Question

2. Determine a equação da reta tangente `a curva f(x) = x2 + 2x − 3 nos pontos (−2,−3) e (1,0)

Answer 1

$\large\boxed{\begin{array}{l}\sf f(x)=x^2+2x-3~~em~P(-2,-3):\\\sf f'(x)=2x+2\\\sf f'(-2)=2\cdot(-2)+2=-4+2\\\sf f'(-2)=-2\\\sf y=y_0+f'(x_0)\cdot(x-x_0)\\\sf y=-3-2(x-(-2))\\\sf y=-3-2(x+2)\\\sf y=-3-2x-4\\\sf y=-2x-7\end{array}}$

$\Large\boxed{\begin{array}{l}\sf f(x)=x^2+2x-3~~em~~P(1,0)\\\sf f'(x)=2x+2\\\sf f'(1)=2\cdot1+2=4\\\sf y=y_0+f'(x_0)(x-x_0)\\\sf y=0+4(x-1)\\\sf y=4x-4\end{array}}$