In the given figue vertices of `DeltaABC` lie on `y=f(x)=ax^(2)+bx+c`. The `DeltaAB`
Channel: Doubtnut
Date: 2020-05-24

In the given figue vertices of `DeltaABC` lie on `y=f(x)=ax^(2)+bx+c`. The `DeltaAB` is right angled isosceles triangle whose hypotenuse `AC=4sqrt(2)` units.  img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/ARH_SKG_ALG_C02_E08_007_Q01.png" width="80%"`y=f(x)` is given by