Question

determine uma equação da reta tangente ao gráfico da função f(x) = 2x^2 +3 que seja paralela a reta y=8x+3

Answer 1

Olá

Se as retas são paralelas, isso quer dizer que as derivadas são iguais.

Então temos que igualar as derivadas das funções para obtermos x0.

$\displaystyle \mathsf{f(x)=2x^2+3~~~~~~~ ~~ ~~~~~~~~~~~~~~~y=8x+3}\\\\\\\mathsf{f'(x)=4x^{2-1}+0}~~~~~~~~~~\longrightarrow~~~~ ~~\mathsf{f'(x)=4x}\\\\\\\mathsf{y'=8x^{1-1}+0}~~~~~~~~~~~~~~\longrightarrow~~~~~ \mathsf{y'=8}\\\\\\\text{Igualando as derivadas}\\\\\\\mathsf{4x_ {0}=8}\\\\\mathsf{x_{0}= \frac{8}{4} }\\\\\\\boxed{\mathsf{x_{0}=2}}$


Substituindo o x0 = 2 em f(x)


$\displaystyle \mathsf{x_{0}=2}\\\\\mathsf{f(x)=2x^2+3}\\\\\mathsf{f(2)=2\cdot (2)^2+3}\\\\\boxed{\mathsf{f(2)=11}}$


Os pontos que encontramos são

$\mathsf{p=(2,11)}$


A equação da reta tangente é dada por

$\displaystyle \mathsf{y-y(x_0)=f'(x_0)\cdot (x-x_{0})}\\\\\\\mathsf{y(x_0)=11~~~~ ~~~~ ~~~~~~~~ ~f'(x_0)=8~~~~~ ~~~~~~ ~~~ x_0=2}\\\\\\\mathsf{y-11=8\cdot (x-2)}\\\\\mathsf{y=8x-16+11}\\\\\boxed{\boxed{\mathsf{y=8x-5}}}~~~~~~~ \Longleftarrow~~~~ \text{Equac\~ao da reta tangente a curva dada e}\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ ~~~~\text{paralela a reta dada}$