Question

Dada a função f(x) = x²- 6x + 10.
Determine o valor de:

a)f (- 3)
b)f (- 2)
c)f (- 1)
d)f (0)
e)f (4)
f) x quando f(x)=0​

Answer 1

Dada a função f(x) = x² - 6x + 10

a)

f(-3) = (-3)² - 6(-3) + 10

f(-3) = 9 + 18 + 10

f(-3) = 37

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b)

f(-2) = (-2)² - 6(-2) + 10

f(-2) = 4 + 12 + 10

f(-2) = 26

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c)

f(-1) = (-1)² - 6(-1) + 10

f(-1) = 1 + 6 + 10

f(-1) = 17

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d)

f(0) = (0)² - 6(0) + 10

f(0) = 0 - 0 + 10

f(0) = 10

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e)

f(4) = (4)² - 6(4) + 10

f(4) = 16 - 24 + 10

f(4) = 2

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f)

- f(x) = x² - 6x + 10

Bhask: a = 1, b = - 6, c = 10

x = [ - b +- √(b² - 4ac) ]/(2a)

x = [ 6 +- √((-6)² - 4*1*10) ]/(2*1)

x = [ 6 +- √(36 - 40) ]/2

x = [ 6 +- √(-4) ]/2

Como √(-4) NÃO é um número real, n existe x real para f(x) = 0

Answer 2

Letra A

$\sf f (x) = {x}^{2} - 6x + 10$

$\sf f ( - 3) = {( - 3)}^{2} - 6\cdot ( - 3) + 10$

$\sf f( - 3) = 9 + 18 + 10$

$\red{\sf f( - 3) = 37}$

Letra B

$\sf f (x) = {x}^{2} - 6x + 10$

$\sf f ( - 2) = {( - 2)}^{2} - 6\cdot ( - 2) + 10$

$\sf f ( - 2) = 4 + 12 + 10$

$\red{\sf f ( - 2) = 26}$

Letra C

$\sf f (x) = {x}^{2} - 6x + 10$

$\sf f ( - 1) = {( - 1)}^{2} - 6\cdot ( - 1) + 10$

$\sf f ( - 1) = 1 + 6 + 10$

$\red{\sf f ( - 1) = 17}$

Letra D

$\sf f (x) = {x}^{2} - 6x + 10$

$\sf f (0) = {0}^{2} - 6\cdot 0 + 10$

$\sf f (0) = 0 + 0 + 10$

$\red{\sf f (0) = 10}$

Letra E

$\sf f (x) = {x}^{2} - 6x + 10$

$\sf f (4) = {4}^{2} - 6\cdot 4 + 10$

$\sf f (4) = 16 - 24 + 10$

$\sf f (4) = 26 - 24$

$\red{\sf f (4) = 2}$

Letra F

$\sf f (x) = {x}^{2} - 6x + 10$

$\sf 0= {x}^{2} - 6x + 10$

$\sf {x}^{2} - 6x + 10 = 0$

$\sf \Delta = {b}^{2} - 4ac$

$\sf \Delta = {( - 6)}^{2} - 4\cdot 1\cdot 10$

$\sf \Delta = 36 - 40$

$\sf \Delta = - 4$

$\sf x _{½}= \dfrac{ -b \pm \: \sqrt{\Delta} }{2a}$

$\sf x _{½}= \dfrac{6 \pm \: \sqrt{ - 4} }{2\cdot 1}$

$\sf x _{½}= \dfrac{6 \pm \: \sqrt{ -1 \cdot 4} }{2}$

$\sf x _{½}= \dfrac{6 \pm \: \sqrt{ -1 } \cdot \sqrt{4} }{2}$

$\sf x _{½}= \dfrac{6 \pm \: i \cdot 2}{2}$

$\sf x _{½}= \dfrac{6 \pm \: 2i}{2}$

$\sf x _{1}= \dfrac{6 - 2i}{2} = \red{3 - i}$

$\sf x _{2}= \dfrac{6 + 2i}{2} = \red{3 + i}$