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 find the equation of the tangent to the curve \( y = x^3 - 3x + 1 \) at the point where \( x = 2 \), we need to follow these steps:

1. First, compute the derivative of \( y \) with respect to \( x \) to find the slope of the tangent line. The derivative of \( y = x^3 - 3x + 1 \) is \( y' = 3x^2 - 3 \).
2. Evaluate the derivative at \( x = 2 \) to find the slope of the tangent line at this point: \( y'(2) = 3(2)^2 - 3 = 12 - 3 = 9 \).
3. The slope of the tangent line is \( 9 \).
4. Next, find the coordinates of the point on the curve when \( x = 2 \). Substitute \( x = 2 \) into the original equation: \( y = (2)^3 - 3(2) + 1 = 8 - 6 + 1 = 3 \). Thus, the point is \( (2, 3) \).
5. Using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is \((2, 3)\) and \(m = 9\), the equation becomes:
   
   \( y - 3 = 9(x - 2) \)
6. Simplify the equation:
   
   \( y - 3 = 9x - 18 \)
   \( y = 9x - 18 + 3 \)
   \( y = 9x - 15 \)

Therefore, the equation of the tangent line is \( y = 9x - 15 \), which matches the second option.

Answer 2

Step 1: Calculate \( y \) at \( x=2 \) 
\[ y = 2^3 - 3 \times 2 + 1 = 8 - 6 + 1 = 3 \] So the point of tangency is \( (2, 3) \).

Step 2: Find derivative \( \frac{dy}{dx} \) 
\[ \frac{dy}{dx} = 3x^2 - 3 \]

Step 3: Find slope at \( x=2 \) 
\[ m = 3(2)^2 - 3 = 3 \times 4 - 3 = 12 - 3 = 9 \]

Step 4: Write tangent line equation 
Using point-slope form: \[ y - y_1 = m(x - x_1) \] Substitute: \[ y - 3 = 9(x - 2) \implies y = 9x - 18 + 3 = 9x - 15 \]