Dertermine as cordenadas do vértice da barabola de cada função
A) f(x)= x²-2x+5
B) f(x)= 3x²+12x+8
C) f(x)= -2x²-4x+7
D) f(x)= -2x²-3
Soluções para a tarefa
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
/2 , 3/4)
Explicação passo-a-passo:
As coordenadas do vértice de uma parábola ax^2+bx+cax
2
+bx+c são dadas por
x_v = \frac{-b}{2a}x
v
=
2a
−b
y_v = \frac{-\Delta}{4a}y
v
=
4a
−Δ
a) f(x) = x^2-2x+3f(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.x
v
=
2a
−b
=
2⋅1
−(−2)
=
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.y
v
=
4⋅a
−Δ
=
4⋅1
−((−2)
2
−4⋅1⋅3)
=
4
−(4−12)
=
4
12−4
=
4
8
=2.
Portanto,
V = (1,2).
b) f(x)=4x^2+4f(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.x
v
=
2a
−b
=
2⋅4
−4
=
8
−4
=−1/2.
y_v=\frac{-\Delta}{4\cdot a}=\frac{-(4^2-4\cdot4\cdot 0)}{4\cdot 4}=\frac{-16}{16}=1.y
v
=
4⋅a
−Δ
=
4⋅4
−(4
2
−4⋅4⋅0)
=
16
−16
=1.
Portanto,
V = (-1/2,1).
c) f(x)=2x^2-12x+7f(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}=3x
v
=
2a
−b
=
2⋅2
−(−12)
=
4
12
=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.y
v
=
4⋅a
−Δ
=
4⋅2
−((−12)
2
−4⋅2⋅7)
=
8
−(144−56)
=
8
−88
=−11.
Portanto,
V = (3,-11).
d) f(x)=3x^2-2x+1f(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).x
v
=
2a
−b
=
2⋅3
−(−2)
=
6
2
=(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.y
v
=
4⋅a
−Δ
=
4⋅3
−((−2)
2
−4⋅3⋅1)
=
12
−(4−12)
=
4
12−4
=
12
8
=2/3.
Portanto,
V = (1/3,2/3).
e) f(x) = 1/3x^2-10/3x+5f(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.x
v
=
2a
−b
=
2⋅1/3
−(−10/3)
=
2/3
10/3
=
3
10
⋅
2
3
=
2
10
=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}.y
v
=
4⋅a
−Δ
=
4⋅1/3
−((−10/3)
2
−4⋅1/3⋅5)
=
4/3
−(100/9−20/3)
=
4/3
−(100/9−60/9)
=
4/3
−40/9
=
9
−40
⋅
4
3
=
36
−120
=
3
−10
.
Portanto,
V = (5,-10/3).
) f(x) = x^2-\sqrt{3}xf(x)=x
2
−
3
x
Temos a = 1, b = -\sqrt{3}−
3
e c = 0. Assim,
x_v=\frac{-b}{2a} = \frac{-(-\sqrt{3})}{2\cdot1}=\frac{\sqrt{3}}{2}x
v
=
2a
−b
=
2⋅1
−(−
3
)
=
2
3
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}y
v
=
4⋅a
−Δ
=
4⋅a
−(b
2
−4⋅a⋅c)
=
4⋅1
−((−
3
)
2
−4⋅1⋅0)
=
4
−(3−0)
=
4
−3
Portanto,
V = (\sqrt{3}/2
3
/2 ,-3/4).