If the curve `y=a x^2-6x+b` pass through `(0,2)` and has its tangent parallel to the x-axis at `x=3/2,` then find the values of `a and bdot`

Options

A. A. a=2, b=-2
B. B. a=2, b=2
C. C. a=-2, b=2
D. D. a=-2, b=-2

Answer

B

Explanation

(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`