Odpowiedź:
[tex]a)\ \ f(x)=2x^2+3x\\\\a=2\ \ ,\ \ b=3\ \ ,\ \ c=0\\\\\\\Delta=b^2-4ac\\\\\Delta=3^2-4\cdot2\cdot0=9-0=9\\\\p=\frac{-b}{2a}=\frac{-3}{2\cdot2}=-\frac{3}{4}\\\\q=\frac{-\Delta}{4a}=\frac{-9}{4\cdot2}=-\frac{9}{8}\\\\W(-\frac{3}{4},-\frac{9}{8})\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=2(x-(-\frac{3}{4}))^2+(-\frac{9}{8})\\\\f(x)=2(x+\frac{3}{4})^2-\frac{9}{8}[/tex]
[tex]b)\ \ f(x)=x^2-4\\\\a=1\ \ ,\ \ b=0\ \ ,\ \ c=-4\\\\\Delta=b^2-4ac\\\\\Delta=0^2-4\cdot1\cdot(-4)=0+16=16\\\\\\p=\frac{-b}{2a}=\frac{0}{2\cdot1}=\frac{0}{2}=0\\\\q=\frac{-\Delta}{4a}=\frac{-16}{4\cdot1}=\frac{-16}{4}=-4\\\\W(0,-4)\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=1(x-0)^2+(-4)\\\\f(x)=(x-0)^2-4[/tex]
[tex]c)\ \ f(x)=-x^2+10x-25\\\\a=-1\ \ ,\ \ b=10\ \ ,\ \ c=-25\\\\\Delta=b^2-4ac\\\\\Delta=10^2-4\cdot(-1)\cdot(-25)=100-100=0\\\\\\p=\frac{-b}{2a}=\frac{-10}{2\cdot(-1)}=\frac{-10}{-2}=5\\\\q=\frac{-\Delta}{4a}=\frac{0}{4\cdot(-1)}=\frac{0}{-4}=0\\\\W(5,-4)\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=-1(x-5)^2+0\\\\f(x)=-(x-5)^2[/tex]
[tex]d)\ \ f(x)=x^2-6x+5\\\\a=1\ \ ,\ \ b=-6\ \ ,\ \ c=5\\\\\Delta=b^2-4ac\\\\\Delta=(-6)^2-4\cdot1\cdot5=36-20=16\\\\\\p=\frac{-b}{2a}=\frac{-(-6)}{2\cdot1}=\frac{6}{2}=3\\\\q=\frac{-\Delta}{4a}=\frac{-16}{4\cdot1}=\frac{-16}{4}=-4\\\\W(3,-4)\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=1(x-3)^2+(-4)\\\\f(x)=(x-3)^2-4[/tex]
[tex]e)\ \ f(x)=4x^2-x+1\\\\a=4\ \ ,\ \ b=-1\ \ ,\ \ c=1\\\\\Delta=b^2-4ac\\\\\Delta=(-1)^2-4\cdot4\cdot1=1-16=-15\\\\\\p=\frac{-b}{2a}=\frac{-(-1)}{2\cdot4}=\frac{1}{8}\\\\q=\frac{-\Delta}{4a}=\frac{-(-15)}{4\cdot4}=\frac{15}{16}\\\\W(\frac{1}{8},\frac{15}{16})\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=4(x-\frac{1}{8})^2+\frac{15}{16}[/tex]
[tex]f)\ \ f(x)=\frac{1}{2}x^2+2x-3\\\\a=\frac{1}{2}\ \ ,\ \ b=2\ \ ,\ \ c=-3\\\\\Delta=b^2-4ac\\\\\Delta=2^2-4\cdot\frac{1}{2}\cdot(-3)=4-2\cdot(-3)=4+6=10\\\\\\p=\frac{-b}{2a}=\frac{-2}{2\cdot\frac{1}{2}}=\frac{-2}{1}=-2\\\\q=\frac{-\Delta}{4a}=\frac{-10}{4\cdot\frac{1}{2}}=\frac{-10}{2}=-5\\\\W(-2,-5)\\\\\\f(x)=a(x-p)^2+q\\\\f(x)=\frac{1}{2}(x-(-2))^2+(-5)\\\\f(x)=\frac{1}{2}(x+2)^2-5[/tex]
