1 szkoły ponadpodstawowej

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MATeMAtyka 1. Zakres podstawowy. Po gimnazjum

a) f(x)=2x2−6x=2x(x−3)

Rozwiązanie 1

a) Wykorzystując podane współrzędne, zapisujemy jakie wartości osiąga funkcja dla danych argumentów:  ⎩<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.316em' style='width:0.8889em' viewBox='0 0 888.89 316' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V316 H384z M384 0 H504 V316 H384z'/></svg>⎨<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.316em' style='width:0.8889em' viewBox='0 0 888.89 316' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V316 H384z M384 0 H504 V316 H384z'/></svg>⎧​f(0)=3f(1)=0f(3)=0​ Mając miejsca zerowe możemy zapisać postać iloczynową:

y=f(x)   →SOY​   y=f(−x)   Przekształcając funkcję symetrycznie względem osi Y zamieniamy x na -x.

a)   3x2−1≥0   ∣:3   x2−31​≥0   (x−31​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
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c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5,25c-5.7,9.3,-9.8,16,-12.5,20
s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13,-13s76,-122,76,-122s77,-121,77,-121
s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081
l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​)(x+31​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
c-1.3,-0.7,-38.5,-172,-111.5,-514c-73,-342,-109.8,-513.3,-110.5,-514
c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5,25c-5.7,9.3,-9.8,16,-12.5,20
s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13,-13s76,-122,76,-122s77,-121,77,-121
s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081
l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​)≥0

a) (2x+3)2≤(x−3)2       ∣−(x−3)2 (2x+3)2−(x−3)2≤0           ∣a2−b2=(a−b)(a+b) [(2x+3)−(x−3)]⋅[(2x+3)+(x−3)]≤0 [2x+3−x+3]⋅[2x+3+x−3]≤0

x2−x−6≤0     Δ=(−1)2−4⋅1⋅(−6)=   1+24=25   Δ<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​=25<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​=5

a) f(x)=2x2+4x=2(x2+2x)=        =2(x2+2x+1−1)=        =2((x+1)2−1)=

a)   a+b=4   −a   b=4−a

(6+x), (12+x)  - wymiary nowego wybiegu (w metrach), oczywiście x jest liczbą dodatnią   P1​=6⋅12=72 P2​=(6+x)⋅(12+x)=6(12+x)+x(12+x)=     =72+6x+12x+x2=x2+18x+72