Salut a)(x-3)^2-(3x-2)^2 => a^2-b^2=>(a-b)(a+b) =>(x-3+3x+2)(x-3-3x-2) =>(4x-5)(-2x-1) forme 1 b)(x-3)^2-(3x-2)^2 =>x^2-6x+9-9x^2+12x-4 =>-8x^2+6x+5 forme 2 c) forme canonique a(x-alpha)^2+beta alpha=-b/2a =>-6/-16 alpha=3/8 beta=f(alpha)=f(3/8) beta=49/8 la forme canonique est -8(x-(3/8))^2+49/8 forme 3
a) forme 1 car (4*0-5)(-2*0-1)=5 b)forme 3 car -8((3/8)-(3/8))^2+49/8 = 49/8 c)forme 3 car alpha et beta representent le sommet de la fonction S(3/8 ; 49/8) d) forme 1 car 4x-5=0 =>x=5/4 et -2x-1=0 => x=-1/2 e)forme 2 car -8x^2+6x+5=5 =>-8x^2+6x=0 en factorisant -2x(4x-3) soit x=0 et x=3/4 f) forme 1 car 4x-5<=0 =>x<+5/4 -2x-1<=0 =>x>=-1/2 ps: x^2 c'est x au carre
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Saluta)(x-3)^2-(3x-2)^2 => a^2-b^2=>(a-b)(a+b)
=>(x-3+3x+2)(x-3-3x-2)
=>(4x-5)(-2x-1) forme 1
b)(x-3)^2-(3x-2)^2
=>x^2-6x+9-9x^2+12x-4
=>-8x^2+6x+5 forme 2
c) forme canonique a(x-alpha)^2+beta
alpha=-b/2a =>-6/-16 alpha=3/8
beta=f(alpha)=f(3/8) beta=49/8
la forme canonique est -8(x-(3/8))^2+49/8 forme 3
a) forme 1 car (4*0-5)(-2*0-1)=5
b)forme 3 car -8((3/8)-(3/8))^2+49/8 = 49/8
c)forme 3 car alpha et beta representent le sommet de la fonction
S(3/8 ; 49/8)
d) forme 1 car 4x-5=0 =>x=5/4 et -2x-1=0 => x=-1/2
e)forme 2 car -8x^2+6x+5=5 =>-8x^2+6x=0
en factorisant -2x(4x-3) soit x=0 et x=3/4
f) forme 1 car 4x-5<=0 =>x<+5/4
-2x-1<=0 =>x>=-1/2
ps: x^2 c'est x au carre
1)
a) f(x) = (x - 3)² - (3x - 2)² = (x - 3 - 3x + 2)(x - 3 + 3x - 2)
= (- 2x - 1)(4x - 5) = - (2x +1)(4x - 5) .
b) f(x) = - (2x +1)(4x - 5) = - (8x² - 10x + 4x - 5)
= - (8x² - 6x - 5) = - 8x² + 6x + 5 .
c) f(x) = - 8x² + 6x + 5 = - 8(x² - 6/8 x - 5/8)
= - 8(x² - 2 * 3/8 * x + 9/64 - 9/64 - 5/8)
= - 8((x - 3/8)² - 9/64 - 40/64)
= - 8((x - 3/8)² - 49/64)
= - 8(x - 3/8)² + 49/8 .