Question

Determine as coordenadas do vértice da parábola correspondente a cada função. a) f(x)=x² -2x+3 b) f(x)=4x²+4 c) f(x)=2x²-12x+7 d) f(x)=3x²-2x+1 e) f(x)=⅓x²-10/3x+5 f) f(x)=x²-√3x

Answer 1

Resposta:

a) V = (1, 2)

b) V = (-1/2, -1)

c) V = (3, -11)

d) V = (1/3, 2/3)

e) V = (5, 10/3)

f) V = ($\sqrt{3}/2$, 3/4)

Explicação passo-a-passo:

As coordenadas do vértice de uma parábola $ax^2+bx+c$ são dadas por

• $x_v = \frac{-b}{2a}$

• $y_v = \frac{-\Delta}{4a}$

a) $f(x) = x^2-2x+3$

Temos a=1, b=-2 e c=3. Assim,

• $x_v=\frac{-b}{2a} = \frac{-(-2)}{2\cdot1}=\frac{2}{2}=1.$

• $y_v=\frac{-\Delta}{4\cdot a}=\frac{-((-2)^2-4\cdot 1\cdot 3)}{4\cdot 1}=\frac{-(4-12)}{4}=\frac{12-4}{4}=\frac{8}{4}=2.$

Portanto,

V = (1,2).

b) $f(x)=4x^2+4$

Temos a=4, b=4 e c=0. Assim,

• $x_v=\frac{-b}{2a} = \frac{-4}{2\cdot4}=\frac{-4}{8}=-1/2.$

• $y_v=\frac{-\Delta}{4\cdot a}=\frac{-(4^2-4\cdot4\cdot 0)}{4\cdot 4}=\frac{-16}{16}=1.$

Portanto,

V = (-1/2,1).

c) $f(x)=2x^2-12x+7$

Temos a=2, b=-12 e c=7. Assim,

• $x_v=\frac{-b}{2a} = \frac{-(-12)}{2\cdot2}=\frac{12}{4}=3$

• $y_v=\frac{-\Delta}{4\cdot a}=\frac{-((-12)^2-4\cdot2\cdot 7)}{4\cdot 2}=\frac{-(144-56)}{8}=\frac{-88}{8}=-11.$

Portanto,

V = (3,-11).

d) $f(x)=3x^2-2x+1$

Temos a=3, b=-2 e c=1. Assim,

• $x_v=\frac{-b}{2a} = \frac{-(-2)}{2\cdot3}=\frac{2}{6}=(1/3).$
• $y_v=\frac{-\Delta}{4\cdot a}=\frac{-((-2)^2-4\cdot3\cdot1)}{4\cdot3}=\frac{-(4-12)}{12}=\frac{12-4}{4}=\frac{8}{12}=2/3.$

Portanto,

V = (1/3,2/3).

e) $f(x) = 1/3x^2-10/3x+5$

Temos a=1/3, b=-10/3 e c=5. Assim,

• $x_v=\frac{-b}{2a} = \frac{-(-10/3)}{2\cdot1/3}=\frac{10/3}{2/3}=\frac{10}{3}\cdot\frac{3}{2}=\frac{10}{2}=5.$

• $y_v=\frac{-\Delta}{4\cdot a}=\frac{-((-10/3)^2-4\cdot1/3\cdot5)}{4\cdot1/3}=\frac{-(100/9-20/3)}{4/3}=\frac{-(100/9-60/9)}{4/3}=\frac{-40/9}{4/3}=\frac{-40}{9}\cdot \frac{3}{4}=\frac{-120}{36}=\frac{-10}{3}.$

Portanto,

V = (5,-10/3).

) $f(x) = x^2-\sqrt{3}x$

Temos a = 1, b = $-\sqrt{3}$ e c = 0. Assim,

• $x_v=\frac{-b}{2a} = \frac{-(-\sqrt{3})}{2\cdot1}=\frac{\sqrt{3}}{2}$

• $y_v=\frac{-\Delta}{4\cdot a}=\frac{-(b^2-4\cdot a \cdot c)}{4\cdot a}=\frac{-((-\sqrt{3})^2-4\cdot 1 \cdot 0)}{4\cdot 1}=\frac{-(3-0)}{4}=\frac{-3}{4}$

Portanto,

V = ($\sqrt{3}/2$,-3/4).