At what point does the curve x^2-2x+y=0 have a slope of -2.
y = -x^2+2x
y' = -2x+2
so, when does -2x+2 = -2?
To find at what point the curve \(x^2 - 2x + y = 0\) has a slope of -2, we need to differentiate the equation with respect to x and solve for \(x\) at which the derivative is equal to -2.
First, let's find the derivative of the equation \(x^2 - 2x + y = 0\) with respect to x. We consider \(y\) as a function of \(x\):
\(\frac{d}{dx}(x^2 - 2x + y) = \frac{d}{dx}(0)\)
Taking the derivative term by term, we get:
\(2x - 2 + \frac{dy}{dx} = 0\)
Simplifying further, we have:
\(\frac{dy}{dx} = 2 - 2x\)
Now, we set the derivative equal to -2 and solve for \(x\):
\(\frac{dy}{dx} = -2\)
\(2 - 2x = -2\)
Adding 2 to both sides:
\(2 - 2x + 2 = -2 + 2\)
\(4 - 2x = 0\)
Subtracting 4 from both sides:
\(4 - 2x - 4 = 0 - 4\)
\(-2x = -4\)
Dividing both sides by -2:
\(\frac{-2x}{-2} = \frac{-4}{-2}\)
\(x = 2\)
So, the curve \(x^2 - 2x + y = 0\) has a slope of -2 at the point \(x = 2\).
To find the corresponding value of \(y\), substitute \(x = 2\) into the original equation:
\((2)^2 - 2(2) + y = 0\)
\(4 - 4 + y = 0\)
\(y = 0\)
Therefore, the curve has a slope of -2 at the point \((x, y) = (2, 0)\).
To find the point on the curve with a slope of -2, we need to find the derivative of the equation and solve for the x-coordinate that gives a derivative of -2.
Step 1: Differentiate the equation with respect to x.
Taking the derivative of the equation x^2 - 2x + y = 0 with respect to x should give us the derivative dy/dx.
dy/dx (2x - 2) + dy/dx(y) = 0
Step 2: Solve for dy/dx.
Simplifying the equation, we get:
2x - 2 + dy/dx(y) = 0
dy/dx(y) = -2x + 2
Step 3: Set dy/dx equal to -2.
To find the x-coordinate where the slope is -2, we set -2x + 2 (-2) = -2 and solve for x.
-4x + 4 = -2
-4x = -6
x = 6/4
x = 3/2
Therefore, the curve x^2 - 2x + y = 0 has a slope of -2 at the point (3/2, y), where y is a value that satisfies the equation.