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AleksAgata [21]
3 years ago
12

¿cuál es el vértice de una parábola que corta los ejes en los puntos (-1,0) , (5,0), (0,-10) ?​

Mathematics
1 answer:
kvasek [131]3 years ago
3 0

Answer:

vertex: (2,-18)

Step-by-step explanation:

y = ax² + bx + c

(-1,0) : a - b + c = 0       ...(1)

(5,0) : 25a +5b +c = 0  ...(2)

(0,-10): 0a + 0b + c = -10        c=-10

(1) x5: 5a - 5b + 5c = 0  ...(3)

(2)+(3): 30a + 6c = 0          30a = -6c = 60        a = 2

(1): 2 - b -10 = 0     b = -8

Equation: y = 2x² - 8x -10 = 2 (x² -4x +4) - 18 = 2(x-2)² -18

equation: y = a(x-h)²+k     (h,k): vertex

vertex: (2,-18)

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How do you do this question?
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C. y₂ = (1 + (t/n))²

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∫▒〖1st .2nd dx=1st∫▒〖2nd dx〗-∫▒〖(derivative of 1st) dx∫▒〖2nd dx〗〗〗

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derivative of arctan(x)=1/(1+x^2 )

∫▒〖1 dx〗=x

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∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-∫▒〖(1/(1+x^2 ))dx.x〗…………Eq1

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Let 1+x^2=u

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1/2 ∫▒(2xdx/u) =1/2  ln⁡(1+x^2 )+C

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∫▒〖arctan⁡(x).1 dx=arctan⁡(x).x〗-1/2  ln⁡(1+x^2 )+C  (required soultion)

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