[tex]f(x)=x^2+2x-3\\\Delta=2^2+12=4+12=16\\\sqrt{\Delta}=4\\x_1=\frac{-2-4}2=\frac{-6}{2}=-3\\x_2=\frac{-2+4}2=\frac22=1\\\\\\a=1 | a > 0 \text{ramiona skierowane w gore}\\\\\\q=\frac{-\Delta}{4a}=\frac{-16}{4}=-4\\p=\frac{-2}{2}=-1\\\\y_{min}=-4 \text{ dla x=-1}\\\\f(2)=2^2+2*2-3=4+4-3=8-3=5\\\\f_{max}=5 \text{ dla x=2}[/tex]
