Question

If the parabola y = αx² - 6x + β passes through the point (0, 2) and has its tangent at x = \(\frac{3}{2}\) parellel to x axis, then

• A. α = 2, β = -2
• B. α = -2, β = 2
• C. α = 2, β = 2
• D. α = -2, β = -2

Answer 1

The Correct Option is C

The equation of the parabola is given by: \[ y = \alpha x^2 - 6x + \beta \] Since the parabola passes through the point \( (0, 2) \), substitute \( x = 0 \) and \( y = 2 \) into the equation: \[ 2 = \alpha(0)^2 - 6(0) + \beta \quad \Rightarrow \quad \beta = 2 \] Now, we need the slope of the tangent at \( x = \frac{3}{2} \) to be 0 (since the tangent is parallel to the x-axis). The derivative of the equation \( y = \alpha x^2 - 6x + \beta \) is: \[ \frac{dy}{dx} = 2\alpha x - 6 \] At \( x = \frac{3}{2} \), set the derivative equal to 0: \[ 0 = 2\alpha \left(\frac{3}{2}\right) - 6 \] Solving for \( \alpha \): \[ 0 = 3\alpha - 6 \quad \Rightarrow \quad 3\alpha = 6 \quad \Rightarrow \quad \alpha = 2 \] Therefore, the values of \( \alpha \) and \( \beta \) are \( \alpha = 2 \) and \( \beta = 2 \).

The correct answer is (C) : α = 2, β = 2.

Answer 2

Given the parabola \( y = \alpha x^2 - 6x + \beta \), it passes through the point (0, 2). Therefore:

\( 2 = \alpha (0)^2 - 6(0) + \beta \)

\( \beta = 2 \)

The derivative of the parabola with respect to x is:

\( \frac{dy}{dx} = 2\alpha x - 6 \)

The tangent at \( x = \frac{3}{2} \) is parallel to the x-axis, which means the slope \( \frac{dy}{dx} \) at \( x = \frac{3}{2} \) is 0:

\( 0 = 2\alpha \left( \frac{3}{2} \right) - 6 \)

\( 0 = 3\alpha - 6 \)

\( 3\alpha = 6 \)

\( \alpha = 2 \)

Therefore, \( \alpha = 2 \) and \( \beta = 2 \).