1.y=-4x^2+20x+5
2. y=-3x^2-3x-9/4
Find the max /min value
Maths 唔該plz help?
2015-08-23 12:33 pm
回答 (2)
2015-08-23 12:48 pm
1.y=-4x^2+20x+5
= -4(x - 5x + 25/4) + 5 + 25
= -4(x - 5/2)^2 + 30
max = 30 if x = 5/2
2. y=-3x^2-3x-9/4
= -3(x^2 + x + 1/4) - 9/4 + 3/4
= -3(x + 1/2)^2 - 3/2
max = -3/2 if x = -1/2
= -4(x - 5x + 25/4) + 5 + 25
= -4(x - 5/2)^2 + 30
max = 30 if x = 5/2
2. y=-3x^2-3x-9/4
= -3(x^2 + x + 1/4) - 9/4 + 3/4
= -3(x + 1/2)^2 - 3/2
max = -3/2 if x = -1/2
2015-08-25 3:22 am
1.y=-4x^2+20x+5
y' = -8x +20 = 0
=> x = -5/2 input the equation we get y = 30.....max
2. y=-3x^2-3x-9/4
y' = -6x-3 = 0
=> x = -1/2 input the equation we get y = -3/2.....max
y' = -8x +20 = 0
=> x = -5/2 input the equation we get y = 30.....max
2. y=-3x^2-3x-9/4
y' = -6x-3 = 0
=> x = -1/2 input the equation we get y = -3/2.....max
收錄日期: 2021-04-30 19:53:32
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150823043344AAwvKNn