1 szkoły ponadpodstawowej

I liceum

I technikum

MATeMAtyka 1. Zakres podstawowy. Po gimnazjum

a)   dla x=−2   (−2)2−3⋅(−2)−6=   4+6−6=4

Rozwiązanie 1

a)   dla x=−21​   4⋅(−21​)2−2⋅(−21​)+3=   =4⋅41​+1+3=1+1+3=5

a)   f(−2)=−(−2)2+6⋅(−2)+1=            =−4−12+1=−15   g(−2)=2⋅(−2)2+10⋅(−2)−5=            =2⋅4−20−5=   8−25=−17   h(−2)=−3⋅(−2)2−21​⋅(−2)+4=            =   −3⋅4+1+4=−12+5=−7

a)   f(−3)=(−3)2−2⋅(−3)−1=            =9+6−1=14   f(−2)=(−2)2−2⋅(−2)−1=            =4+4−1=7   f(−1)=(−1)2−2⋅(−1)−1=            =1+2−1=2

punkt A   f(−1)=2⋅(−1)2−2<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>​⋅(−1)−22<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>​=            =   2⋅1+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>​−22<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>​=   2−2<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>​=2−32<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>​   Punkt A nie należy do wykresu funkcji f.

a)   f(−3)=21​⋅(−3)2+(−3)−1=            =21​⋅9−3−1=   29​−4=421​−4=21​   f(1)=21​⋅12+1−1=   21​   f(−3)=f(1)

a)   f(−2)=(−2)2−(−2)−2=            =   4+2−2=4   f(−1)=(−1)2−(−1)−2=            =1+1−2=0