![y=(x+9)^2\cdot (x-5)-5\; \; ,\; \; x\in [-19;-5\; ]\\\\\\y'=2(x+9)(x-5)+(x+9)^2=(x+9)\cdot (2x-10+x+9)=\\\\=(x+9)(3x-1)=0\; \; \to \; \; \; x_1=-9\in [-19;-5\; ]\; ,\; x_2=\frac{1}{3}\notin [-19;-5\; ]\\\\\\y(-19)=(-19+9)^2(-19-5)-5=-2405\\\\y(-9)=(-9+9)^2(-9-5)-5=-5\\\\y(-5)=(-5+9)^2(-5-5)-5=-165\\\\y_{naibolshee}=y(-9)=-5](/tpl/images/1078/1598/9cac3.png)
ответ:
d=b^2-4ac=(-1)^2-4*1*(-72)=1+288=\sqrt{289}
289
=17
х1=\frac{-b- \sqrt{d} }{2a} = \frac{1-17}{2} = \frac{-16}{2} =-8
2a
−b−
d
=
2
1−17
=
2
−16
=−8
х2=\frac{-b+ \sqrt{d} }{2a} = \frac{1+17}{2} = \frac{18}{2} = 9
2a
−b+
d
=
2
1+17
=
2
18
=9
ответ: -8 и 9
d=b^2-4ac=7^2-4*(-4)*(-3)=49-48=\sqrt{1} =1
1
=1
х1=\frac{-b- \sqrt{d} }{2a} = \frac{-7-1}{2*(-4)} = \frac{-8}{-8} =1
2a
−b−
d
=
2∗(−4)
−7−1
=
−8
−8
=1
х2=\frac{-b+ \sqrt{d} }{2a} = \frac{-7+1}{(-8)} = \frac{-6}{-8} =0,75
2a
−b+
d
=
(−8)
−7+1
=
−8
−6
=0,75
