The slope of the normal to the curve \(y = 2x^2\) at \(x = 1\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → −1/4
The slope of the tangent is the derivative:
\( \displaystyle \frac{dy}{dx} = 4x \)
\(\text{At } x = 1:\quad m_{\text{tangent}} = 4\)
The normal is perpendicular to the tangent, so its slope is the negative reciprocal:
\( \displaystyle m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{4} \)
Why the other options are incorrect
Option (1) 4 — Incorrect. This is the slope of the tangent, not of the normal.
Option (2) −4 — Incorrect. Only the sign of the tangent slope has been changed; the reciprocal is also needed.
Option (3) 1/2 — Incorrect. The slope of the normal at x = 1 is −1/4, not 1/2.