Odpowiedź:
równanie paraboli:
[tex]y=ax^2+bx+c[/tex]
podstawiamy dane 2 punkty W(-1,8) oraz P(0,15/2) pamiętając jak zapisywane są punkty: P(x,y) wychodzi nam układ równań:
[tex]\left \{ {{8=(-1)^2\cdot a+b \cdot (-1)+c} \atop {\frac{15}{2} =(0)^2\cdot a+b \cdot (0)+c}} \right. \\\left \{ {{8=a-b+c} \atop {\frac{15}{2} =c}} \right. \\a-b=8-\frac{15}{2} =\frac{1}{2}[/tex]
z równania na wierzchołek paraboli:
[tex]p=\frac{-b}{2a} \\-1=\frac{-b}{2a}\\ b=2a[/tex]
podstawiając do tego pierwszego równania:
[tex]a-2a=\frac{1}{2} \\a=-\frac{1}{2}\\b=-1[/tex]
więc: ostatecznie b = -1
[tex]f(x)=ax^{2} +bx+c\\Dane ~~z~~zadania:\\\\W=(-1,8)~~wspolrzedne~~wierzcholka~~paraboli\\P=(0,\frac{15}{2} )~~punkt ,~~ktory~~~P\in f(x)~~\land~~przecina ~~OY\\Dane~~z~~wykreu:\\Odczytac~~mozna~~miejsca~~zerowe:\\M_{0} =\{ -5,~~3\}~~\Rightarrow ~~A=(-5,0) ~~\land~~B=(3,0) ~~\land ~~A,B\in f(x)\\\\P=(0,\frac{15}{2} )~~\land~~P\in f(x)~~\Rightarrow ~~0^{2} a+0b+c=\frac{15}{2}~~\Rightarrow ~~c=\frac{15}{2} \\\\[/tex]
[tex]A=(-5,0)~~A\in f(x)~~\Rightarrow ~~(-5)^{2} a-5b+c=0~~\land ~~c=\frac{15}{2} ~~\Rightarrow~~25a-5b+\frac{15}{2} =0\\\\B=(3,0)~~A\in f(x)~~\Rightarrow ~~3^{2} a+3b+c=0~~\land ~~c=\frac{15}{2} ~~\Rightarrow~~9a+3b+\frac{15}{2} =0\\\\Mamy ~~dwa~~rownania~~i~~dwie~~niewiadome~~a~~oraz~~b\\25a-5b+\frac{15}{2} =0~~\mid \cdot 2~~\Rightarrow ~~50a-10b+15=0~~\mid \div 5~~\Rightarrow ~~10a-2b+3=0\\\\9a+3b+\frac{15}{2} =0~~\mid \cdot 2~~\Rightarrow ~~18a+6b+15=0~~\mid \cdot 3~~\Rightarrow ~~6a+2b+5=0\\\\[/tex]
[tex]\left \{ {{10a-2b+3=0} \atop {6a+2b+5=0}} \right. ~~~\mid + \\\\10a-2b+3+6a+2b+5=0\\\\16a=-8~~\mid \div 16\\\\a=-\dfrac{1}{2} \\\\10a-2b+3=0~~\land ~~a=-\dfrac{1}{2}~~\Rightarrow~~10\cdot (-\dfrac{1}{2}) -2b +3=0\\\\-5-2b+3=0\\-2b-2=0\\-2b=2~~\mid \div (-2)\\b=-1\\\\f(x)=ax^{2} +bx+c~~\land ~~a=-\dfrac{1}{2} ~~\land~~b=-1~~\land ~~c=\dfrac{15}{2} =7\dfrac{1}{2} \\\\Szukana ~~funkcja~~kwadratowa:~~\\\\f(x)=-\dfrac{1}{2} x^{2} -x+7\dfrac{1}{2}[/tex]
[tex]Sprawdzam:\\\\f(x)=-\dfrac{1}{2} x^{2} -x+7\dfrac{1}{2}\\\\f(0)=7\dfrac{1}{2}~~\Rightarrow ~~P\in f(x)\\\\f(-1)=-\dfrac{1}{2} (-1)^{2} -(-1)+7\dfrac{1}{2}=-\dfrac{1}{2} +1+7\dfrac{1}{2}=8~~\Rightarrow~~W\in f(x)\\\\f(-5)=-\dfrac{1}{2} (-5)^{2} -(-5)+7\dfrac{1}{2}=-\dfrac{25}{2} +5+7\dfrac{1}{2}=-12\dfrac{1}{2} x^{2} +5+7\dfrac{1}{2}=0~~\Rightarrow~~A\in f(x)\\\\f(3)=-\dfrac{1}{2} 3^{2} -3+7\dfrac{1}{2}=-4\dfrac{1}{2} -3+7\dfrac{1}{2}=0~~\Rightarrow~~B\in f(x)[/tex]
