Question

Em relação à função do 2º grau f(x) = -x2 - 6x - 5, calcule:

a) A(s) sua(s) raiz(es).

b) O seu vértice.​

Answer 1

a) (-5, 0) e (-1, 0)

b) (-3, 4)

$\Large{\sf{\Huge ————————————————}}$

$\sf \red{f(x) } \green{ = } \blue{-x^{2} } \pink{ - } \purple{ 6x} \orange{ - } \red{ 5} \\ \sf \green{ 0 } \blue{ =} \pink{ -x^{2} } \purple{- } \orange{ 6x } \red{ - } \green{ 5 } \\ \sf \blue{ -x^{2} } \pink{ - } \purple{ 6x } \orange{ - } \red{ 5 } \green{ = } \blue{ 0 }$

Delta

$\sf { \Delta } \green{ =} \blue{ (6)^{2} } \pink{ - } \purple{ 4} \orange{ ( 1) } \red{ (5) }\\ \sf \Delta = \green{36 } \blue{ - } \pink{ 4} \purple{ (1) } \orange{(5) }\\ \sf \Delta =\red{ 36 } \green{ - } \blue{ 4 } \pink{ (5) } \\ \sf \Delta = \purple{36 } \orange{- } \red{20 } \\ \sf \Delta = \green{ 16 }$

Bhaskara

$\sf \red{ \frac{-(6) \pm \sqrt{16}}{2.1}} \\ \sf \pink{ \frac{-6 \pm 4}{2} } \\ \sf \purple{\frac{-6+4}{2} } \orange{= } \red{ \frac{-2}{2} } \green{ = } \Huge\boxed{\large \blue{ \sf -1 }} \\ \sf \pink{\frac{-6-4}{2} } \purple{ = } \orange{\frac{-10}{5} } \red{ =}\Huge\boxed{\large \green{\sf -5 }}$

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Vértice

$\sf \red{f(x) } \green{ = } \blue{ -x^{2} } \pink{ - } \purple{6x } \orange{ - } \red{ 5 } \\ \sf \green{ } \blue{ } x \pink{ = -\frac{-6}{2(-1)} } \\ \sf x \purple{ = \frac{-6}{-2}} \\ \sf x \orange{ = -3 } \\ \\ \sf \red{ f(-3) } \green{ = } \blue{ -x^{2} } \pink{- } \purple{ 6x} \orange{ - } \red{5 } \\ \sf \green{f(-3) = -(-3)^{2} } \blue{ - } \pink{ 6(-3) } \purple{ - } \orange{ 5 } \\ \sf \red{ f(-3) } \green{ = } \blue{ -6 } \pink{ - } \purple{ (-18) } \orange{ -} \red{ 5 }\\ \sf \green{ -9} \blue{ + } \pink{ 18 } \purple{ -5 } \\ \sf \orange{ 9} \red{ - } \green{ 5 }\\ \sf \blue{4 }$