Tangent Lines and Derivatives: Videos & Practice Problems

The tangent line touches a curve at one point, contrasting with the secant line, which intersects at two points. To find the slope of a tangent line, use the limit as x approaches a specific value c: $lim\left(f\right(x)-f(c\left)\right)/(x-c))$. This slope represents the instantaneous rate of change, also known as the derivative. To find the equation of the tangent line, apply point-slope form using the calculated slope and the point of tangency.

Slopes of Tangent Lines

In calculus, understanding the concepts of secant and tangent lines is crucial for analyzing functions. A secant line intersects a curve at two distinct points, while a tangent line touches the curve at just one point. This distinction is fundamental, as it leads to different methods of calculating slopes.

To illustrate, consider the function \( f(x) = x^2 \). To find the slope of the tangent line at \( x = 1 \), we focus on a single point rather than two. The slope of the tangent line can be determined using the limit definition of the derivative, which is expressed as:

\[m = \lim_{{x \to c}} \frac{f(x) - f(c)}{x - c}\]

In this case, \( c = 1 \). Thus, we substitute into the formula:

\[m = \lim_{{x \to 1}} \frac{f(x) - f(1)}{x - 1}\]

Calculating \( f(1) \) gives us \( 1^2 = 1 \), leading to:

\[m = \lim_{{x \to 1}} \frac{x^2 - 1}{x - 1}\]

To simplify, we recognize that \( x^2 - 1 \) can be factored as \( (x - 1)(x + 1) \). This allows us to cancel the \( (x - 1) \) term, resulting in:

\[m = \lim_{{x \to 1}} (x + 1)\]

Now, substituting \( x = 1 \) yields:

\[m = 1 + 1 = 2\]

This value, 2, represents the slope of the tangent line at the point \( (1, 1) \). In contrast, if we were to calculate the slope of a secant line between two points, we would find a different value, illustrating that the secant line provides an average rate of change, while the tangent line gives the instantaneous rate of change.

The slope of the tangent line is also referred to as the derivative of the function at that point. Thus, the derivative encapsulates the concept of instantaneous rate of change, which is a key idea in calculus.

Given the function $f(x)=4x^2-1$, calculate the slope of the tangent line at $x=-3$.

• A. 36
• B. -36
• C. -24
• D. 24

Answer: (C)

Given the function $f(x)=x^2-10x+2$, calculate the slope of the tangent line at $x=2$.

• A. -6
• B. 6
• C. 10
• D. -10

Answer: (A)

Given the function $f(x)=x^2+100$, calculate the slope of the tangent line at $x=0$.

• A. 0
• B. 100
• C. -100
• D. Infinite (vertical line)

Answer: (A)

Equations of Tangent Lines

Understanding tangent lines is crucial in calculus, as they represent the slope of a function at a specific point. A tangent line touches the curve of a function at exactly one point, and calculating its slope involves using limits. To find the equation of a tangent line, we can follow a systematic approach that builds on previously learned concepts.

Consider the function \( f(x) = 3x^2 - 4 \) and the task of finding the equation of the tangent line at \( x = -2 \). The first step is to determine the corresponding \( y \)-value, denoted as \( f(c) \), where \( c \) is the point of interest. By substituting \( c = -2 \) into the function, we calculate:

\( f(-2) = 3(-2)^2 - 4 = 3(4) - 4 = 12 - 4 = 8 \).

Now that we have \( c = -2 \) and \( f(c) = 8 \), we can proceed to find the slope of the tangent line using the limit definition of the derivative:

\( m = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \).

Substituting our values, we have:

\( m = \lim_{x \to -2} \frac{3x^2 - 4 - 8}{x + 2} = \lim_{x \to -2} \frac{3x^2 - 12}{x + 2} \).

Next, we simplify the expression. Factoring out a 3 gives:

\( m = \lim_{x \to -2} \frac{3(x^2 - 4)}{x + 2} = \lim_{x \to -2} \frac{3(x + 2)(x - 2)}{x + 2} \).

We can cancel \( x + 2 \) from the numerator and denominator, leading to:

\( m = \lim_{x \to -2} 3(x - 2) \).

Now substituting \( x = -2 \) gives:

\( m = 3(-2 - 2) = 3(-4) = -12 \).

