The tangent to the curve $y=x^3-11 x+5$ at $x=2$ is parallel to :
(A) $y=x+1$
(B) $y=-x+2$
(C) $2 y=x+3$
(D) $2 y=2 x+4$
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
$y = x^3 - 11x + 5$
so $y' = 3x^2 - 11$ (slope of curve is derivative)
at x = 2 $\left.y'\right]_{x=2}= 3(2)^2 - 11 = 12 - 11 = 1$
slope = 1 → it is parallel to line having slope (1)
| lines | derivative | slope = derivative |
| y = x + 1 | y' = 1 | 1 |
| y = -x + 2 | y' = -1 | -1 |
| 2y = x + 3 | y' = $\frac{1}{2}$ | $\frac{1}{2}$ |
| 2y = 2x + 4 | y' = 1 | 1 |
lines with some slopes are parallel to curve at that point