Question

1) f (x)=x2-12x+35

a) f (3)=
b) f (-2)=
c) f (10)=
d) f (0)=
e) zeros da função
f) o vértice da parábola
g) o gráfico da função

2) f (x)=x2-x-20

a) f (-3)=
b) f (2)=
c) f (-8)=
d) f (0)=
e) os zeros da função
f) o vértice da parábola
g) o gráfico da função

3) f (x)=x2-8x-12

a) f (5)=
b) f (-1)=
c) f (12)=
d) f (0)=
e) os zeros da função
f) o vértice da parábola
g) o gráfico da função

Answer 1

1) $f(x)=x^2-12x+35$

a) $f(3)=3^2-12\cdot3+35=9-36+35=8$

b) $f(-2)=(-2)^2-12\cdot(-2)+35=4+24+35=63$

c) $f(10)=10^2-12\cdot10+35=100-120+35=15$

d) $f(0)=0^2-12\cdot0+35=0+0+35=35$

e) $x^2-12x+35=0$

$\Delta=(-12)^2-4\cdot1\cdot35=144-140=4$

$x=\dfrac{-(-12)\pm\sqrt{4}}{2\cdot1}=\dfrac{12\pm2}{2}=6\pm1$

$x'=6+1=7$

$x"=6-1=5$

f) $x_v=\dfrac{-b}{2a}=\dfrac{-(-12)}{2\cdot1}=\dfrac{12}{2}=6$

$y_v=\dfrac{-\Delta}{4a}=\dfrac{-4}{4\cdot1}=\dfrac{-4}{4}=-1$

g) Está na imagem em anexo


2) $f(x)=x^2-x-20$

a) $f(-3)=(-3)^2-(-3)-20=9+3-20=-8$

b) $f(2)=2^2-2-20=4-2-20=-18$

c) $f(-8)=(-8)^2-(-8)-20=64+8-20=52$

d) $f(0)=0^2-0-20=0-0-20=-20$

e) $x^2-x-20=0$

$\Delta=(-1)^2-4\cdot1\cdot(-20)=1+80=81$

$x=\dfrac{-(-1)\pm\sqrt{81}}{2\cdot1}=\dfrac{1\pm9}{2}$

$x'=\dfrac{1+9}{2}=\dfrac{10}{2}=5$

$x"=\dfrac{1-9}{2}=\dfrac{-8}{2}=-4$

f) $x_v=\dfrac{-b}{2a}=\dfrac{-(-1)}{2}=\dfrac{1}{2}$

$y_v=\dfrac{-\Delta}{4a}=\dfrac{-81}{4}$

g) Está em anexo


3) $f(x)=x^2-8x-12$

a) $f(5)=5^2-8\cdot5-12=25-40-12=-27$

b) $f(-1)=(-1)^2-8\cdot(-1)-12=1+8-12=-3$

c) $f(12)=12^2-8\cdot12-12=144-96-12=36$

d) $f(0)=0^2-8\cdot0-12=0-0-12=-12$

e) $x^2-8x-12=0$

$\Delta=(-8)^2-4\cdot1\cdot(-12)=64+48=112$

$x=\dfrac{-(-8)\pm\sqrt{112}}{2\cdot1}=\dfrac{8\pm4\sqrt{7}}{2}=4\pm2\sqrt{7}$

$x'=4+2\sqrt{7}$

$x"=4-2\sqrt{7}$

f) $x_v=\dfrac{-b}{2a}=\dfrac{-(-8)}{2}=4$

$y_v=\dfrac{-\Delta}{4a}=\dfrac{-112}{4}=-28$

g) Em anexo