Question

Dê as coordenadas do vértice da função f(x)= - x²+7x-10

Answer 1

-x²+7x-10
Bhaskara
Δ=(b)²-4.(a).(c)
Δ=49-40
Δ = 9

Xv = -b/(2.a)
Xv = -7/-2
Xv = 7/2

Yv = -Δ/(4.a)
Yv = -9/-4
Yv = 9/4

Resposta: as coordenadas são (7/2 , 9/4)

Espero ter ajudado ^-^

Answer 2

✅ Após resolver os cálculos, concluímos que o vértice da referida função quadrática é:

        $\Large\displaystyle\begin{gathered} V = \Bigg(\frac{7}{2}, \frac{9}{4} \Bigg)\end{gathered}$

Seja a função quadrática:

      $\Large\displaystyle\begin{gathered} f(x) = -x^{2} + 7x - 10\end{gathered}$

Cujos coeficientes são:

            $\Large\begin{cases}a = -1\\ b = 7\\c = -10\end{cases}$

Para calcular o vértice da função devemos fazer:

      $\Large\displaystyle\begin{gathered}V = (X_{V}, Y_{V}) \end{gathered}$

          $\Large\displaystyle\begin{gathered}= \Bigg(-\frac{b}{2a}, -\frac{\Delta}{4a} \Bigg) \end{gathered}$

          $\Large\displaystyle\text{ \begin{gathered}= \Bigg[-\frac{b}{2a}, -\frac{(b^{2} - 4ac)}{4a} \Bigg] \end{gathered} }$

          $\Large\displaystyle\text{ \begin{gathered}= \Bigg[\frac{-7}{2\cdot(-1)}, -\frac{(7^{2} - 4\cdot(-1)\cdot(-10))}{4a} \Bigg] \end{gathered} }$

          $\Large\displaystyle\begin{gathered}= \Bigg[\frac{-7}{-2}, -\frac{(49 - 40)}{-4} \Bigg] \end{gathered}$

          $\Large\displaystyle\begin{gathered}= \Bigg[\frac{7}{2}, \frac{9}{4} \Bigg] \end{gathered}$

✅ Portanto, o vértice da função é:

      $\Large\displaystyle\begin{gathered} V = \Bigg(\frac{7}{2}, \frac{9}{4} \Bigg)\end{gathered}$    

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