Question

Construa o gráfico das funções abaixo. atribuindo a x os valores: -2,-1.0.1, 2.
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Answer 1

Resposta:

9)

A construção do gráfico, vc localiza os pontos A,B,C,D e E no plano Cartesiano e liga todos os pontos. Isso em cada alternativa.

Com base no exemplo da letra a) que está anexado acima construa os demais gráficos das outras alternativas.

a)

$y = {x}^{2} + 2x - 3 \\ \\ f( - 2) = ( - 2 {)}^{2} + 2( - 2) - 3 \\ \\ f( - 2) = \red{ 4 - 4 }- 3 \\ \\ f( - 2) = - 3 \\ \\ \Large \boxed{ A( - 2, - 3)}$

$y = {x}^{2} + 2x - 3 \\ \\ x = - 2 ,- 1,0,1,2 \\ \\ f( - 1) = {( - 1)}^{2} + 2( - 1) - 3 \\ \\ f( - 1) = 1 - 2 - 3 \\ \\ f( - 1) = - 4 \\ \\ \Large \boxed{B ( - 1, - 4)}$

$f(0) = {(0)}^{2} + 2(0) - 3 \\ \\ f(0) = - 3 \\ \\ \Large \boxed{ C(0, - 3)}$

$f(1) = {(1)}^{2} + 2(1) - 3 \\ \\ f(1) = 1 + 2 - 3 \\ \\ f(1) = 3 - 3 \\ \\ f(1) = 0 \\ \\ \Large \boxed{D(1,0)}$

$f(2) = {(2)}^{2} + 2(2) - 3 \\ \\ f(2) = 4 + 4 - 3 \\ \\ f(2) = 8 - 3 \\ \\ f(2) = 5 \\ \\ \Large \boxed{E(2,5)}$

$y = {x}^{2} + 2x - 3$

$\Large \boxed{\bf V \left( \frac{ - b}{2a} , \frac{ - \Delta}{4a} \right) }\\ \\ y = {x}^{2} + 2x - 3 \\ \\ a = 1 \: \\ b = 2 \\ \: c = - 3 \\ \\\Large \boxed{ \bf \Delta = {b}^{2} - 4ac }\\ \\ \Delta = ( {2)}^{2} - 4(1)( - 3) \\ \\ \Delta = 4 + 12 \\ \\ \Large \boxed{\Delta = 16} \\ \\ xv = - \frac{2}{2(1)} \\ \\ xv = - \frac{ 2 }{2} \\ \\ \Large \boxed{ xv = - 1} \\ \\ yv = \frac{ - 16}{4(1)} \\ \\ yv = \frac{ - 16}{4} \\ \\ \Large \boxed{ yv = - 4} \\ \\ \Large{ \boxed{ \green{ V \bf( - 1, - 4)}}}$

b)

$y = {x}^{2} + 2x \\ \\ f(2) = {( - 2)}^{2} + 2( - 2) \\ \\ f( - 2) = 4 - 4 \\ \\ f( - 2) = 0 \\ \\ \Large \boxed{A( - 2,0)} \\ \\ f( - 1) = ( - {1)}^{2} + 2( - 1) \\ \\ f( - 1) = 1 - 2 \\ \\ { f( - 1) = - 1} \\ \\ \Large \boxed{B( - 1,- 1) }\\ \\ f(0) = ( {0)}^{2} + 2(0) \\ \\ f(0) = 0 \\ \\ \Large \boxed{ C(0,0)} \\ \\ f(1) = ( {1)}^{2} + 2(1) \\ \\ f(1 )= 1 + 2 \\ \\ { f(1) = 3} \\ \\ \Large \boxed{D(1,3)} \\ \\ f(2) = ( {2)}^{2} + 2(2) \\ \\ f(2) = 4 + 4 \\ \\ f(2) = 8 \\ \\ \Large \boxed{E(2,8)}$

$y = {x}^{2} + 2x \\ \\ xv = \frac{ - 2}{2(1)} \\ \\ \large \boxed{xv = - 1} \\ \\yv = f( - 1) \\ \\ yv = ( - 1) {}^{2} + 2( - 1) \\ \\ yv = 1 - 2 \\ \\ \Large \boxed{ yv = - 1} \\ \\ \Large \boxed{ \green{ \: V \bf( - 1, - 1)}}$

c)

$y = {x}^{2} - 2x + 1 \\ \\ f( - 2) = ( { - 2)}^{2} - 2( - 2) + 1 \\ \\ f( - 2) = 4 + 4 + 1 \\ \\ f( - 2) = 9 \\ \\ \Large \boxed{A (- 2,9)}$

$f( - 1) = ( { - 1)}^{2} - 2( - 1) + 1 \\ f( - 1) = 1 + 2 + 1 \\ \\ f( - 1) = 4 \\ \\ \Large \boxed{B( - 1,4)}$

