Answer:
a. -2x^2 + 3x+5=y
b.-x^2 +5x-4=y
c.-x^2 +6x-12=y
Step-by-step explanation:
a.y=-2x^2 +bx+c
the points P and Q lie on curve
[tex]\left \{ {{-b+c=2} \atop {b+c=8}} \right.[/tex]
=> b=3, c=5
b. delta = b^2 +4c
x1= [tex]\frac{-b+\sqrt{b^{2}+4c } }{-2}[/tex] =4
=> [tex]b-\sqrt{4c+b^{2} }[/tex]=8
x2=[tex]\frac{-b-\sqrt{b^{2}+4c } }{-2}[/tex]=1
=>2=b+[tex]\sqrt{b^{2}+4c }[/tex]
suppose : [tex]\sqrt{b^{2}+4c }[/tex] = a
=> [tex]\left \{ {{a+b=2} \atop {-a+b=8}} \right.[/tex]
=> a=-3
b=5
a=-3 => [tex]\sqrt{b^{2}+4c }[/tex] =-3 => b^2 +4c =9 =>5^2 +4c=9 => c=-4.
c.vertex (3;-3)
[tex]\frac{-b}{-2}[/tex]=3 => b =6
[tex]\frac{-D}{4a}[/tex]=[tex]\frac{-b^{2}+4ac}{-4}[/tex] =-3 =>-b^2+4ac=12 => c=-12.
