Question

If the parabola $y=\alpha x^{2}-6 x+\beta$ passes through the point $(0,2)$ and has its tangent at $x=\frac{3}{2}$ parallel to $x$ axis, then

• A. $\alpha=2, \beta=-2$
• B. $\alpha=-2, \beta=2$
• C. $\alpha=2, \beta=2$
• D. $\alpha=-2, \beta=-2$

Answer 1

Answer: (C)

$y=\alpha x^{2}-6 x+\beta$ passes through $(0,2)$
$2=\beta$
$\frac{d y}{d x}=2 \alpha x-6$
$\left(\frac{d y}{d x}\right)_{x=\frac{3}{2}}=0$
$2 \alpha\left(\frac{3}{2}\right)-6=0$
$3 \alpha=6$
$\alpha=2$