Question

a parábola y= ax² + bx + c passa pelos pontos A(1, 2), B(0, 2) e C (-1,3) determine as coordenadas do vértice dessa parábola. ​

Answer 1

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$\sf f(x)=ax^2+bx+c\\\sf f(0)=a\cdot0^2+b\cdot0+c\\\sf c=2\\\sf f(1)=a\cdot 1^2+b\cdot1+2\\\sf a+b+2=2\\\sf a+b=2-2\\\sf a+b=0\\\sf f(-1)=a\cdot(-1)^2+b\cdot(-1)+4\\\sf a-b+2=3\\\sf a-b=3-2\\\sf a-b=1\\+\underline{\begin{cases}\sf a+\diagup\!\!\!b=0\\\sf a-\diagup\!\!\!b=1\end{cases}}\\\sf 2a=1\\\sf a=\dfrac{1}{2}\\\sf a+b=0\\\sf\dfrac{1}{2}+b=0\cdot2\\\sf1+2b=0\\\sf 2b=-1\\\sf b=-\dfrac{1}{2}\\\sf f(x)=\dfrac{1}{2}x^2-\dfrac{1}{2}x+2$

$\sf\Delta=b^2-4ac\\\sf\Delta=\bigg(-\dfrac{1}{2}\bigg)^2-4\cdot\bigg(\dfrac{1}{2}\bigg)\cdot2\\\sf\Delta=\dfrac{1}{4}-4\\\sf\Delta=\dfrac{1-16}{4}\\\sf\Delta=-\dfrac{15}{4}$

$\sf x_V=-\dfrac{b}{2a}\\\sf x_V=-\dfrac{-\frac{1}{2}}{\diagup\!\!\!2\cdot\bigg(\frac{1}{\diagup\!\!\!2}\bigg)}\\\sf x_V=\frac{1}{2}\\\sf y_V=-\dfrac{\Delta}{4a}\\\sf y_V=-\dfrac{-\frac{15}{4}}{\diagup\!\!\!4\cdot\bigg(\frac{1}{\diagup\!\!\!2}\bigg)}\\\sf y_V=-\dfrac{\frac{15}{4}}{2}\\\sf y_V=\dfrac{15}{4}\cdot\dfrac{1}{2}\\\sf y_V=\dfrac{15}{8}\\\huge\boxed{\boxed{\boxed{\boxed{\sf V\bigg(\dfrac{1}{2},\dfrac{15}{8}\bigg)}}}}$