Find The Tangent Line: A Complete Guide With Examples

Hey guys! Let's dive into a classic calculus problem: finding the slope and equation of a tangent line. We'll be using the limit definition of the derivative to do it, and I'll break down each step to make it super clear. So, grab your pencils, and let's get started!

Understanding the Problem

Our mission, should we choose to accept it, is to figure out the slope of the tangent line to the curve $y=3x^2 + 2x + 1$ at the specific point (2, 17). We also need to find the equation of this tangent line. Remember, the tangent line touches the curve at a single point and represents the instantaneous rate of change of the function at that point. This is where calculus shines!

So, to recap, here's what we're aiming to find:

• The slope (m) of the tangent line at x = 2.
• The equation of the tangent line.

We are given the function $f(x) = 3x^2 + 2x + 1$, and the point (2, 17). The point is correct now. Originally, it was (2,5), but let us prove this point does not belong to the graph. If we evaluate the function at x = 2, we get: $f(2) = 3(2)^2 + 2(2) + 1 = 12 + 4 + 1 = 17$. Therefore, the point is (2, 17). We will use the formula $m=rac{\Delta y}{\Delta x}=\lim_{\Delta x \longrightarrow 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$. This is the limit definition of the derivative, which gives us the slope of the tangent line. Let's get to work!

The Derivative: Our Secret Weapon

The derivative, my friends, is our key to unlocking the slope of that tangent line. It's the instantaneous rate of change of a function at a specific point. In essence, it tells us how the y-value is changing with respect to the x-value at any given moment. The formula given to us is the definition of the derivative, which is exactly what we'll be using. The derivative of a function $f(x)$ is denoted as $f'(x)$ or $\frac{dy}{dx}$. In our case, we will use the definition and we can also evaluate the derivative at a specific point like x = 2, which is denoted as $f'(2)$.

Step-by-Step Solution

Alright, let's roll up our sleeves and find that slope and equation! We'll follow these steps:

1. Find f(x + Δx). This involves substituting (x + Δx) into our function.
2. Calculate f(x + Δx) - f(x). This is the difference quotient's numerator.
3. Divide by Δx
4. Take the limit as Δx approaches 0. This gives us the derivative, f'(x).
5. Evaluate f'(2). That's how we find the slope at x = 2.
6. Find the equation of the tangent line. Using the point-slope form.

Let's break down each step in detail.

Step 1: Finding f(x + Δx)

We start with our function: $f(x) = 3x^2 + 2x + 1$. Now, we need to find $f(x + \Delta x)$. This means we replace every 'x' in our function with '(x + Δx)'. Watch closely, guys:

$$f(x + \Delta x) = 3(x + \Delta x)^2 + 2(x + \Delta x) + 1$$

Now, let's expand this. Remember your algebra rules! We need to expand $(x + \Delta x)^2$, which equals $x^2 + 2x\Delta x + (\Delta x)^2$.

So, substituting this in, we get:

$$f(x + \Delta x) = 3(x^2 + 2x\Delta x + (\Delta x)^2) + 2(x + \Delta x) + 1$$

Simplifying further:

$$f(x + \Delta x) = 3x^2 + 6x\Delta x + 3(\Delta x)^2 + 2x + 2\Delta x + 1$$

Step 2: Calculate f(x + Δx) - f(x)

Next up, we need to subtract our original function, $f(x)$, from $f(x + \Delta x)$. This is where things start to get interesting, because we want the difference between the function at a point and a very close point.

$$f(x + \Delta x) - f(x) = (3x^2 + 6x\Delta x + 3(\Delta x)^2 + 2x + 2\Delta x + 1) - (3x^2 + 2x + 1)$$

Notice how we're subtracting the entire function, so we need to make sure we subtract each term. Now, let's simplify by combining like terms:

$$f(x + \Delta x) - f(x) = 6x\Delta x + 3(\Delta x)^2 + 2\Delta x$$

Step 3: Divide by Δx

We're getting closer! Now, we divide the result from Step 2 by Δx:

$$\frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{6x\Delta x + 3(\Delta x)^2 + 2\Delta x}{\Delta x}$$

We can factor out a Δx from the numerator:

$$\frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{\Delta x(6x + 3\Delta x + 2)}{\Delta x}$$

Now, we can cancel out the Δx from the numerator and the denominator:

$$\frac{f(x + \Delta x) - f(x)}{\Delta x} = 6x + 3\Delta x + 2$$

Step 4: Take the Limit as Δx Approaches 0

Here comes the final step to find the derivative. We take the limit as Δx approaches 0:

$$f'(x) = \lim_{\Delta x \longrightarrow 0} (6x + 3\Delta x + 2)$$

As Δx gets infinitely small (approaches 0), the term $3\Delta x$ becomes 0.

Therefore, our derivative is:

$$f'(x) = 6x + 2$$

This is a super important result! This gives us the slope of the tangent line at any point x on our curve. We now have a general formula for the slope.

Step 5: Evaluate f'(2)

Now that we have the derivative, $f'(x) = 6x + 2$, we can find the slope of the tangent line at the specific point x = 2. We just plug in '2' for 'x':

$$f'(2) = 6(2) + 2 = 12 + 2 = 14$$

So, the slope of the tangent line at the point (2, 17) is 14. We're almost there, guys!

Step 6: Find the Equation of the Tangent Line

We have the slope (m = 14) and a point on the line (2, 17). We can use the point-slope form of a line, which is: $y - y_1 = m(x - x_1)$

Where: $(x_1, y_1)$ is our point (2, 17), and 'm' is the slope, which is 14.

Plugging in the values, we get:

$$y - 17 = 14(x - 2)$$

Now, let's simplify this equation into slope-intercept form (y = mx + b) for a nice and clean look.

$$y - 17 = 14x - 28$$

Add 17 to both sides:

$$y = 14x - 11$$

And there you have it! The equation of the tangent line to the curve $y = 3x^2 + 2x + 1$ at the point (2, 17) is $y = 14x - 11$.

Conclusion: We Did It!

We successfully found the slope of the tangent line (14) and its equation (y = 14x - 11). This process, though it may seem like a lot of steps, is fundamental to understanding calculus and how functions change. Well done, guys! Keep practicing, and you'll become tangent line masters in no time! Remember, the key is to break down the problem into smaller, manageable steps, and to understand the concepts behind each step. Now, go forth and conquer those curves!