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`
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
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Correct Answer - 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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