• Matéria: Matemática
  • Autor: daniellyketily075
  • Perguntado 3 anos atrás

Determine as coordenadas do vértice da função? ta aí: ^​

Anexos:

Respostas

respondido por: franciscosuassuna12
1

(Xv,Yv)= (1, 3)

f(x) =  - x {}^{2}  + 2x + 2

a =  - 1 \:  \:  \: b = 2 \:  \:  \:  \: e \:  \:  \: c = 2

xv =  \frac{ - b}{2a}  =  \frac-2/-2=1</p><p>[tex]yv =  \frac{ - (b {}^{2} - 4ac) }{4a}  =  \frac{ - (2 {}^{2} - 4. - 1.2) }{4. - 1}  =  \frac{ - (4 + 8)}{ - 4}  =  \frac{ - 12}{ - 4}  = 3

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respondido por: CyberKirito
3

\Large\boxed{\begin{array}{l}\sf f(x)=-x^2+2x+2\\\sf\Delta=b^2-4ac\\\sf\Delta=2^2-4\cdot(-1)\cdot2\\\sf\Delta= 4+8\\\sf\Delta=12\\\sf x_V=-\dfrac{b}{2a}\\\\\sf x_V=-\dfrac{\backslash\!\!\!2}{\backslash\!\!\!2\cdot(-1)}=1\\\\\sf y_V=-\dfrac{\Delta}{4a}\\\\\sf y_V=-\dfrac{12}{4\cdot(-1)}=3\\\\\sf\red{V(1,3)}\end{array}}

Anexos:
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