f(x)=x2y=x2f(x)=x^2\\y=x^2f(x)=x2y=x2
now according to given condition
y−5=13×(x−6)2y−5=13×(x2+36−12x)y−5=x23+12−4xy-5=\frac{1}{3}\times(x-6)^2\\y-5=\frac{1}{3}\times(x^2+36-12x)\\y-5=\frac{x^2}{3}+12-4xy−5=31×(x−6)2y−5=31×(x2+36−12x)y−5=3x2+12−4x
y=x23+17−4xf(x)=x23+17−4xy=\frac{x^2}{3}+17-4x\\f(x)=\frac{x^2}{3}+17-4xy=3x2+17−4xf(x)=3x2+17−4x
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