Question

Найти координату точки , в которойНайти координату точки , в которой касательная к параболе y=x(в квадрате)-x-12 образует с осью OX угол 45 градусов

Answer 1

Производная, тангенс угла наклона касательнйой и угловой коэффициент-это одно и то же.

Answer 2

Finding the Coordinate of the Point where the Tangent to the Parabola Forms a 45-Degree Angle with the x-Axis

To find the coordinate of the point where the tangent to the parabola y = x^2 - x - 12 forms a 45-degree angle with the x-axis, we need to determine the point of tangency.

The slope of the tangent line to a curve at a given point is equal to the derivative of the curve at that point. Therefore, we can find the slope of the tangent line to the parabola at the point of tangency by taking the derivative of the equation y = x^2 - x - 12 with respect to x.

Taking the derivative of the equation, we get:

dy/dx = 2x - 1 To find the point of tangency, we need to find the x-coordinate where the slope of the tangent line is equal to the tangent of a 45-degree angle. The tangent of a 45-degree angle is equal to 1.

Setting the derivative equal to 1, we have:

2x - 1 = 1

Solving for x, we get:

2x = 2

x = 1

Now that we have the x-coordinate of the point of tangency, we can substitute it back into the original equation y = x^2 - x - 12 to find the y-coordinate.

Substituting x = 1 into the equation, we have:

y = 1^2 - 1 - 12

y = -12

Therefore, the coordinate of the point where the tangent to the parabola forms a 45-degree angle with the x-axis is (1, -12).

I hope this helps! Let me know if you have any further questions.