Question

f(x)=(x-1)(x+3)
determine os valores de 'a' 'b' e 'c'
encontre as coordenadas do vértice da parábola

Answer 1

$f(x)=(x-1)(x+3)=x^2+2x-3 \\ a=1 \\ b=2 \\ c=-3$

$Xv= -\frac{b}{2a} \\ Xv=-\frac{2}{2}=-1$

$Yv=f(Xv)=(-1)^2+2.(-1)-3\\=1-2-3=-4$

$V(Xv,Yv)\\ V(-1,-4)$

Answer 2

Explicação passo-a-passo:

Função Quadrática:

$\mathsf{f_{(x)}~=~\Big(x-1\Big)\Big(x+3\Big) }$

$\mathsf{f_{(x)}~=~x^2+3x-x-3 }$

$\mathsf{{\color{blue}{f_{(x)}~=~x^2+2x-3 }} }$

$\mathsf{Coeficientes}\begin{cases} ~\mathsf{a~=~1} \\ \\ \mathsf{b~=~2} \\ \\ \mathsf{c~=~-3} \end{cases}$

Coordenadas do Vértice:

$\mathsf{X_{(v)}~=~-\dfrac{b}{2a} }$

$\mathsf{X_{(v)}~=~-\dfrac{2}{2.1} }$

$\mathsf{X_{(v)}~=~-\dfrac{2}{2} }$

$\boxed{\mathsf{X_{(v)}~=~-1}}}}$

$\mathsf{Y_{(v)}~=~-\dfrac{\Delta}{4a} }$

$\mathsf{Y_{(v)}~=~-\dfrac{\Big(b^2-4ac\Big)}{4a} }$

$\mathsf{Y_{(v)}~=~-\dfrac{\Big(2^2-4.1.(-3)\Big)}{4.1}~=~-\dfrac{16}{4}~=~{-4}}$