which equation is the inverse of $y = 16x^2 + 1$?

$y = \pm \sqrt{\frac{x}{16} - 1}$

$y = \frac{\pm \sqrt{x - 1}}{16}$

$y = \frac{\pm \sqrt{x}}{4} - \frac{1}{4}$

$y = \frac{\pm \sqrt{x - 1}}{4}$

Question

which equation is the inverse of $y = 16x^2 + 1$?

$y = \pm \sqrt{\frac{x}{16} - 1}$

$y = \frac{\pm \sqrt{x - 1}}{16}$

$y = \frac{\pm \sqrt{x}}{4} - \frac{1}{4}$

$y = \frac{\pm \sqrt{x - 1}}{4}$

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" ] } ## Swap variables to set up the inverse relation $$ x = 16y^2 + 1 $$ ## Isolate the quadratic term $$ \begin{aligned} x - 1 &= 16y^2 \ y^2 &= \frac{x - 1}{16} \end{aligned} $$ ## Solve for y by taking the square root $$ y = \pm\sqrt{\frac{x - 1}{16}} = \frac{\pm\sqrt{x - 1}}{4} $$

(A) $y = \pm\sqrt{\frac{x}{16} - 1}$

(B) $y = \frac{\pm\sqrt{x - 1}}{16}$

(C) $y = \frac{\pm\sqrt{x}}{4} - \frac{1}{4}$

(D) $y = \frac{\pm\sqrt{x - 1}}{4}$

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

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Swap variables to set up the inverse relation

Isolate the quadratic term

Solve for y by taking the square root

Answer: