Question

zad. Oblicz współrzędne wierzchołka. W wykresu funkcji kwadratowej: a0 y=x do potęgi 2 +x-4 b) y= x do potęgi 2 -3x+2 c) y= x do potęgi 2 +4x+1 d) y= 1/2 x do potęgi2 -x+1 e) y= -x do potęgi 2 +3x+1 f) y= 2x do potęgi 2 + x - 1/2

Answer 1

a)

y = x^2 + x - 4

a = 1, b = 1 , c = - 4

p = -b/(2a) = -1/(-2) = 1/2

q = (1/2)^2 + 1/2 - 4 = 1/4 + 2/4 - 4 =3/4 - 4 = - 3  1/4

zatem

W = (p; q) = ( 1/2;  - 3  1/4)

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b)

y = x^2 - 3x + 2

p = 3/(2) =  3/2 = 1,5

delta = b^2 - 4ac = (-3)^2 - 4*1*2 = 9 - 8 = 1

q = - delta/(4a) = -1/4

zatem

W = ( p;q ) = (  3/2;  - 1/4 )

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c)

y = x^2 + 4x + 1

p = -4/2 = - 2

q = (-2)^2 +4*(-2) + 1 = 4 - 8 + 1 = - 3

zatem

W = (p;q) = ( -2; - 3)

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d)

y = (1/2) x^2 - x + 1

p = 1/( 2*(1/2)) = 1/1 = 1

q = (1/2)*1^2 - 1 + 1 = (1/2)*1 = 1/2

zatem

W = ( p;q) = ( 1 ; 1/2)

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e)

y = - x^2 +3x + 1

a = - 1,  b = 3 ,  c = 1

p = -3/(-2) = 3/2

q = - (3/2)^2 + 3*(3/2) + 1 = - 9/4 + 9/2 + 1 = -9/4 + 18/4 + 4/4 = 13/4

q = 3  1/4

zatem

W = ( p; q ) = ( 3/2 ;   3   1/4)

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f)

y = 2 x^2 + x - 1/2

p = - 1/4

delta = b^2 - 4ac = 1^2 - 4*2*(-1/2) = 1 + 4 = 5

q = - delta/(4a) = - 5/8

zatem

W = ( p; q) = ( - 1/4 ;  - 5/8 )

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q   obliczano  dwoma sposobami

1) jako wartośc funkcji dla x = p

2) ze wzoru

q = - delta/(4a), gdzie   delta = b^2 - 4ac

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