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loris [4]
3 years ago
8

What are the vertex, axis of symmetry, maximum or minimum value, and range of y = 3x^2 + 6x − 1?

Mathematics
1 answer:
Bess [88]3 years ago
6 0
The vertex is when x=-b/2a, which is also the line of symmetry. 
in this case, b=6, a=3, so x=-6/6=-1
when x=-1. y=3-6-1=-4

another way to do it is to write the equation into vertex form by making a square:
3x²+6x-1
3(x²+2x-1/3)
3(x²+2x+1-1-1/3))
3(x²+2x+1-4/3))
notice the bold part makes a square:
3(x+1)²-4

Either way, 
the vertex is (-1, -4)
the axis of symmetry is x=-1
the coefficient of x² is 3, a positive number, so this parabola opens upward. the function has a minimum value of y=-4
the range is all real number larger than -4
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x = 2000

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I don't know to find the answer to 8. Can someone explain to me?
lapo4ka [179]
Perhaps the easiest way to find the midpoint between two given points is to average their coordinates: add them up and divide by 2.

A) The midpoint C' of AB is
.. (A +B)/2 = ((0, 0) +(m, n))/2 = ((0 +m)/2, (0 +n)/2) = (m/2, n/2) = C'
The midpoint B' is
.. (A +C)/2 = ((0, 0) +(p, 0))/2 = (p/2, 0) = B'
The midpoint A' is
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B) The slope of the line between (x1, y1) and (x2, y2) is given by
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C) We know the line goes through A = (0, 0), so we can write the point-slope form of the equation for AA' as
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D) To show the point lies on the line, we can substitute its coordinates for x and y and see if we get something that looks true.
.. (x, y) = ((m+p)/3, n/3)
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The purpose of the exercise is to show that all three medians of a triangle intersect in a single point.
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ololo11 [35]

Answer:

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Answer:

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