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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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Answer» Correct Answer - 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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