Question

Напишите уравнение касательной к графику функции (За ответ) 1) у = 2 + х2 , х0 = 1

2) у = х2 - 2х, х0 = 2

3) у = х2 + 2х -1, х0 = -3

Answer 1

$f(x) = y(x0) + y'(x0) \times (x - x0)$

уравнение касательной

1.

$y = 2 + {x}^{2}$

$y(1) = 2 + 1 = 3$

$y '= 2x \\ y'(1) = 2$

$f(x) = 3 + 2(x - 1) = 3 + 2x - 2 \\ f(x) = 2x + 1$

2.

$y = {x}^{2} - 2x$

$y(2) = {2}^{2} - 2 \times 2 = 0$

$y '= 2x - 2$

$y'(2) = 4 - 2 = 2$

$f(x) = 0 + 2(x - 2) = 2x - 4$

3.

$y = {x}^{2} - 2x + 1$

$y( - 3) = 9 + 6 + 1 = 16$

$y '= 2x - 2$

$y'( - 3) = - 6 - 2 = - 8$

$f(x) = 16 - 8(x + 3) = 16 - 8x - 24 \\ f(x) = - 8x - 8$