Find the coordinates of the point on the curve y = x
3 − 3x
2 + 6x + 2 at which
the gradient is 3
Online, use ^ for powers, not line-feeds!
y = x^3 - 3x^2 + 6x + 2
y' = 3x^2 - 6x + 6
so, when is y'=3?
3x^2-6x+6 = 3
3x^2-6x+3 = 0
3(x-1)^2 = 0
x = 1
since y(1) = 6, the tangent line there is y-6=3(x-1)
see the graphs at
https://www.wolframalpha.com/input/?i=plot+x%5E3+-+3x%5E2+%2B+6x+%2B+2%2C+y%3D3%28x-1%29%2B6
Well, it seems like you're trying to find a point on the given curve where the gradient is 3. Let me clown around a bit to help you out!
To find the point where the gradient is 3, we need to find the derivative of the curve and solve for the x-coordinate. So, let's get down to some mathematical giggles!
First, we find the derivative of the curve y = x³ - 3x² + 6x + 2. Applying the power rule, we get:
dy/dx = 3x² - 6x + 6
Now, we need to set the derivative equal to 3 and solve for x:
3x² - 6x + 6 = 3
To make math more amusing, let's rearrange the equation:
3x² - 6x + 3 = 0
Divide everything by 3 because sharing is caring:
x² - 2x + 1 = 0
And this equation tells us that (x - 1)² = 0, which leads us to the hilarious conclusion that x = 1.
Now that we know the x-coordinate, we can find the corresponding y-coordinate by plugging it back into the original equation:
y = 1³ - 3(1)² + 6(1) + 2
Guess what? It turns out that y = 8!
So, the coordinates of the point where the gradient is 3 on the curve y = x³ - 3x² + 6x + 2 are (1, 8).
I hope my clownish calculations didn't make you dizzy!
To find the point on the curve where the gradient is 3, we need to find the derivative of the curve first.
Given the equation y = x^3 - 3x^2 + 6x + 2, we can find the derivative (dy/dx) by differentiating each term with respect to x.
dy/dx = d/dx (x^3 - 3x^2 + 6x + 2)
= 3x^2 - 6x + 6
Now, we set dy/dx equal to 3 and solve for x:
3x^2 - 6x + 6 = 3
Subtracting 3 from both sides:
3x^2 - 6x + 3 = 0
Divide both sides by 3:
x^2 - 2x + 1 = 0
Factoring this quadratic equation:
(x - 1)^2 = 0
Taking the square root of both sides:
x - 1 = 0
Solving for x:
x = 1
Now that we have the x-coordinate, we can substitute it back into the original equation to find the y-coordinate:
y = 1^3 - 3(1)^2 + 6(1) + 2
= 1 - 3 + 6 + 2
= 6
Therefore, the point on the curve where the gradient is 3 is (1, 6).
To find the coordinates of the point on the curve at which the gradient is 3, we need to find the derivative of the curve and set it equal to 3.
First, let's find the derivative of the curve y = x^3 - 3x^2 + 6x + 2. The derivative will give us the gradient of the curve at any point. Taking the derivative with respect to x, we get:
dy/dx = 3x^2 - 6x + 6
Next, we set the derivative equal to 3 and solve for x:
3x^2 - 6x + 6 = 3
Subtracting 3 from both sides:
3x^2 - 6x + 3 = 0
Now, we can solve this quadratic equation. Factoring out a common factor of 3:
3(x^2 - 2x + 1) = 0
(x^2 - 2x + 1) = 0
(x - 1)^2 = 0
Taking the square root of both sides:
x - 1 = 0
x = 1
So, the value of x at which the gradient is 3 is x = 1.
Now, to find the y-coordinate, we substitute this value of x into the equation y = x^3 - 3x^2 + 6x + 2:
y = 1^3 - 3(1)^2 + 6(1) + 2
y = 1 - 3 + 6 + 2
y = 6
Therefore, the point on the curve where the gradient is 3 is (1, 6).