Question

Determine a equação da reta tangente a curva y= x² - 3x + 6, no seu ponto P(-1 , 10)

Answer 1

$\text{Calcule a derivada dessa fun\c{c}\~ao em P = -1}\\\\\\ f'~(p)= \lim_{h\to0}~ \dfrac{f(p+h)-f(p)}{h}~~^\Rightarrow~~\dfrac{f(h-1)-f(-1)}{h}\\\\\\=\dfrac{[(h-1)^2-3\cdot (h-1)+6]-[(-1)^2-3\cdot (-1)+6)]}{h}\\\\\\=\dfrac{(h^2-2h+1^2-3h+3+6)-(1+3+6)}{h}\\\\=\dfrac{h^2-5h+10\hspace{-8}\diagup-10\hspace{-8}\diagup}{h}\\\\\\= \dfrac{h^2-5h}{h}\\\\\\=\dfrac{h\hspace{-8}\diagup(h-5)}{h\hspace{-8}\diagup}\\\\\\=1(h-5)~~~~~~(\lim_{h \to 0})\\\\\\=-5$


$\text{A equa\c{c}\~ao da reta tangente a f \'e dada por:}\\\\\\ y-f(p)=f'~(p)\cdot(x-p)\\\\\\y-10=-5\cdot(x-(-1))\\\\\\y-10=-5\cdot (x+1)\\\\\\\ y-10=-5x-5\\\\\\ y=-5x+5\\\\\\ \text{Logo a equa\c{c}\~ao da reta \'e:}~~\large\boxed{\boxed{ y = -5 x + 5 }}$