1 szkoły ponadpodstawowej

I liceum

I technikum

MATeMAtyka 1. Zakres podstawowy. Po gimnazjum

Rozwiązanie 1

a) x2−50=0   ∣+50    x2=50    x1​=50<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
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M834 80h400000v40h-400000z'/></svg>​=25⋅2<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
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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
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M834 80h400000v40h-400000z'/></svg>​⋅2<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
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​=52<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
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​    x2​=−50<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
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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
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c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​=−52<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
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l0 -0
c5.3,-9.3,12,-14,20,-14
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​     b) 2x2−7=0   ∣+7    2x2=7   ∣:2

Oś symetrii paraboli to pionowa prosta przechodząca przez wierzchołek paraboli. Aby obliczyć równanie tej osi (x=a) wystarczy policzyć średnią arytmetyczną z miejsc zerowych     a)   x2−8x=0   x(x−8)=0   x=0   lub   x−8=0                    x=8   x1​=0,  x2​=8   ⇒   x=20+8​=28​=4   - oś symetrii paraboli       b)   x2+6x=0

a)   x2+4x−21=0   Δ=42−4⋅1⋅(−21)=   16+84=100   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
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M834 80h400000v40h-400000z'/></svg>​=100<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
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M834 80h400000v40h-400000z'/></svg>​=10

Skorzystamy z tego, że pierwsza współrzędna wierzchołka paraboli to średnia arytmetyczna miejsc zerowych funkcji kwadratowej, której wykresem jest ta parabola.  xw​=2x1​+x2​​    ∣x1​=3   xw​=23+x2​​   ∣⋅2

Punkt przecięcia paraboli z osią Y ma współrzędne (0, f(0)), ten punkt na potrzeby zadania oznaczymy A.  Punkty przecięcia paraboli z osią X to miejsca zerowe, o współrzędnych (x, 0), oznaczymy je B i C.

a) a=1,  b=0,  c=−7 xw​=−2ab​=−20​=0 yw​=f(xw​)=f(0)=02−7=−7 Współczynnik a jest dodatni, więc ramiona paraboli są skierowane w górę, osiągana jest wartość najmniejsza,  czyli -7.

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