Question

Найдите угол между касательными к графику функции: f(x)=x^2-3x+2,проведенными в точках пересечения этого графика с осью абцисс.Связано с производной.

Answer 1

y=x^2-3x+2

1) Находим точки пересечения графика функции с осью Ох:

     х^2-3x+2=0

     x1=1, x2=2

    (1;0) и (2;0) - искомые точки

 

2) Находим уравнение касательной к графику функции в точке х=1

    y`(x)=(x^2-3x+2)`=2x-3

    y`(1)=2*1-3=-1   k1=-1

    y(1)=1^2-3*1+2=1-3+2=0

    y=0+(-1)(x-1)=-x+1 -уравнение касательной в точке х=1

 

3) Находим уравнение касательной к графику функции в точке х=2

    y`(2)=2*2-3=4-3=1  k2=1

    y(2)=2^2-3*2+2=4-6+2=0

    y=0+1(x-2)=x-2 -уравнение касательной в точке х=2

 

4) Коэффициент угла наклона первой касательной k1=-1, а второй касательной k2=1,

    следовательно, касательные взаимно перпендикулярны,

    т.е.угол между ними равен 90 градусов.

 

Answer 2

Finding the Angle between Tangents at the Points of Intersection with the x-axis

To find the angle between the tangents at the points of intersection of the graph of the function f(x) = x^2 - 3x + 2 with the x-axis, we need to determine the coordinates of the points of intersection and then calculate the slopes of the tangents at those points.

Let's start by finding the points of intersection of the graph with the x-axis. The graph intersects the x-axis when the y-coordinate is equal to zero. So, we can set the function equal to zero and solve for x:

x^2 - 3x + 2 = 0

To solve this quadratic equation, we can factor it or use the quadratic formula. Factoring the equation, we have:

(x - 1)(x - 2) = 0

This gives us two solutions: x = 1 and x = 2. Therefore, the graph intersects the x-axis at the points (1, 0) and (2, 0).

Now, let's calculate the slopes of the tangents at these points. The slope of a tangent line to a curve at a given point can be found using the derivative of the function at that point. The derivative of f(x) = x^2 - 3x + 2 is:

f'(x) = 2x - 3

To find the slopes of the tangents at the points of intersection, we substitute the x-values of the points into the derivative:

At x = 1: f'(1) = 2(1) - 3 = -1

At x = 2: f'(2) = 2(2) - 3 = 1

Now that we have the slopes of the tangents, we can calculate the angle between them. The angle between two lines with slopes m1 and m2 is given by the formula:

angle = arctan(|(m1 - m2) / (1 + m1 * m2)|)

Substituting the slopes we found, we get:

angle = arctan(|(-1 - 1) / (1 + (-1) * 1)|) = arctan(2/0) = π/2

Therefore, the angle between the tangents at the points of intersection with the x-axis is π/2 radians or 90 degrees.

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