Question

65 pontos pra quem conseguir responder todas essas ai tem q ser resposta completa !!

Answer 1

1) $f(x)=x^2-x-12$

a) $f(-2)=(-2)^2-(-2)-12=4+2-12=-4$

b) $f(3)=3^2-3-12=9-3-12=-6$

c) $f(0)=0^2-0-12=-12$

d) $f(5)=5^2-5-12=25-5-12=8$

e) $f(-2/5)=(-2/5)^2-(-2/5)-12=0,16+0,4-12=-11,44$


2)

a) $y=x^2+x-2$

$y=0$

$x^2+x-2=0$

$\Delta=1^2-4\cdot1\cdot(-2)=1+8=9$

$x=\dfrac{-1\pm\sqrt{9}}{2}=\dfrac{-1\pm3}{2}$

$x'=\dfrac{-1+3}{2}=\dfrac{2}{2}=1$

$x"=\dfrac{-1-3}{2}=\dfrac{-4}{2}=-2$


b) $y=-2x^2-10x-8$

$y=0$

$-2x^2-10x-8=0$

$-x^2-5x-4=0$

$\Delta=(-5)^2-4\cdot(-1)\cdot(-4)=25-16=9$

$x=\dfrac{-(-5)\pm\sqrt{9}}{2\cdot(-1)}=\dfrac{5\pm3}{-2}$.

$x'=\dfrac{5+3}{-2}=\dfrac{8}{-2}=-4$

$x"=\dfrac{5-3}{-2}=\dfrac{2}{-2}=-1$


3)

a) $y=x^2+x-2$

$x_v=\dfrac{-b}{2a}$

$x_v=\dfrac{-1}{2\cdot1}=\dfrac{-1}{2}$

$y_v=\dfrac{-\Delta}{4a}$

$\Delta=9$

$y_v=\dfrac{-9}{4}$


b) $y=-2x^2-10x-8$

$x_v=\dfrac{-(-10)}{2\cdot(-2)}=\dfrac{10}{-4}=\dfrac{-5}{2}$.

$\Delta=(-10)^2-4\cdot(-2)(-8)=100-64=36$

$y_v=\dfrac{-36}{4\cdot(-2)}=\dfrac{-36}{-8}=\dfrac{9}{2}$


4)

a) $y=x^2-2bx-2$

$y=10$, se $x=2$:

$2^2-4b-2=10$

$4b=4-2-10$

$4b=-8$

$b=-2$

b) $y=2x^2+bx+c$

$y=13$, se $x=1$ e $c-b=5$, -> $c=b+5$

$2+b+c=13$

$2+b+b+5=13$

$2b=6$

$b=3$ e $c=8$.


5) $y=x^2+x-2$

$x'=1$ e $x"=-2$

$x_v=\dfrac{-1}{2}$ e $y_v=\dfrac{-9}{4}$.