91Ó°ÊÓ

The slope of a line is a measure of its steepness. In the equation of a line in slope-intercept form, expressed as \( y = mx + b \), \( m \) represents the slope. It tells us how much \( y \) increases (or decreases) when \( x \) increases by 1 unit. For a line, the slope is constant, meaning the line is straight. Determining the slope is crucial when identifying lines that are parallel or perpendicular. Parallel lines share the same slope, which means they never intersect. In our problem, the given line \( x - 2y = 2 \) was rearranged to \( y = \frac{1}{2}x - 1 \) to find its slope of \( \frac{1}{2} \). This slope helps us ascertain which tangent lines to a curve are parallel to this line.

Differentiation is a fundamental concept in calculus used to find the rate at which a function is changing at any given point. This rate of change is called the derivative. In our context, the derivative of a function gives us the slope of the tangent line at any point along the curve. For the function \( y = \frac{x-1}{x+1} \), we need the derivative to find the tangent's slope. The derivative is calculated using the quotient rule because our function is a fraction (a quotient of two functions). The derivative \( y' = \frac{2}{(x+1)^2} \) tells us the slope at each point \( x \) on the curve.

The point-slope form is a way to write the equation of a line when you know a point on the line and the slope. This form is written as \( y - y_1 = m(x - x_1) \), where \( m \) is the slope, and \((x_1, y_1)\) is the given point. This format is especially useful for quickly writing the equation of a tangent line once you have the slope.For the points of tangency \((1, 0)\) and \((-3, 2)\), along with the slope \( \frac{1}{2} \), we used the point-slope form to derive the equations of the tangent lines. Substituting these into the point-slope formula gave us \( y = \frac{1}{2}x - \frac{1}{2} \) and \( y = \frac{1}{2}x + \frac{7}{2} \) respectively, perfectly describing the tangent lines at the specified points.

The quotient rule is used in calculus to find the derivative of a ratio of two functions. If you have a function \( y = \frac{u}{v} \), where both \( u \) and \( v \) are differentiable functions of \( x \), the derivative \( y' \) is found using:\[ y' = \frac{u'v - uv'}{v^2} \] In our exercise, we need this rule because \( y = \frac{x-1}{x+1} \) is a quotient of the functions \( u = x-1 \) and \( v = x+1 \). Differentiating each part, we apply the quotient rule to get the derivative \( y' = \frac{2}{(x+1)^2} \). This derived slope equation is essential for finding points of tangency where lines are parallel to a given line, making it a crucial step in this calculus problem.