Question

The equation of the tangent to the curve $y = x^3 - 6x + 5$ at $(2, 1)$ is:

• A. $6x - y - 11 = 0$
• B. $6x - y - 13 = 0$
• C. $6x + y + 11 = 0$
• D. $6x - y + 11 = 0$
• E. $x - 6y - 11 = 0$

Hint:

Always check if the point $(2, 1)$ satisfies your final equation. $6(2) - 1 - 11 = 12 - 12 = 0$. Since it satisfies the equation, your calculation is likely correct.

Answer 1

The Correct Option is A

Concept: The equation of a tangent line at a point $(x_1, y_1)$ is given by $y - y_1 = m(x - x_1)$, where the slope $m$ is the value of the derivative $\frac{dy}{dx}$ at that point.

Step 1: Find the derivative of the curve.
Given $y = x^3 - 6x + 5$: \[ \frac{dy}{dx} = 3x^2 - 6 \]

Step 2: Calculate the slope at $(2, 1)$.
Substitute $x = 2$ into the derivative: \[ m = 3(2)^2 - 6 = 3(4) - 6 = 12 - 6 = 6 \]

Step 3: Form the equation of the tangent line.
Using point-slope form with $m = 6$ and $(x_1, y_1) = (2, 1)$: \[ y - 1 = 6(x - 2) \] \[ y - 1 = 6x - 12 \] Rearrange into the standard form $Ax + By + C = 0$: \[ 6x - y - 12 + 1 = 0 \] \[ 6x - y - 11 = 0 \]