Răspuns:
Explicație pas cu pas:
Exercitiul 12
a) f(x)=[tex]-x^{2} +4x[/tex]
Gf∩O×=> y=0
f(x)=y=> [tex]-x^{2} +4x[/tex]=0=> x(-x+4)=0=> x=0 A(0,0)
-x+4=0=> x=4 B(4,0)
Gf∩Oy=> x=0 f(0)=0
b)f(x)=[tex]x^{2}[/tex]-7x+10
Gf∩O×=> y=0
f(x)=y=>[tex]x^{2}[/tex]-7x+10=0 d=[tex]b^{2} -4*a*c[/tex] => d=[tex](-7)^{2}[/tex]-4*10*1=49-40=9
[tex]x_{1} =\frac{-b+\sqrt[]{d} }{2*a}[/tex] [tex]x_{1}=\frac{-(-7)+\sqrt[]{9} }{2*1}[/tex]=5 A(5,0)
[tex]x_{2} =\frac{-b+\sqrt[]{d} }{2*a}=\frac{-(-7)-\sqrt{9} }{2*1}[/tex]=2 B(2,0)
Gf∩Oy=> x=0 f(0)=10 C(0,10)
c)f(x)=[tex]x^{2}[/tex]-4x-12
Gf∩O×=> y=0
f(x)=y=>[tex]x^{2}[/tex]-4x-12=0 d=[tex]b^{2} -4*a*c[/tex] => d=[tex](-4)^{2}[/tex]-4*(-12)*1=16+48=64
[tex]x_{1} =\frac{-b+\sqrt[]{d} }{2*a}[/tex] =[tex]\frac{-(-4)+\sqrt{64} }{2*1}[/tex]=6 A(6,0)
[tex]x_{2}=\frac{-b-\sqrt{d} }{2*a}=\frac{-(-4)-\sqrt{64} }{2*1}=-2[/tex] B(-2,0)
Gf∩Oy=> x=0 f(0)=-12 C(0,-12)
d)f(x)=3[tex]x^{2}[/tex]-9x+7
Gf∩O×=> y=0
f(x)=y=>3[tex]x^{2}[/tex]-9x+7=0 d=[tex]b^{2} -4*a*c[/tex] => d=[tex]3^{2}[/tex]-4*3*7=9-84=-75<0 ecuatia nu admite solutii reale
Gf∩Oy=> x=0 f(0)=7 A(0,7)
e)f(x)=-4[tex]x^{2}[/tex]+17x-15
Gf∩O×=> y=0
f(x)=y=>-4[tex]x^{2}[/tex]+17x+-15=0 d=[tex]b^{2} -4*a*c[/tex] => d=[tex]17^{2}[/tex]-4*(-4)*(-15)=289-240=49
[tex]x_{1}=\frac{-b+\sqrt{d} }{2*a}=\frac{-17+\sqrt{49} }{2*(-4)}=\frac{5}{4} A(\frac{5}{4},0)[/tex]
[tex]x_{2}=\frac{-b-\sqrt{d} }{2*a}=\frac{-17-\sqrt{49} }{2*(-4)}=3 B(3,0)[/tex]
Gf∩Oy=> x=0 f(0)=-15 C(0,-15)
f)f(x)=9[tex]x^{2}[/tex]+12x+4
f(x)=y=>9[tex]x^{2}[/tex]+12x+4=0 d=[tex]b^{2} -4*a*c[/tex] => d=[tex]12^{2}[/tex]-4*4*9=144-144=0
[tex]x_{1}=x_{2}=\frac{-b}{2*a}=\frac{-12}{2*9}=\frac{-2}{3} A(\frac{-2}{3},0)[/tex]
Gf∩Oy=> x=0 f(0)=4 B(0,4)