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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=-2B. a=2, b=2C. a=-2, b=2D. a=-2, b=-2 |
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Answer» 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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