If the parabola y=α x2-6 x+β passes through the point (0,2) and has its tangent at x=(3/2) parallel to x axis, then

Question

If the parabola y=α x2-6 x+β passes through the point (0,2) and has its tangent at x=(3/2) parallel to x axis, then (A) α=2, β=-2 (B) α=-2, β=2 (C) α=2, β=2 (D) α=-2, β=-2

Tardigrade Admin · Accepted Answer

y=α x2-6 x+β passes through (0,2) 2=β (d y/d x)=2 α x-6 ((d y/d x))x=(3/2)=0 2 α((3/2))-6=0 3 α=6 α=2

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

$α = 2, β = - 2$

$α = - 2, β = 2$

$α = 2, β = 2$

$α = - 2, β = - 2$

$y = α x^{2} - 6 x + β$ passes through $(0, 2)$
$2 = β$
$\frac{d y}{d x} = 2 αx - 6$
$(\frac{d y}{d x})_{x = \frac{3}{2}} = 0$
$2 α (\frac{3}{2}) - 6 = 0$
$3 α = 6$
$α = 2$

Q. If the parabola y=αx2−6x+β passes through the point (0,2) and has its tangent at x=23​ parallel to x axis, then

α=2,β=−2

α=−2,β=2

α=2,β=2

α=−2,β=−2