91Ó°ÊÓ

Derivatives are a cornerstone of calculus and help us understand how functions change. Essentially, the derivative of a function gives us the slope of the tangent line to the curve at any given point. This slope tells us how steep the curve is at that point. For a quadratic function like \( y = x^2 + 4x - 2 \), finding the derivative involves differentiating each term individually using power rules:

• The derivative of \( x^2 \) is \( 2x \).
• The derivative of \( 4x \) is 4.
• The derivative of a constant term like \(-2\) is 0, since constants do not change.

Thus, the derivative of the function is \( \frac{dy}{dx} = 2x + 4 \). This formula allows us to calculate the slope of the tangent line at any point \( x \). Understanding this concept is crucial for determining how quickly or slowly the function's value is changing at any point along the curve.

The tangent line to a curve at a specific point is the straight line that just 'touches' the curve at that point and has the same slope as the curve there. The concept of a tangent line is useful for approximation and provides a linear estimation of the curve near the point of tangency.
To find the equation of the tangent line to the curve \( y = x^2 + 4x - 2 \) at \( x = -3 \), we first found the slope using the derivative: \( \frac{dy}{dx} = 2x + 4 \). At \( x = -3 \), the slope is \(-2\).
With this slope and knowing the coordinates \((-3, -5)\) where the curve meets \( x = -3 \), we can use the point-slope form \( y - y_1 = m(x - x_1) \) to write the tangent line's equation:

• Substitute the slope \( m = -2 \).
• Use the point \((-3, -5)\).

Ultimately, this tool helps in understanding the direction the curve is heading near any point.

A normal line to a curve at a given point is perpendicular to the tangent line at the same point. This means the product of their slopes is \(-1\). For a tangent line with a slope \(-2\), the normal line's slope is the negative reciprocal, which is \(\frac{1}{2}\).
To find the equation of the normal line, we apply the point-slope form of a line once more, using this new slope and the same point of intersection with the curve.

• Slope: Use \( \frac{1}{2} \).
• Point: \((-3, -5)\).

The equation of the normal line serves a dual purpose in this exercise: it provides a perpendicular reference at \((-3, -5)\), and it is useful in examining further intersections of the line with the curve.

Quadratic equations are polynomial equations of degree two, generally in the form \( ax^2 + bx + c = 0 \). Solving a quadratic equation can reveal important properties about the relationships between elements in problems like the one given.
In this exercise, we use the quadratic equation \( 2x^2 + 7x - 3 = 0 \) to find the intersections of the normal line with the curve again. The quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), allows us to solve for \( x \):

• Identify \( a = 2, b = 7, \) and \( c = -3 \).
• Plug these values into the quadratic formula.

Solving gives two roots, but one already corresponds to the point \((x = -3)\). The other represents where the normal line meets the curve again. This method outlines how quadratic equations can be applied to find points of intersection in geometry, providing deeper insights into the relationships between curves and lines.