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Podaj współrzędne wierchołka paraboli któa jest wykresem funkcji
y=-3-7
B) y=
C) y=
D) y=
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współrzedne paraboli to
p=-b/2a i q=-delta/4a
b] y=3x^2-7
p=0
delta=0^2-4*(-3)*(-7)=0+84=84
q=-84/4*(-3)=84/12=7
w=(0,7)
c]
y=3/7(x+5)^2+1/5
3/7(x^2+10x+25)+1/5 =0 /*35
15(x^2+10x+25)+7
15x^2+150x+375+7=0
15x^2+150x+382=0
Δ=22500-4*15*382=22500-22920=-420
p=-150/30=-5
q=420/60=7
W(-5,7)
d]
y=4x^2+10x-6
delta=10+-4*4*6=100+96=196
p=-10/2*4=-10/8=-5/4
q=-196/4*4=-196/16=-12,25
w=(-5/4 ; -12 1/4)
prosze bardzo
wierzchołek W( p,q)
p= -b/2a
q = -Δ/4a
B)
y= - 3x² - 7
-3x² - 7 = 0
Δ = b² -4ac
∆ = 0 – 4 * (-3)*(-7) = -84
p = -0/2 = 0
q= 84/4*(-3) = 84/-12 = -7
W(0,7)
C)
y = 3/7(x+5)² + 1/5
3/7(x+5)² + 1/5 = 0 /*35
15(x+5)² + 7 = 0
15(x² + 10x + 25) + 7 = 0
15x² + 150x + 375 + 7 = 0
15x² + 150x + 382 = 0
Δ = b² -4ac
Δ =150² -4*15*382 = 22500 – 22920 = -420
p=-b/2a
p = -150/2*15 = -150/30 = -5
q=-∆/4a
q = 420/4*15=420/60 =7
W (-5,7)
D)
y = 4x² +10x – 6
4x² +10x – 6 = 0
Δ = b² -4ac
Δ = 10² -4*4* (-6) = 100 + 96 =196
p = -b/2a
p =-10/2*4 = -10/8 = 1,25
q=-∆/4a
q= -196/4*4 = -196/16 = 12,25
W (1,25 ; 12,25)