Find the x coordinate, h, of the vertex of the graph of the following function:?
回答 (3)
2016-06-21 2:18 pm
Well,
f(x) = -3x^2 - x - 1
therefore
f '(x) = -6x - 1
h is given for the point where the tangent is horizontal
therefore for
f '(x) = 0
6x + 1 = 0 ==> x = -1/6
h = -1/6
hope it' ll help !!
f(x) = -3x^2 - x - 1
therefore
f '(x) = -6x - 1
h is given for the point where the tangent is horizontal
therefore for
f '(x) = 0
6x + 1 = 0 ==> x = -1/6
h = -1/6
hope it' ll help !!
2016-06-21 2:29 pm
(h, k) is the vertex of f(x) = a(x - h) + k
The given function :
f(x) = -3x² - x - 1
f(x) = -3[x² + (1/3)x] - 1
f(x) = -3[x² + 2(1/6)x + (1/3)²] + 3(1/6)² - 1
f(x) = -3[x² + (1/6)]² - (11/12)
The vertex is (-1/6, -11/12)
The x-coordinate of the vertex, h = -1/6
The given function :
f(x) = -3x² - x - 1
f(x) = -3[x² + (1/3)x] - 1
f(x) = -3[x² + 2(1/6)x + (1/3)²] + 3(1/6)² - 1
f(x) = -3[x² + (1/6)]² - (11/12)
The vertex is (-1/6, -11/12)
The x-coordinate of the vertex, h = -1/6
2016-06-21 2:34 pm
y = ax^2 + bx + c is the parabola
y = -3x^2-1x-1
x-coordinate of vertex is -b/2a
= -(-1) / (2)(-3)
= +1 / (2)(-3) = -1/ 6
y = -3x^2-1x-1
x-coordinate of vertex is -b/2a
= -(-1) / (2)(-3)
= +1 / (2)(-3) = -1/ 6
收錄日期: 2021-04-18 15:14:35
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