+\frac{7}{2}} \)
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\( y=-3\pm\sqrt{5x+\frac{7}{2}} \) (the fourth option: \( y=-3\pm\sqrt{5x+\frac{7}{2}} \))