Question

Прямая y=5x+1 является касательной к графику функции y=x^3+0.5x^2+3x-0.5 найдите ординату точки косания подробнее пожалуйста со штрихами

Answer 1

Решение подробное сфоткала, здесь писать очень неудобно. Надеюсь, будет понятно.

Answer 2

Finding the Point of Tangency

To determine if the line y = 5x + 1 is a tangent to the graph of the function y = x^3 + 0.5x^2 + 3x - 0.5, we need to find the point of tangency.

The point of tangency occurs when the slope of the line y = 5x + 1 is equal to the slope of the function y = x^3 + 0.5x^2 + 3x - 0.5 at that point.

Let's find the derivative of the function y = x^3 + 0.5x^2 + 3x - 0.5 to find its slope at any given point.

The derivative of y = x^3 + 0.5x^2 + 3x - 0.5 is: y' = 3x^2 + x + 3

Now, let's set the derivative equal to the slope of the line y = 5x + 1 and solve for x:

3x^2 + x + 3 = 5

Simplifying the equation, we get: 3x^2 + x - 2 = 0

To solve this quadratic equation, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)

For our equation, a = 3, b = 1, and c = -2. Plugging these values into the quadratic formula, we get:

x = (-1 ± √(1^2 - 4 * 3 * -2)) / (2 * 3)

Simplifying further, we have:

x = (-1 ± √(1 + 24)) / 6 x = (-1 ± √25) / 6

This gives us two possible values for x:

x1 = (-1 + √25) / 6 x2 = (-1 - √25) / 6

Simplifying these expressions, we find:

x1 = 2/3 x2 = -1

Now that we have the x-values, we can find the corresponding y-values by substituting them into the original function y = x^3 + 0.5x^2 + 3x - 0.5:

For x1 = 2/3: y1 = (2/3)^3 + 0.5(2/3)^2 + 3(2/3) - 0.5

For x2 = -1: y2 = (-1)^3 + 0.5(-1)^2 + 3(-1) - 0.5

Calculating these values, we find:

y1 = 1.5185 y2 = -0.5

Therefore, the point of tangency is (2/3, 1.5185) or (-1, -0.5).

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