Question

Find the x coordinates of the points where the curve with equation y = x³ - 4x² + 5x + 4 has a gradient of 1

Answer 1

To find the x coordinates of the points where the curve with equation y = x³ - 4x² + 5x + 4 has a gradient of 1, we need to find the derivative of the equation and set it equal to 1. The x coordinates are x = 2 and x = 2/3.

Step-by-step explanation:

To find the x coordinates of the points where the curve with equation y = x³ - 4x² + 5x + 4 has a gradient of 1, we need to find the derivative of the equation and set it equal to 1. The derivative of y with respect to x is given by dy/dx = 3x² - 8x + 5. Setting this equal to 1, we have 3x² - 8x + 5 = 1. Simplifying, we get 3x² - 8x + 4 = 0. Factoring this quadratic equation, we have (3x - 2)(x - 2) = 0. Solving for x gives us two possible x coordinates: x = 2 and x = 2/3.

Answer 2

Step-by-step explanation:

To find the x-coordinates of the points where the curve with the equation y = x³ - 4x² + 5x + 4 has a gradient of 1, we need to find the points where the derivative of the equation y = x³ - 4x² + 5x + 4 is equal to 1. The derivative of the equation y = x³ - 4x² + 5x + 4 is given by:

dy/dx = 3x² - 8x + 5

So, we need to find the solutions of the equation:

3x² - 8x + 5 = 1

This is a quadratic equation that can be solved by factoring, completing the square, or using the quadratic formula. To use the quadratic formula, we can write:

x = (-b ± √(b² - 4ac)) / (2a)

Where a = 3, b = -8, and c = 4. Plugging in these values, we get:

x = (-(-8) ± √((-8)² - 4 * 3 * 4)) / (2 * 3)

x = (8 ± √(64 - 36)) / 6

x = (8 ± √(28)) / 6

x = (8 ± 2√7) / 6

So, the two solutions are:

x = (8 + 2√7) / 6 and x = (8 - 2√7) / 6

These are the x-coordinates of the points where the curve with the equation y = x³ - 4x² + 5x + 4 has a gradient of 1.