Question

F (x)=x^2-2x Xo =1 касательная нормали ​

Answer 1

Ответ: производная равна у'=2*х-2. При этом у(1)=1-2=-1, у'(1)=2-2=0. Уравнение касательной у(1)-у'(1)*(х-1) равно -1 (не зависит от х).

Уравнение нормали х=1 перпендикулярно оси ох. .

Пошаговое объяснение:

Answer 2

To find the equation of the tangent and normal lines to the curve $F(x) = x^2 - 2x$ at the point $x_0 = 1$, we need to determine the slope of the tangent and normal lines at that point.

The slope of the tangent line at a given point can be found by taking the derivative of the function $F(x)$ and evaluating it at that point. The derivative of $F(x)$ is the rate of change of the function with respect to $x$.

Let's find the derivative of $F(x)$: $F'(x) = \frac{d}{dx}(x^2 - 2x)$ Using the power rule of differentiation, we get: $F'(x) = 2x - 2$

Now, let's evaluate the derivative at $x = x_0 = 1$ to find the slope of the tangent line: $m_{\text{tangent}} = F'(1) = 2(1) - 2 = 0$

The slope of the tangent line at $x = 1$ is 0. This means the tangent line is horizontal.

To find the slope of the normal line, we take the negative reciprocal of the tangent line's slope. Since the tangent line has a slope of 0, the normal line's slope will be undefined.

The equation of the tangent line is a horizontal line passing through the point $(1, F(1))$. Let's find the value of $F(1)$: $F(1) = (1)^2 - 2(1) = 1 - 2 = -1$

Therefore, the point of tangency is $(1, -1)$, and the equation of the tangent line is: $y = -1$

Since the slope of the normal line is undefined, we need to determine the equation using the point-slope form.

The point-slope form of a line is given by $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is a point on the line, and $m$ is the slope.

Let's use the point of tangency $(1, -1)$ to write the equation of the normal line: $y - (-1) = \text{undefined}(x - 1)$

Since the slope is undefined, the equation simplifies to: $x = 1$

Therefore, the equation of the normal line is $x = 1$.

In summary, the equation of the tangent line to the curve $F(x) = x^2 - 2x$ at the point $x = 1$ is $y = -1$, and the equation of the normal line is $x = 1$.