2) Organize um quadro da função quadrática
Soluções para a tarefa
Resposta: wkashfahsjd
.
Explicação passo-a-passo:
Resposta:
y = ax² + bx + cy=ax²+bx+c
delta
\Delta^2=b^2 - 4acΔ
2
=b
2
−4ac
vértice
V_x = \frac{-b}{2a}V
x
=
2a
−b
V_y = \frac{-\Delta^2}{4a}V
y
=
4a
−Δ
2
a)
y= x^2- 1y=x
2
−1
a=1a=1
b=0b=0
c=-1c=−1
delta
\Delta^2=0^2-4\times1\times-1Δ
2
=0
2
−4×1×−1
\Delta^2=4Δ
2
=4
vértice
V_x = \frac{-0}{2\times1}V
x
=
2×1
−0
V_x=0V
x
=0
V_y=\frac{-4}{4\times1}V
y
=
4×1
−4
V_y=\frac{-4}{4}=-1V
y
=
4
−4
=−1
b)
y=-x^2y=−x
2
a=-1a=−1
b=0b=0
c=0c=0
delta
\Delta^2=0^2-4\times1\times0Δ
2
=0
2
−4×1×0
\Delta^2=0Δ
2
=0
vértice
V_x=\frac{0}{2\times1}V
x
=
2×1
0
V_x=0V
x
=0
V_y=\frac{0}{4\times1}V
y
=
4×1
0
V_x=0V
x
=0
c)
y=x^2+2x-8y=x
2
+2x−8
a=1a=1
b=2b=2
c=-8c=−8
delta
\Delta^2=2^2-4\time1\times-8Δ
2
=2
2
−4\time1×−8
\Delta^2=36Δ
2
=36
vértice
V_x=\frac{-2}{2\times1}V
x
=
2×1
−2
V_x=\frac{-2}{2}=-1V
x
=
2
−2
=−1
V_y=\frac{36}{4\times1}V
y
=
4×1
36
V_y=\frac{36}{4}=9V
y
=
4
36
=9
d)
y=x^2-2xy=x
2
−2x
a=1a=1
b=-2b=−2
c=0c=0
delta
\Delta=-2^2-4\times1\times0Δ=−2
2
−4×1×0
\Delta=4Δ=4
vértice
V_x=\frac{-(-2)}{2\times1}V
x
=
2×1
−(−2)
V_x=\frac{-2}{2}=-1V
x
=
2
−2
=−1
V_y = \frac{-4}{4\times1}V
y
=
4×1
−4
V_y = \frac{-4}{4}=-1V
y
=
4
−4
=−1
e)
y=x^2-2x+4y=x
2
−2x+4
a=-1a=−1
b=-2b=−2
c=4c=4
delta
\Delta^2=-2^2-4\times-1\times4Δ
2
=−2
2
−4×−1×4
\Delta^2=20Δ
2
=20
vértice
V_x = \frac{-(-2)}{2\times-1}V
x
=
2×−1
−(−2)
V_x = \frac{2}{-2}=-1V
x
=
−2
2
=−1
V_y = \frac{-20}{4\times-1}V
y
=
4×−1
−20
V_y = \frac{-20}{-4}=5V
y
=
−4
−20
=5
f)
y= -x^2+6x-9y=−x
2
+6x−9
a=-1a=−1
b=6b=6
c=-9c=−9
delta
\Delta^2=6^2 - 4\times1\times-9Δ
2
=6
2
−4×1×−9
\Delta^2=72Δ
2
=72
vértice
V_x=\frac{-6}{2\times-1}V
x
=
2×−1
−6
V_x=\frac{-6}{-2}=3V
x
=
−2
−6
=3
V_y=\frac{-72}{4\times-1}V
y
=
4×−1
−72
V_y=\frac{-72}{-4}=18V
y
=
−4
−72
=18
A tabela eu vou ficar devendo...