Question

Calcule as coordenadas do vertices das parábola abaixo: A-x²-10x+9=0

B-x²+8x+16=0

C-x²-6x+9=0

D-x²-x-6=0

Answer 1

a)

$x^2-10x+9=0\\\Delta=b^2-4ac\\\Delta=(-10)^2-4.1.9\\\Delta=100-36\\\Delta=64\\\\X_v=\dfrac{-b}{2a}=\dfrac{-(-10)}{2.1}=\dfrac{10}{2}=5\\\\Y_v=\dfrac{-\Delta}{4a}=\dfrac{-64}{4.1}=-16\\\\V=(5,-16)$

b)

$x^2+8x+16=0\\\Delta=b^2-4ac\\\Delta=8^2-4.1.16\\\Delta=64-64\\\Delta=0\\\\X_v=\dfrac{-b}{2a}=\dfrac{-8}{2.1}=\dfrac{-8}{2}=-4\\\\Y_v=\dfrac{-\Delta}{4a}=\dfrac{-0}{4.1}=0\\\\V=(-4,0)$

c)

$x^2-6x+9=0\\\Delta=b^2-4ac\\\Delta=(-6)^2-4.1.9\\\Delta=36-36\\\Delta=0\\\\X_v=\dfrac{-b}{2a}=\dfrac{-(-6)}{2.1}=\dfrac{6}{2}=3\\\\Y_v=\dfrac{-\Delta}{4a}=\dfrac{-0}{4.1}=0\\\\V=(3,0)$

d)Resposta:

Explicação passo-a-passo:

$X_v=\dfrac{-b}{2a}\qquad Y_v=\dfrac{-\Delta}{4a}$

$x^2-x-6=0\\\Delta=b^2-4ac\\\Delta=(-1)^2-4.1.(-6)\\\Delta=1+24\\\Delta=25\\\\X_v=\dfrac{-b}{2a}=\dfrac{-(-1)}{2.1}=\dfrac{1}{2}\\\\Y_v=\dfrac{-\Delta}{4a}=\dfrac{-25}{4.1}=\dfrac{-25}{4}\\\\V=\Bigg(\dfrac{1}{2},-\dfrac{25}{4}\Bigg)$