$f(0) = {(0)}^{2} - 2(0) + 1 \\ \\ f(0) = 1 \\ \\ \Large \boxed{C(0,1)}$

$f(1) = {(1)}^{2} - 2(1) + 1 \\ \\ f(1) = 1 - 2 + 1 \\ \\ f(1) = \red{+ 2 - 2} \\ \\ \: f(1) = 0 \\ \\ \Large \boxed{D(1,0)}$

$f(2) = ( {2)}^{2} - 2(2) + 1 \\ \\ f(2) = \red{ 4 - 4 } + 1 \\ \\ f(2) = 1 \\ \\ \Large \boxed{ E(2,1)}$

$y = {x}^{2} - 2x + 1 \\ \\ xv = \frac{ - ( - 2)}{2(1)} \\ \\ xv = \frac{2}{2} \\ \\ \Large \boxed{ xv = 1} \\ \\ yv = f(1) \\ \\ yv = {1}^{2} - 2(1) + 1 \\ \\ yv = 1 - 2 + 1 \\ \\ yv = - 1 + 1 \\ \\ \Large \boxed{ yv = 0} \\ \\ \Large \boxed{ \green{ V \bf \: (1,0)}}$

d)

$y = - {x}^{2} + 4x - 3 \\ \\ f (- 2) = - ( - 2) {}^{2} + 4( - 2) - 3 \\ \\ f( - 2) = - 4 - 4 - 3 \\ \\ f( - 2) = - 11 \\ \\ \Large \boxed{A( - 2,- 11)}$

$f( - 1) = - ( - 1) {}^{2} + 4( - 1) - 3 \\ \\ f( - 1) = - 1 - 4 - 3 \\ \\ f( - 1) = - 8 \\ \\ \Large \boxed{ B( - 1, - 8)}$

$f(0) = - ( {0)}^{2} + 4(0) - 3 \\ \\ f(0) = - 3 \\ \\ \Large \boxed{ C(0, - 3)}$

$f(1) = - ( {1)}^{2} + 4(1) - 3 \\ \\ f(1) = - 1 + 4 - 3 \\ \\ f(1) = \red{ + 3 - 3} \\ \\ f(1) = 0 \\ \\ \Large \boxed{ D(1,0)}$

$f(2) = - ( {2)}^{2} + 4(2) - 3 \\ \\ f(2) = - 4 + 8 - 3 \\ \\ f(2) = + 4- 3 \\ \\ f(2) = 1 \\ \\ \Large \boxed{E(2,1)}$

$y = - {x}^{2} + 4x - 3 \\ \\ xv = \frac{ - 4}{2( - 1)} \\ \\ xv = \frac{ - 4}{ - 2} \\ \\ \Large \boxed{xv = 2} \\ \\ yv = f(2) \\ \\ yv = - (2) {}^{2} + 4(2) - 3 \\ \\ yv = - 4 + 8 - 3 \\ \\ \Large \boxed{ yv = 1} \\ \\ \Large \boxed{ \green{ V \bf(2,1)}}$

e)

$y = {x}^{2} \\ \\ f( - 2) = {( - 2)}^{2} \\ \\ f( - 2) = 4 \\ \\ \Large \boxed{A( - 2,4)}$

$f( - 1) = ( { - 1)}^{2} \\ \\ f( - 1) = 1 \\ \\ \Large \boxed{B (- 1,1)}$

$f(0) = (0) {}^{2} \\ \\ f(0) = 0 \\ \\ \Large \boxed{C(0,0)}$

$f(1) = (1) {}^{2} \\ \\ f(1) = 1 \\ \\ \Large \boxed{ D(1,1)}$

$f(2) = {(2)}^{2} \\ \\ f(2) = 4 \\ \\ \Large \boxed{ E(2,4)}$

$y = {x}^{2} \\ \\ xv = \frac{ - 0}{2(1)} \\ \\ \Large \boxed{ xv = 0} \\ \\ yv = f(0) \\ \\ yv = ( {0)}^{2} \\ \\ \Large \boxed{ yv = 0} \\ \\ \Large \boxed{ \green{V \bf(0,0)}}$

f)

$y = - {x}^{2} \\ \\ xv = \frac{ - 0}{2( - 1)} \\ \\ \Large \boxed{ xv = 0} \\ \\ yv = f(0) \\ \\ yv = - (0) {}^{2} \\ \\ \Large\boxed{ yv = 0} \\ \\ \Large \boxed{ \green{ V \bf (0,0)}} \\ \\ \Large \boxed{ \underline{ \blue{ \bf \: Bons \: Estudos!} \: \bf \: 20/05/2021}}$