Question

Find the point of intersection of tangents drawn at points (0,1) and (2,3) on $y=2x^2-3x+1$.

• A. The point of intersection is (1, -2).
• B. The point of intersection is (1, 2).
• C. The point of intersection is (-1, -2).
• D. The point of intersection is (-1, 2).

Answer 1

Answer: (A)

The point of intersection is (1, -2).

Answer 2

1. Find the derivative of the curve: The given curve is $y = 2x^2 - 3x + 1$. The derivative is $\frac{dy}{dx} = 4x - 3$.

2. Find the equation of the tangent at (0,1): The slope of the tangent at $x=0$ is $m_1 = 4(0) - 3 = -3$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - 1 = -3(x - 0)$ $y = -3x + 1$ (Equation 1)

3. Find the equation of the tangent at (2,3): The slope of the tangent at $x=2$ is $m_2 = 4(2) - 3 = 8 - 3 = 5$. Using the point-slope form $y - y_1 = m(x - x_1)$: $y - 3 = 5(x - 2)$ $y - 3 = 5x - 10$ $y = 5x - 7$ (Equation 2)

4. Find the point of intersection: Equate Equation 1 and Equation 2: $-3x + 1 = 5x - 7$ $8x = 8$ $x = 1$ Substitute $x=1$ into Equation 1: $y = -3(1) + 1 = -2$. The point of intersection is (1, -2).