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Обчисли кутовий коефіцієнт дотичної до графіка функції f(x)=x^(2) в точці з абсцисою x_(0)=-4. Відповідь:

Обчисли кутовий коефіцієнт дотичної до графіка функції f(x)=x2 f(x)=x^{2} в точці з абсцисою x0=−4 x_{0}=-4 .\newlineВідповідь:

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Q. Обчисли кутовий коефіцієнт дотичної до графіка функції f(x)=x2 f(x)=x^{2} в точці з абсцисою x0=−4 x_{0}=-4 .\newlineВідповідь:
  1. Identify function and point: Identify the function and the point at which we need to find the slope of the tangent. f(x)=x2f(x) = x^2 and x0=−4x_0 = -4.
  2. Find derivative of function: Find the derivative of the function f(x)=x2f(x) = x^2, which represents the slope of the tangent at any point xx.f′(x)=ddx(x2)=2xf'(x) = \frac{d}{dx}(x^2) = 2x.
  3. Evaluate derivative at point: Evaluate the derivative at the point x0=−4x_0 = -4 to find the slope of the tangent at that point.f′(−4)=2(−4)=−8f'(-4) = 2(-4) = -8.
  4. Calculate slope of tangent: The slope of the tangent to the graph of the function at x0=−4x_0 = -4 is −8-8.

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