which equation is the inverse of $(x-4)^2 - \frac{2}{3} = 6y - 12$?

$y = \frac{1}{6}x^2 - \frac{4}{3}x + \frac{43}{9}$

$y = 4 \pm \sqrt{6x - \frac{34}{3}}$

$y = -4 \pm \sqrt{6x - \frac{34}{3}}$

$-(x-4)^2 - \frac{2}{3} = -6y + 12$

Question

which equation is the inverse of $(x-4)^2 - \frac{2}{3} = 6y - 12$?

$y = \frac{1}{6}x^2 - \frac{4}{3}x + \frac{43}{9}$

$y = 4 \pm \sqrt{6x - \frac{34}{3}}$

$y = -4 \pm \sqrt{6x - \frac{34}{3}}$

$-(x-4)^2 - \frac{2}{3} = -6y + 12$

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Finding Inverses Algebraically" ], "new_concepts": [], "current_concepts": [ "Finding Inverses Algebraically", "Inverse Functions" ] } ## Swap variables to set up the inverse relation $$ (y-4)^2 - \frac{2}{3} = 6x - 12 $$ ## Isolate the squared term $$ \begin{aligned} (y-4)^2 &= 6x - 12 + \frac{2}{3} \ (y-4)^2 &= 6x - \frac{34}{3} \end{aligned} $$ ## Solve for y $$ \begin{aligned} y-4 &= \pm\sqrt{6x - \frac{34}{3}} \ y &= 4 \pm\sqrt{6x - \frac{34}{3}} \end{aligned} $$

(A) $y = \frac{1}{6}x^2 - \frac{4}{3}x + \frac{43}{9}$

(B) $y = 4 \pm \sqrt{6x - \frac{34}{3}}$

(C) $y = -4 \pm \sqrt{6x - \frac{34}{3}}$

(D) $-(x-4)^2 - \frac{2}{3} = -6y + 12$

{ "subject": "Mathematics", "question_type": "Multiple Choice", "knowledge_point": [ "Mathematics", "Algebra", "Finding Inverses Algebraically" ] }

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QUESTION IMAGE

Question

Explanation:

Swap variables to set up the inverse relation

Isolate the squared term

Solve for y

Answer: