encontrar a equação da reta taangente y=4x-3x² (2, -4)
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Boa noite Maluzinha!
Solução!
Vamos derivar a função para encontramos o coeficiente angular da reta.
![y=4x-3 x^{2} \\\\\\
y'=4-6x\\\\\\
m=4-6x\\\\\\
A(2,-4)\\\\\\
Substituindo~~x ~~na~~ func\~ao ~~derivada!
m=4-6x\\\\\\
m=4-6(2)\\\\\\
m=4-12\\\\\
\boxed{m=-8}](https://tex.z-dn.net/?f=y%3D4x-3+x%5E%7B2%7D+%5C%5C%5C%5C%5C%5C%0Ay%27%3D4-6x%5C%5C%5C%5C%5C%5C%0Am%3D4-6x%5C%5C%5C%5C%5C%5C%0AA%282%2C-4%29%5C%5C%5C%5C%5C%5C%0ASubstituindo%7E%7Ex+%7E%7Ena%7E%7E+func%5C%7Eao+%7E%7Ederivada%21%0Am%3D4-6x%5C%5C%5C%5C%5C%5C%0Am%3D4-6%282%29%5C%5C%5C%5C%5C%5C%0Am%3D4-12%5C%5C%5C%5C%5C%0A%5Cboxed%7Bm%3D-8%7D "y=4x-3 x^{2} \\\\\\
y'=4-6x\\\\\\
m=4-6x\\\\\\
A(2,-4)\\\\\\
Substituindo~~x ~~na~~ func\~ao ~~derivada!
m=4-6x\\\\\\
m=4-6(2)\\\\\\
m=4-12\\\\\
\boxed{m=-8}")
![Eq:da~~reta!\\\\\\\\
y-y _{A}=m(x- x_{A})\\\\\\\
y+4= -8(x-2)\\\\\
y+4=-8x+16\\\\\
8x+y+4-16=0\\\\\
\boxed{Resposta:Eq~~da~~reta~~8x+y-12= 0}](https://tex.z-dn.net/?f=Eq%3Ada%7E%7Ereta%21%5C%5C%5C%5C%5C%5C%5C%5C%0Ay-y+_%7BA%7D%3Dm%28x-+x_%7BA%7D%29%5C%5C%5C%5C%5C%5C%5C%0Ay%2B4%3D+-8%28x-2%29%5C%5C%5C%5C%5C%0Ay%2B4%3D-8x%2B16%5C%5C%5C%5C%5C%0A8x%2By%2B4-16%3D0%5C%5C%5C%5C%5C%0A%5Cboxed%7BResposta%3AEq%7E%7Eda%7E%7Ereta%7E%7E8x%2By-12%3D+0%7D "Eq:da~~reta!\\\\\\\\
y-y _{A}=m(x- x_{A})\\\\\\\
y+4= -8(x-2)\\\\\
y+4=-8x+16\\\\\
8x+y+4-16=0\\\\\
\boxed{Resposta:Eq~~da~~reta~~8x+y-12= 0}")
Boa noite!
Bons estudos!
Solução!
Vamos derivar a função para encontramos o coeficiente angular da reta.
Boa noite!
Bons estudos!
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