Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. \(m=f^{\prime}(x)=3 x^{2}+8 x\) gives the slope of the tangent line to the graph of \(f(x)=x^{3}+4 x^{2}-5\) at any number \(x\). Find an equation of the tangent line to \(f\) at \(x=-2\).

Short answer

The equation of the tangent line is \( y = -4x - 5 \).

Step-by-step solution

- Evaluate the slope at x = -2

Given the derivative, which represents the slope of the tangent line, is: \[ f'(x) = 3x^2 + 8x \]Substitute \( x = -2 \) into the derivative to find the slope at this point:\[ m = 3(-2)^2 + 8(-2) = 3(4) - 16 = 12 - 16 = -4 \]So, the slope of the tangent line at \( x = -2 \) is \( m = -4 \).

- Evaluate the function at x = -2

To find the point of tangency, substitute \( x = -2 \) into the original function \( f(x) \):\[ f(x) = x^3 + 4x^2 - 5 \]\[ f(-2) = (-2)^3 + 4(-2)^2 - 5 = -8 + 16 - 5 = 3 \]Therefore, the point of tangency is \( (-2, 3) \).

- Use the point-slope form to write the equation of the tangent line

The point-slope form of a line is \( y - y_1 = m(x - x_1) \), where \((x_1, y_1)\) is the point of tangency and \( m \) is the slope.Substitute \( (-2, 3) \) for \((x_1, y_1)\) and \( -4 \) for \( m \):\[ y - 3 = -4(x + 2) \]Simplify the equation:\[ y - 3 = -4x - 8 \]Add 3 to both sides:\[ y = -4x - 5 \]Hence, the equation of the tangent line is \( y = -4x - 5 \).

Key concepts

derivatives

Derivatives are a fundamental concept in calculus. They represent the rate at which a function is changing at any given point. More formally, if you have a function \( f(x) \), its derivative is denoted by \( f'(x) \) and is defined as the limit of the difference quotient when the increment approaches zero: \[ f'(x) = \frac{df(x)}{dx} = \text{lim}_{\triangle x \to 0} \frac{f(x + \triangle x) - f(x)}{\triangle x} \].

In simpler terms, the derivative gives you the slope of the tangent line to the function at any given point. For example, in our exercise, the derivative given is \( f'(x) = 3x^2 + 8x \). This tells us how steep the function \( f(x) = x^3 + 4x^2 - 5 \) is at any particular value of \( x \).

To find the slope at a specific point, like \( x = -2 \), you simply substitute \( x \) into the derivative: \[ f'(-2) = 3(-2)^2 + 8(-2) = 12 - 16 = -4 \]. The slope of the tangent line at \( x = -2 \) is thus \(-4 \).

point-slope form

The point-slope form of a line equation is a convenient way to write the equation of a line when you know the slope and one point on the line. The equation is given by: \[ y - y_1 = m(x - x_1) \], where \( m \) is the slope and \( (x_1, y_1) \) is the point through which the line passes.

This form is especially useful for finding the equation of a tangent line to a curve at a specific point. Once you have both the slope from the derivative and the coordinates of the tangent point, you can substitute them into the point-slope form to get the equation of the tangent line.

For instance, in our exercise, we have the slope \(-4 \) and the point \((-2, 3) \). Plugging these into the point-slope form, we get: \[ y - 3 = -4(x + 2) \].

slope of a tangent line

The slope of the tangent line is a crucial idea when studying the behavior of functions. It represents how fast the function's value is changing at a particular point.

To find the slope of the tangent line at a particular point, you first need the derivative of the function. The value of the derivative at the point of interest gives you the slope. In our exercise, the derivative of the function \( f(x) = x^3 + 4x^2 - 5 \) is \( f'(x) = 3x^2 + 8x \). At \( x = -2 \):

\[ f'(-2) = 3(-2)^2 + 8(-2) = 12 - 16 = -4 \]

So, the slope of the tangent line to the curve at \( x = -2 \) is \(-4 \). This means that at \( x = -2 \), the function is decreasing.

evaluating functions

Evaluating functions is the process of finding the output of a function for a given input. This is essential for determining points on the graph or verifying the coordinates of the tangent point.

To find the value of a function at a specific input, simply substitute the input into the function. For example, for \( f(x) = x^3 + 4x^2 - 5 \), if we want to evaluate the function at \( x = -2 \):

\[ f(-2) = (-2)^3 + 4(-2)^2 - 5 = -8 + 16 - 5 = 3 \]

This tells us that the function has a value of \( 3 \) when \( x = -2 \). Therefore, the point of tangency on the function is \((-2, 3) \).