Slope of the Tangent to the Curve y = x² - x at the Point where the Line y = 2 Cuts the Curve in the First Quadrant

To find the slope of the tangent to the curve y = x² - x at the point where the line y = 2 cuts the curve in the first quadrant, we need to follow a few steps:

Step 1: Find the x-coordinate of the point of intersection
To find the x-coordinate of the point where the line y = 2 cuts the curve y = x² - x, we can equate the two equations and solve for x.
Substituting y = 2 in the equation y = x² - x, we get:
2 = x² - x

Rearranging the equation, we have:
x² - x - 2 = 0

Factoring the quadratic equation, we get:
(x - 2)(x + 1) = 0

Setting each factor equal to zero, we have two possible values for x:
x - 2 = 0 or x + 1 = 0

Solving for x, we get:
x = 2 or x = -1

Since we are looking for the point of intersection in the first quadrant, we choose x = 2 as the x-coordinate of the point.

Step 2: Find the y-coordinate of the point of intersection
To find the y-coordinate of the point where the line y = 2 cuts the curve y = x² - x, we substitute x = 2 into the equation y = x² - x:
y = (2)² - 2
y = 4 - 2
y = 2

Therefore, the point of intersection is (2, 2).

Step 3: Find the derivative of the curve
To find the slope of the tangent to the curve at the point (2, 2), we need to find the derivative of the curve y = x² - x. Taking the derivative with respect to x, we get:
dy/dx = 2x - 1

Step 4: Substitute the x-coordinate into the derivative
Substituting x = 2 into the derivative, we have:
dy/dx = 2(2) - 1
dy/dx = 4 - 1
dy/dx = 3

Therefore, the slope of the tangent to the curve y = x² - x at the point where the line y = 2 cuts the curve in the first quadrant is 3.