Find the maximum or minimum value of the quadratic function and the corresponding
value of x by the algebraic method.
1) y=x^2 -3x-1
2) y=4x^2 -12x+3
Max and Min value of quadratic
2009-11-14 10:42 pm
回答 (2)
2009-11-14 10:50 pm
✔ 最佳答案
We shall use the method of completing square.1. y = x^2 - 3x - 1
y = [x^2 - 2(3/2)x + (3/2)^2] - (3/2)^2 - 1
y = (x - 3/2)^2 - 13/4
Minimum of y is obtained when x = 3/2, and the value is -13/4
2. y = 4x^2 - 12x + 3
y = 4(x^2 - 3x) + 3
y = 4[x^2 - 2(3/2)x + (3/2)^2] - 4(3/2)^2 + 3
y = 4(x - 3/2)^2 - 6
Minimum of y is obtained when x = 3/2, and the value is -6.
參考: Physics king
2009-11-14 10:53 pm
1)
y
= x^2 - 3x - 1
= x^2 - (2)(3/2)x +(3/2)^2 - (3/2)^2 - 1
= (x - 3/2)^2 - (3/2)^2 - 1
= (x - 3/2)^2 - 13/4
Minimum value of y = -13/4 when x = 3/2
2)
y
= 4x^2 - 12x + 3
= 4(x^2 - 3x) + 3
= 4(x^2 -(2)(3/2)x + (3/2)^2) - (4)(3/2)^2 + 3
= 4(x - 3/2)^2 - 6
Minimum value of y = -6 when x = 3/2
y
= x^2 - 3x - 1
= x^2 - (2)(3/2)x +(3/2)^2 - (3/2)^2 - 1
= (x - 3/2)^2 - (3/2)^2 - 1
= (x - 3/2)^2 - 13/4
Minimum value of y = -13/4 when x = 3/2
2)
y
= 4x^2 - 12x + 3
= 4(x^2 - 3x) + 3
= 4(x^2 -(2)(3/2)x + (3/2)^2) - (4)(3/2)^2 + 3
= 4(x - 3/2)^2 - 6
Minimum value of y = -6 when x = 3/2
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