The slope of the normal to the curve y=2x^2-4 at P(1, −2) is:

The correct answer is Option (3) → \(-\displaystyle\frac{1}{4}\)

The slope of the tangent is the derivative:

    \( \displaystyle \frac{dy}{dx}=4x \)

At P(1, −2):

    \(\text{slope of tangent}=4(1)=4\)

The normal is perpendicular to the tangent, so its slope is the negative reciprocal:

    \( \displaystyle \text{slope of normal}=-\frac{1}{4} \)

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 is changed; the reciprocal is missing.

• Option (4) 0 — Incorrect. A normal of slope 0 would need a vertical tangent, but the tangent slope here is 4.