With the slope \( m = -12 \) and the point \( (-2, 8) \), we can use the point-slope form of the equation of a line:

\( y - y_1 = m(x - x_1) \),

where \( (x_1, y_1) = (-2, 8) \). Plugging in the values, we have:

\( y - 8 = -12(x + 2) \).

Distributing the slope results in:

\( y - 8 = -12x - 24 \).

Finally, adding 8 to both sides yields:

\( y = -12x - 16 \).

This equation represents the tangent line to the function at the specified point. Mastering this process of finding tangent lines not only enhances understanding of derivatives but also prepares students for more complex calculus concepts.

Given the function $f(x)=3(x^2-1)$, find the equation of the tangent line at $x=1$.

• A. $y=-6x+6$
• B. $y=-6x-6$
• C. $y=6x+6$
• D. $y=6x-6$

Answer: (D)

Determine Continuity Algebraically Example 1

To find the equation of the tangent line to the function \( f(x) = -4x^2 \) at the point where \( x = -2 \), we start by determining the slope of the tangent line using the limit definition. The slope \( m \) of the tangent line can be expressed as:

\[m = \lim_{{x \to c}} \frac{{f(x) - f(c)}}{{x - c}}\]

In this case, \( c = -2 \). First, we need to calculate \( f(-2) \):

\[f(-2) = -4(-2)^2 = -4(4) = -16\]

Now substituting into the slope formula, we have:

\[m = \lim_{{x \to -2}} \frac{{-4x^2 - (-16)}}{{x - (-2)}} = \lim_{{x \to -2}} \frac{{-4x^2 + 16}}{{x + 2}}\]

Next, we simplify the expression. Factoring the numerator gives:

\[-4(x^2 - 4) = -4(x + 2)(x - 2)\]

Thus, the limit becomes:

\[m = \lim_{{x \to -2}} \frac{{-4(x + 2)(x - 2)}}{{x + 2}}\]

We can cancel \( (x + 2) \) from the numerator and denominator (noting that \( x \neq -2 \) during the limit process):

\[m = \lim_{{x \to -2}} -4(x - 2) = -4(-2 - 2) = -4(-4) = 16\]

Now that we have the slope \( m = 16 \), we can use the point-slope form of the equation of a line, which is:

\[y - y_0 = m(x - x_0)\]

Here, \( (x_0, y_0) = (-2, -16) \). Substituting the values, we get:

\[y - (-16) = 16(x - (-2))\]

This simplifies to:

\[y + 16 = 16(x + 2)\]

Distributing the 16 gives:

\[y + 16 = 16x + 32\]

Finally, isolating \( y \) yields:

\[y = 16x + 32 - 16\]

Thus, the equation of the tangent line is:

\[y = 16x + 16\]

In summary, the slope of the tangent line at \( x = -2 \) is 16, and the equation of the tangent line is \( y = 16x + 16 \).

Here’s what students ask on this topic:

What is the difference between a secant line and a tangent line?

A secant line intersects a curve at two points, while a tangent line touches the curve at exactly one point. The secant line represents the average rate of change between two points, whereas the tangent line represents the instantaneous rate of change at a single point. The slope of the tangent line is also known as the derivative.

How do you find the slope of a tangent line to a function at a given point?

To find the slope of a tangent line to a function at a given point, use the limit definition of the derivative: $\frac{lim\left(f\right(x)-f(c\left)\right)}{(x-c)}$ as x approaches c. This gives the instantaneous rate of change at the point x = c.

What is the equation of the tangent line to a function at a given point?

To find the equation of the tangent line to a function at a given point, first find the slope of the tangent line using the derivative. Then, use the point-slope form of the equation of a line: $y-y$₁=m(x-x₁), where m is the slope and (x₁, y₁) is the point of tangency.

What is the instantaneous rate of change and how is it related to the derivative?

The instantaneous rate of change is the rate at which a function is changing at a specific point. It is represented by the slope of the tangent line at that point. The derivative of a function at a point gives the instantaneous rate of change at that point.

How do you calculate the derivative of a function using the limit definition?

To calculate the derivative of a function using the limit definition, use the formula: $lim(h\to 0)\frac{f(x+h)-f\left(x\right)}{h}$. This formula finds the slope of the tangent line to the function at any point x.