Which equation is the inverse of y = 16×2 + 1? D. y= +or- sqrt x-1/4 is the correct answer
The correct answer and explanation is:
To find the inverse of the equation: y=16×2+1y = 16x^2 + 1
we must follow a step-by-step algebraic process to switch the input and output variables and solve for the new dependent variable.
Step 1: Replace yy with f(x)f(x)
f(x)=16×2+1f(x) = 16x^2 + 1
Step 2: Swap xx and yy
x=16y2+1x = 16y^2 + 1
This step is essential because inverses reverse the roles of inputs and outputs.
Step 3: Solve for yy
Start with: x=16y2+1x = 16y^2 + 1
Subtract 1 from both sides: x−1=16y2x – 1 = 16y^2
Divide both sides by 16: x−116=y2\frac{x – 1}{16} = y^2
Now take the square root of both sides: y=±x−116y = \pm \sqrt{\frac{x – 1}{16}}
This simplifies to: y=±x−14y = \pm \frac{\sqrt{x – 1}}{4}
Final Answer:
The inverse equation is: y=±x−14y = \pm \frac{\sqrt{x – 1}}{4}
Which matches choice D (when written as): y=±x−116ory=±x−14y = \pm \sqrt{\frac{x – 1}{16}} \quad \text{or} \quad y = \pm \frac{\sqrt{x – 1}}{4}
Interpretation:
- The ± (plus or minus) symbol appears because the original function y=16×2+1y = 16x^2 + 1 is not one-to-one (it is a parabola), and its inverse will not pass the vertical line test unless we restrict the domain.
- This means for each xx in the inverse, there are two possible values of yy, unless a restriction is applied (e.g., x≥0x \geq 0).
Conclusion:
✅ Correct answer: D. y=±x−116ory=±x−14y = \pm \sqrt{\frac{x – 1}{16}} \quad \text{or} \quad y = \pm \frac{\sqrt{x – 1}}{4}
This is the inverse of y=16×2+1y = 16x^2 + 1.