PILNE ! Przedstaw funkcję kwadratową w postaci ogólnej. Oblicz wyróżnik trójmianu kwadratowego. Wy.... Question from @Crazy98

January 2019 1 21 Report

PILNE !
Przedstaw funkcję kwadratową w postaci ogólnej.
Oblicz wyróżnik trójmianu kwadratowego. Wyznacz współrzędne wierzchołka paraboli będącego wykresem tego trójmianu.
Proszę o szybkie rozwiązanie tych zadań. Z góry dziękuje :)
Zadania nr 1 i 2 strona 188

luki130193 Zad.1
a)
$y=(x- \frac{1}{2})^2+4$
$y=x^2-x+ \frac{1}{4}+4=x^2-x+4 \frac{1}{4}$
b)
$y=-(x-4)^2-3$
$y=-(x^2-8x+16)-3=-x^2+8x-16-3=-x^2+8x-19$
c)
$y=2(x-3)^2+1$
$y=2(x^2-6x+9)+1=2x^2-12x+18+1=2x^2-12x+19$
d)
$y=-3(x+2)^2-4$
$y=-3(x^2+4x+4)-4=-3x^2-12x-12-4=-3x^2-12x-16$
e)
$y=- \frac{1}{2} (x-2)^2+2$
$y=- \frac{1}{2} (x^2-4x+4)+2 =- \frac{1}{2}x^2+2x-2+2=- \frac{1}{2} x^2+2x$
f)
$y=- \frac{1}{3} (x-3)^2+4$
$y=- \frac{1}{3} (x^2-6x+9)+4 =- \frac{1}{3}x^2+2x-3+4=- \frac{1}{3}x^2+2x+1$

Zad.2
a)
$y=x^2+4x-3$
a = 1 b = 4 c = -3
Δ = $b^2-4ac=4^2-4*1*(-3)=16+12=28$
$p= \frac{-b}{2a} = \frac{-4}{2*1} = \frac{-4}{2} =-2$
$q= \frac{-delta}{4a} = \frac{-28}{4*1}= \frac{-28}{4}=-7$
$W=(p;q)=(-2;-7)$
b)
$y=x^2-2x+6$
a = 1 b = -2 c = 6
Δ = $b^2-4ac=(-2)^2-4*1*6=4-24=-20$
$p= \frac{-b}{2a} = \frac{-(-2)}{2*1}= \frac{2}{2}=1$
$q= \frac{-delta}{4a} = \frac{-(-20)}{4*1} = \frac{20}{4} =5$
$W=(p;q)=(1;5)$
c)
$y =-x^2+6x-1$
a = -1 b = 6 c = -1
Δ = $b^2-4ac=6^2-4*(-1)*(-1)=36-4=32$
$p= \frac{-b}{2a} = \frac{-6}{2*(-1)}= \frac{-6}{-2} =3$
$q= \frac{-delta}{4a} = \frac{-32}{4*(-1)} = \frac{-32}{-4} =8$
$W=(p;q)=(3;8)$
d)
$y=2x^2+4x-5$
a = 2 b = 4 c = -5
Δ = $b^2-4ac=4^2-4*2*(-5)=16+40=56$
$p= \frac{-b}{2a} = \frac{-4}{2*2} = \frac{-4}{4} =-1$
$q= \frac{-delta}{4a} = \frac{-56}{4*2} = \frac{-56}{8} =-7$
$W=(p;q)=(-1;-7)$
e)
$y= -2x^2-x+2$
a = -2 b = -1 c = 2
Δ = $b^2-4ac=(-1)^2-4*(-2)*2=1+16=17$
$p= \frac{-b}{2a} = \frac{-(-1)}{2*(-2)} = \frac{1}{-4} =- \frac{1}{4}$
$q= \frac{-delta}{4a} = \frac{-17}{4*(-2)} = \frac{-17}{-8} = \frac{17}{8} =2 \frac{1}{8}$
$W=(p;q)=(- \frac{1}{4} ;2 \frac{1}{8})$
f)
$y=-3x^2+6x-2$
a = -3 b = 6 c = -2
Δ = $b^2-4ac=6^2-4*(-3)*(-2)=36-24=12$
$p= \frac{-b}{2a} = \frac{-6}{2*(-3)} = \frac{-6}{-6} =1$
$q= \frac{-delta}{4a} = \frac{-12}{4*(-3)} = \frac{-12}{-12} =1$
$W=(p;q)=(1;1)$
g)
$y=3x^2-2x+ \frac{1}{2}$
a = 3 b = -2 $c= \frac{1}{2}$
Δ = $b^2-4ac=(-2)^2-4*3* \frac{1}{2} =4-6=-2$
$p= \frac{-b}{2a} = \frac{-(-2)}{2*3} = \frac{2}{6}= \frac{1}{3}$
$q= \frac{-delta}{4a} = \frac{-(-2)}{4*3} = \frac{2}{12}= \frac{1}{6}$
$W=(p;q)=( \frac{1}{3}; \frac{1}{6})$
h)
$y=- \frac{3}{4}x^2+x+1$
$a = - \frac{3}{4}$ b = 1 c = 1
Δ = $b^2-4ac=1^2-4*(- \frac{3}{4} )*1=1+3=4$
$p= \frac{-b}{2a} = \frac{-1}{2*(- \frac{3}{4}) } = \frac{-1}{- \frac{3}{2} } = \frac{2}{3}$
$q= \frac{-delta}{4a} = \frac{-4}{4*(- \frac{3}{4}) } = \frac{-4}{-3} = \frac{4}{3} =1 \frac{1}{3}$
$W=(p;q)=( \frac{2}{3};1 \frac{1}{3})$
i)
$y=2x^2 - 5$
a = 2 b = 0 c = -5
Δ = $b^2-4ac=-4ac=-4*2*(-5)=40$
$p= \frac{-b}{2a} = \frac{0}{4*2} =0$
$q= \frac{-delta}{4a} = \frac{-40}{4*2} = \frac{-40}{8}=-5$
$W=(p;q)=(0;-5)$

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