Problem 9 - University Calculus: Early Transcendentals

Question

Find an equation for the tangent to the curve at the given point. Then sketch the curve and tangent together. $$y=x^{3}, \quad(-2,-8)$$

Step-by-Step Solution

Answer

The tangent line at \((-2, -8)\) is \(y = 12x + 16\).

Step 1: Find the derivative of the curve

To find the equation of the tangent, we need the derivative of the curve because it represents the slope of the tangent line. The curve given is \( y = x^3 \). Calculate the derivative using the power rule \( \frac{d}{dx}(x^n) = nx^{n-1} \). Thus:\[ \frac{dy}{dx} = 3x^2 \]

Step 2: Evaluate the derivative at the given point

The slope of the tangent line at a specific point is found by substituting the x-coordinate of the point into the derivative. For the point \((-2, -8)\), substitute \(x = -2\) into the derivative:\[ \frac{dy}{dx} \bigg|_{x=-2} = 3(-2)^2 = 3 imes 4 = 12 \]

Step 3: Use the point-slope form to find the equation of the tangent line

With the slope from Step 2 and the given point, use the point-slope equation of a line: \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \((x_1, y_1)\) are the coordinates of the point.Using \( m = 12 \) and \( (-2, -8) \):\[ y - (-8) = 12(x - (-2)) \]Simplify:\[ y + 8 = 12(x + 2) \]Further simplify to:\[ y = 12x + 24 - 8 \]\[ y = 12x + 16 \]

Step 4: Sketch the curve and tangent line

The curve \( y = x^3 \) is a cubic function that passes through the origin \((0,0)\) and is symmetric about the origin. The point \((-2, -8)\) is on this curve. The tangent line was found to be \( y = 12x + 16 \). You can sketch the curve, which has a general S-shape characteristic of cubic equations, and the tangent line, which is a straight line intersecting the curve at \((-2, -8)\) with a slope of 12.

Key Concepts

Understanding Cubic Functions

Cubic functions are polynomial functions of degree three, and they have the general form of \( y = ax^3 + bx^2 + cx + d \). In simple terms, the exponent on the highest degree term is three. In our exercise, the function is \( y = x^3 \), which means:

• It has a coefficient of 1 for the \( x^3 \) term and no \( x^2 \), \( x \), or constant term, simplifying the cubic function.
• Cubic functions tend to have an "S" shape when graphed. Depending on the signs of the coefficients, this shape might be stretched or compressed.

The important properties of cubic functions include:

• Their graphs typically have turning points - places where the graph changes direction from increasing to decreasing or vice versa.
• They can intersect the x-axis at one, two, or three points, known as roots or zeros.
• They are not symmetric over the y-axis like quadratic functions but are symmetric about their point of inflection, where the graph transitions from concave up to concave down or the reverse.

For the cubic function \( y = x^3 \), it passes through the origin, is odd, and has rotational symmetry about the origin.

The Role of Derivative in Tangent Line

The derivative of a function gives us the slope of the tangent line to the curve at any given point. For our cubic function \( y = x^3 \), finding the derivative is essential to determine the equation of the tangent line at the specified point.To find the derivative, we used the power rule: \( \frac{d}{dx}(x^n) = nx^{n-1} \). This means for \( y = x^3 \), the derivative is:

• \( \frac{dy}{dx} = 3x^2 \)

This derivative tells us that, unlike a linear function with a constant slope, the slope of a cubic function changes with \( x \). To find the slope of the tangent line at the point \((-2, -8)\), we substitute \( x = -2 \) into the derivative:

• \( \frac{dy}{dx} \big|_{x=-2} = 3(-2)^2 = 12 \)

This calculated slope indicates how steep the tangent line is at that particular point.

Understanding Slope in Tangent Lines

The slope of a tangent line is a crucial element in determining its equation. We often use the slope-point form to express the tangent line, which is formulated as \( y - y_1 = m(x - x_1) \). Here, \( m \) stands for the slope, and \((x_1, y_1)\) are the coordinates where the tangent touches the curve.In our exercise, the slope \( m \) is found to be 12 at the point \((-2, -8)\). Thus, using the point-slope formula:

• \( y + 8 = 12(x + 2) \)

When simplified, this gives the line equation:

• \( y = 12x + 16 \)

The tangent line's slope tells us how quickly the line rises or falls as we move along the x-axis. A positive slope means the line rises, and a negative slope means it falls. With a slope of 12, this line rises steeply, illustrating the rapid increase in the cubic function \( y = x^3 \) as \( x \) moves away from zero.