Find the coordinates of the point on the curve y=x³-3x²+6x+2 at which the gradient is 3
To find the coordinates of the point on the curve where the gradient is 3, we need to find the derivative of the curve and set it equal to 3.
First, we differentiate the equation of the curve y = x³ - 3x² + 6x + 2 with respect to x:
dy/dx = 3x² - 6x + 6
Next, we set the derivative equal to 3:
3x² - 6x + 6 = 3
Simplifying this equation, we get:
3x² - 6x + 3 = 0
Divide through by 3:
x² - 2x + 1 = 0
This equation can be factored as:
(x - 1)² = 0
Therefore, the value of x is 1.
Now, to find the y-coordinate of the point on the curve, substitute x = 1 into the original equation:
y = 1³ - 3(1)² + 6(1) + 2
y = 1 - 3 + 6 + 2
y = 6
Therefore, the coordinates of the point on the curve where the gradient is 3 are (1, 6).
To find the coordinates of the point on the curve where the gradient is 3, we need to find the derivative of the curve and then solve for x.
Step 1: Calculate the derivative of the curve.
To find the derivative, we differentiate the equation y = x³ - 3x² + 6x + 2 with respect to x.
dy/dx = 3x² - 6x + 6
Step 2: Set the derivative equal to 3 and solve for x.
Setting dy/dx = 3, we get:
3x² - 6x + 6 = 3
Step 3: Simplify and solve for x.
Rearranging the equation, we have:
3x² - 6x + 3 = 0
Dividing the equation by 3, we get:
x² - 2x + 1 = 0
Factoring the quadratic equation, we have:
(x - 1)² = 0
Taking the square root of both sides, we get:
x - 1 = 0
Solving for x, we find:
x = 1
Step 4: Find the value of y.
Now that we have the value of x, we can substitute it back into the original equation to find the corresponding y-coordinate.
y = 1³ - 3(1)² + 6(1) + 2
y = 1 - 3 + 6 + 2
y = 6
The coordinates of the point on the curve where the gradient is 3 are (1, 6).