Mathematics, 20.08.2019 01:40 majj979

What is the vertex and the point of this graph k (×)=[2 (×+4)]^2+3

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The vertex form of a quadratic is given by
y = a(x – h)^2 + k, where (h, k) is the vertex ;
In your case , ( - 4 , 3 ) is the vertex ;

Answer from: luis3863

K(x) = 2(x + 4)² + 3
k(x) = 2(x + 4)(x + 4) + 3
k(x) = 2(x² + 4x + 4x + 16) + 3
k(x) = 2(x² + 8x + 16) + 3
k(x) = 2(x²) + 2(8x) + 2(16) + 3
k(x) = 2x² + 16x + 32 + 3
k(x) = 2x² + 16x + 35
2x² + 16x + 35 = 0
x = -(16) +/- √((16)² - 4(2)(35))
2(2)
x = -16 +/- √(256 - 280)
4
x = -16 +/- √(-24)
4
x = -16 +/- 2i√(6)
4
x = -4 + 0.5i√(6)
x = -4 + 0.5i√(6) x = -4 - 0.5i√(6)
k(x) = 2x² + 16x + 35
k(-4 + 0.5i√(6)) = 2(-4 + 0.5i√(6))² + 16(-4 + 0.5i√(6)) + 35
k(-4 + 0.5i√(6)) = 2(-4 + 0.5i√(6))(-4 + 0.5i√(6)) + 16(-4) + 16(0.5i√(6)) + 35
k(-4 + 0.5i√(6)) = 2(16 - 2i√(6) - 2√(6) + 0.25i²√(36)) - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 2(16 - 4i√(6) + 0.25i²(6)) - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 2(16 - 4i√(6) + 1.5i²) - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 2(16 - 4i√(6) + 1.5(-√(1²)) - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 2(16 - 4i√(6) + 1.5(-√(1 × 1)) - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 2(16 - 4i√(6) + 1.5(-√1) - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 2(16 - 4i√(6) + 1.5(-1)) - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 2(16 - 4i√(6) - 1.5) - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 2(16) - 2(4i√(6)) - 2(1.5) - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 32 - 8i√(6) - 3 - 64 + 8i√(6) + 35
k(-4 + 0.5i√(6)) = 32 - 3 - 64 + 35 - 8i√(6) + 8i√(6)
k(-4 + 0.5i√(6)) = 29 - 64 + 35 + 0i√(6)
k(-4 + 0.5i√(6)) = -35 + 35 + 0
k(-4 + 0.5i√(6)) = 0 + 0
k(-4 + 0,5i√(6)) = 0
(x, k(x)) = (-4 + 0.5i√(6), 0)
or
k(x) = 2x² + 16x + 35
k(-4 - 0.5i√(6)) = 2(-4 - 0.5i√(6))² + 16(-4 - 0.5i√(6)) + 35
k(-4 - 0.5i√(6)) = 2(-4 - 0.5i√(6))(-4 - 0.5i√(6)) + 16(-4) - 16(0.5i√(6)) + 35
k(-4 - 0.5i√(6)) = 2(16 + 2i√(6) + 2i√(6) + 0.25i²√(36)) - 64 - 8i√(6) + 35
k(-4 - 0.5i√(6)) = 2(16 + 4i√(6) + 0.25i²(6)) - 64 - 8i√(6) + 35
k(-4 - 0.5i√(6)) = 2(16 + 4i√(6) + 1.5i²) - 64 - 8i√(6) + 35
k(-4 - 0.5i√(6)) = 2(16 + 4i√(6) + 1.5(-√(1²)) - 64 - 8i√(6) + 35
k(-4 - 0.5i√(6)) = 2(16 + 4i√(6) + 1.5(-√(1 × 1)) - 64 - 8i√(6) + 35
k(-4 - 0.5i√(6)) = 2(16 + 4i√(6) + 1.5(-√(1)) - 64 - 8i√(6) + 35
k(-4 - 0.5i√(6)) = 2(16 + 4i√(6) + 1.5(-1)) - 64 - 8i√(6) + 35
k(-4 - 0.5i√(6)) = 2(16 + 4i√(6) - 1.5) - 64 - 8i√(6) + 35
k(-4 - 0.45i√(6)) = 2(16) + 2(4i√(6)) - 2(1.5) - 64 - 8i√(6) + 35
k(-4 - 0.5i√(6)) = 32 + 8i√(6) - 3 - 64 - 8i√(6) + 35
k(-4 - 0.5i√(6)) = 32 - 3 - 64 + 35 + 8i√(6) - 8i√(6)
k(-4 - 0.5i√(6)) = 29 - 64 + 35 + 0i√(6)
k(-4 - 0.5i√(6)) = -35 + 35 + 0
f(-4 - 0.5i√(6)) = 0 + 0
f(-4 - 0.5i√(6)) = 0
(x, k(x)) = (-4 - 0.5i√(6), 0)

The point of the graph is (-4 + 0.5i√(6), 0), or (-4 + 0.5i√(6), 0) and (-4 - 0.5i√(6),0). The vertex of the graph is (-4, 3).

Answer from: Quest

1935/10,000= 387/2000the answer to the question is c

Answer from: Quest

i think asking google will you get the answer to that question

Find the center of the circle whose equation is x2 + y2 =4

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What is the vertex and the point of this graph k (×)=[2 (×+4)]^2+3...

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