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The tangent to y=ax^(2)+bx+c at (1,-2) i...

The tangent to `y=ax^(2)+bx+c` at `(1,-2)` is parallel to the normal at the point `(-2,2)` on the same curve. Find the value of `3a-b+c`

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The correct Answer is:
2
This curve passes through points `(1,2) & (-2,2)`
Therefore `a+b+c=2` .(i)
`4a-2b+c=2` .(ii)
`impliesa-b=0`
`because3a-b+c=2a+c=a+b+c=2`
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