Question

The point at which the tangent to the curve y = 2x^2 - x + 1 is parallel to y = 3x + 9 will be

• A. (2, 1)
• B. (1, 2)
• C. (3, 9)
• D. (-2, 1)

Answer 1

Answer: (B)

(1, 2)

Answer 2

The slope of the line $y = 3x + 9$ is 3. The derivative of the curve $y = 2x^2 - x + 1$ is $\frac{dy}{dx} = 4x - 1$. For the tangent to be parallel to the line, their slopes must be equal. Thus, $4x - 1 = 3$, which gives $x = 1$. Substituting $x = 1$ into the curve's equation gives $y = 2(1)^2 - 1 + 1 = 2$. The point is $(1, 2)$.