Question

Чому дорівнює кутовий коефіцієнт дотичної до графіка функції y=3x²-8x+1 у точці з абсцисою x=2

Answer 1

Ответ:

решение смотри на фотографии

Answer 2

Calculating the Tangent Line's Slope at a Specific Point

To find the slope of the tangent line to the graph of the function \( y = 3x^2 - 8x + 1 \) at the point with the abscissa \( x = 2 \), we can use the derivative of the function.

The derivative of a function gives the slope of the tangent line to the graph of the function at any point. For the function \( y = 3x^2 - 8x + 1 \), the derivative \( y' \) can be found using the power rule of differentiation.

The power rule states that if \( f(x) = x^n \), then \( f'(x) = nx^{n-1} \).

Finding the Derivative of the Function

The derivative of the function \( y = 3x^2 - 8x + 1 \) is given by: \[ y' = \frac{d}{dx}(3x^2 - 8x + 1) \]

Using the power rule, we can find the derivative as follows: \[ y' = 6x - 8 \]

Calculating the Slope at the Given Point

To find the slope of the tangent line at the point with the abscissa \( x = 2 \), we can substitute \( x = 2 \) into the derivative \( y' = 6x - 8 \).

Substituting \( x = 2 \) into the derivative gives: \[ y' = 6(2) - 8 = 12 - 8 = 4 \]

Therefore, the slope of the tangent line to the graph of the function \( y = 3x^2 - 8x + 1 \) at the point with the abscissa \( x = 2 \) is 4.