Question

Найти угловой коэффициент касательной к графику функции f(x)=2x^2+x в точке x0= -2 пожалуйста

Answer 1

Решение задания смотри на фотографии

Answer 2

Finding the Slope of the Tangent Line

To find the slope of the tangent line to the graph of the function f(x) = 2x^2 + x at the point x0 = -2, we can use the derivative of the function. The derivative gives us the rate of change of the function at any given point.

The derivative of f(x) = 2x^2 + x is given by f'(x) = 4x + 1. To find the slope of the tangent line at x0 = -2, we substitute x = -2 into the derivative:

f'(-2) = 4(-2) + 1 = -8 + 1 = -7.

Therefore, the slope of the tangent line to the graph of f(x) = 2x^2 + x at the point x0 = -2 is -7.