Question

escreva na forma conônica as seguintes funções quadraticas:​

Answer 1

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Forma Canônica de uma função do 2º grau

$\boxed{\boxed{\boxed{\boxed{\sf f(x)=a\cdot\bigg[\bigg(x+\dfrac{b}{2a} \bigg)^2-\dfrac{\Delta}{4a^2}\bigg]}}}}$

$\tt a)~\sf f(x)=x^2+2x-3\\\sf\Delta=b^2-4ac\\\sf\Delta=2^2-4\cdot1\cdot(-3)\\\sf\Delta=4+12\\\sf\Delta=16\\\sf\dfrac{b}{2a}=\dfrac{2}{2\cdot1}=1\\\sf\dfrac{\Delta}{4a^2}=\dfrac{16}{4\cdot1^2}=\dfrac{16}{4}=4\\\sf f(x)=1\cdot\bigg[\bigg(x+1\bigg) ^2-4\bigg]$

$\tt b)~\sf f(x)=2x^2+8x-5\\\sf\Delta=b^2-4ac\\\sf\Delta=8^2-4\cdot2\cdot(-5)\\\sf\Delta=64+40\\\sf\Delta=104\\\sf\dfrac{b}{2a}=\dfrac{8}{2\cdot2}=\dfrac{8}{4}=2\\\sf\dfrac{\Delta}{4a^2}=\dfrac{104}{4\cdot2^2}=\dfrac{104}{16}=\dfrac{13}{2}\\\sf f(x)=2\cdot\bigg[\bigg(x+2\bigg)^2-\dfrac{13}{2}\bigg]$

$\tt c)~\sf f(x)=-x^2+6x+7\\\sf\Delta=b^2-4ac\\\sf\Delta=6^2-4\cdot(-1)\cdot7\\\sf\Delta=36+28\\\sf\Delta=64\\\sf \dfrac{b}{2a}=\dfrac{6}{2\cdot(-1)} =\dfrac{6}{-2}=-3\\\sf\dfrac{\Delta}{4a^2}=\dfrac{64}{4\cdot(-1)^2}=\dfrac{64}{4}=16\\\sf f(x)=-1\bigg[\bigg(x-3\bigg)^2-16\bigg]$

$\tt d)~\sf f(x)= x^2+2x-24\\\sf\Delta=b^2-4ac\\\sf\Delta=2^2-4\cdot1\cdot(-24)\\\sf\Delta=4+96\\\sf\Delta=100\\\sf\dfrac{b}{2a}=\dfrac{\diagdown\!\!\!2}{\diagdown\!\!\!2\cdot2}=\dfrac{1}{2}\\\sf\dfrac{\Delta}{4a^2}=\dfrac{100}{4\cdot1^2}=\dfrac{100}{4}=25\\\sf f(x)=1\cdot\bigg[\bigg(x+\dfrac{1}{2}\bigg)^2-25\bigg]$

$\tt e)~\sf f(x)=10+5x-5x^2\\\sf\Delta=b^2-4ac\\\sf\Delta=5^2-4\cdot(-5)\cdot10\\\sf\Delta=25+200\\\sf\Delta=225\\\sf\dfrac{b}{2a}=\dfrac{\diagup\!\!\!5}{2\cdot(-\diagup\!\!\!5)}=-\dfrac{1}{2}\\\sf\dfrac{\Delta}{4a^2}=\dfrac{225}{4\cdot(-5)^2}=\dfrac{225}{100}=\dfrac{9}{4}\\\sf f(x)=-5\bigg[\bigg(x-\dfrac{1}{2}\bigg)^2-\dfrac{9}{4}\bigg]$

$\tt f)~\sf f(x)=-2x^2+5x-1\\\sf\Delta=b^2-4ac\\\sf\Delta=5^2-4\cdot(-2)\cdot(-1)\\\sf\Delta=25-8\\\sf\Delta=17\\\sf\dfrac{b}{2a}=\dfrac{5}{2\cdot(-2)}=-\dfrac{5}{4}\\\sf\dfrac{\Delta}{4a^2}=\dfrac{17}{4\cdot(-2)^2}=\dfrac{17}{16}\\\sf f(x)=-2\cdot\bigg[\bigg(x-\dfrac{5}{4}\bigg)^2-\dfrac{17}{16}\bigg]$