Question

Find the equation of the tangent to the curve $ y = x^3 - 3x + 1 $ at the point where $ x = 2 $.

• A. \( y = 9x - 19 \)
• B. \( y = 9x - 15 \)
• C. \( y = 13x - 23 \)
• D. \( y = 15x - 25 \)

Hint:

Tip: Always compute derivative carefully and verify point coordinates before finalizing tangent line.

Answer 1

The Correct Option is B

To determine the equation of the tangent to the curve \( y = x^3 - 3x + 1 \) at \( x = 2 \), perform the following steps:

1. Calculate the derivative of \( y \) with respect to \( x \) to obtain the tangent line's slope. The derivative is \( y' = 3x^2 - 3 \).
2. Substitute \( x = 2 \) into the derivative to find the slope at that point: \( y'(2) = 3(2)^2 - 3 = 12 - 3 = 9 \).
3. The slope of the tangent line is \( 9 \).
4. Determine the y-coordinate of the point on the curve when \( x = 2 \) by substituting \( x = 2 \) into the original equation: \( y = (2)^3 - 3(2) + 1 = 8 - 6 + 1 = 3 \). The point is \( (2, 3) \).
5. Apply the point-slope form of a line equation, \( y - y_1 = m(x - x_1) \), with \((x_1, y_1) = (2, 3)\) and \(m = 9\):
   \( y - 3 = 9(x - 2) \)
6. Simplify the equation:
   \( y - 3 = 9x - 18 \)
   \( y = 9x - 18 + 3 \)
   \( y = 9x - 15 \)

The equation of the tangent line is \( y = 9x - 15 \).