Question

Найдите тангенс угла наклона касательной к графику функции f ( x ) = 3x^2+ 2x - 5 , проходящей через его точку с абсциссой x0 =2. помогите плиз Очень прошу

Answer 1

$f(x) = 3x^2+2x-5$

$f'(x) = 6x+2$ 

$tga = f'(x0) = f'(2) = 12+2 = 14$ 

 

Answer 2

Finding the Tangent Line's Slope

To find the slope of the tangent line to the graph of the function f(x) = 3x^2 + 2x - 5 at the point with an x-coordinate of x0 = 2, we can use the derivative of the function. The derivative gives us the slope of the tangent line at any given point on the graph.

The derivative of the function f(x) = 3x^2 + 2x - 5 can be found by applying the power rule of differentiation. The power rule states that if we have a term of the form ax^n, the derivative is given by nx^(n-1).

Applying the power rule to each term of the function f(x) = 3x^2 + 2x - 5, we get:

f'(x) = d/dx (3x^2) + d/dx (2x) - d/dx (5) = 6x + 2 - 0 = 6x + 2

Now that we have the derivative of the function, we can find the slope of the tangent line at x = 2 by substituting x = 2 into the derivative:

f'(2) = 6(2) + 2 = 12 + 2 = 14

Therefore, the slope of the tangent line to the graph of f(x) = 3x^2 + 2x - 5 at the point with an x-coordinate of x0 = 2 is 14.

Please let me know if there's anything else I can help you with!