for the following function y=3x2+12x-8 find the coordinates of the vertex and state the minimum?
回答 (1)
2016-06-22 6:53 pm
y = 3x² + 12x - 8
y = 3(x² + 4x) - 8
y = 3(x² + 4x + 4) - 8 - 3*4
y = 3(x + 2)² - 20
Coordinates of the vertex = (-2, 20)
For any real value of x, 3(x + 2)² ≥ 0
Then, y = 3(x + 2)² - 20 ≥ -20
Maximum of y = -20 at x = -2
y = 3(x² + 4x) - 8
y = 3(x² + 4x + 4) - 8 - 3*4
y = 3(x + 2)² - 20
Coordinates of the vertex = (-2, 20)
For any real value of x, 3(x + 2)² ≥ 0
Then, y = 3(x + 2)² - 20 ≥ -20
Maximum of y = -20 at x = -2
收錄日期: 2021-04-18 15:11:38
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