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If the parabola y=a x^2-6x+b passes thro...

If the parabola `y=a x^2-6x+b` passes through `(0,2)` and has its tangent at `x=3/2` parallel to the x-axis, then `a=2,b=-2` (b) `a=2,b=2` `a=-2,b=2` (d) `a=-2,b=-2`

A

a=2, b=-2

B

a=2, b=2

C

a=-2, b=2

D

a=-2, b=-2

Text Solution

Verified by Experts

The correct Answer is:
B
(2) `y=ax^(2)-6x+b` passes through (0,2). Here,
`2=a(0^(2))-6(0)+b`
`:." "b=2`
Also, `(dy)/(dx)=2ax-6`
`:." "((dy)/(dx))_(x=3//2)=2a((3)/(2))-6`
`or" "3a-6=0`
`:." "a=2`
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