Problem 31 - Calculus Single Variable

Question

A function \(f\) and a point \(P\) are given. Find the slope-intercept form of the equation of the tangent line to the graph of \(f\) at \(P\). $$ f(x)=3 x^{2}+2 x \quad P=(1,5) $$

Step-by-Step Solution

Answer

The equation of the tangent line is \( y = 8x - 3 \).

Step 1: Find the Derivative of the Function

To find the slope of the tangent line, we first need to find the derivative of the function, which gives us the slope of the tangent line at any point. The function given is \( f(x) = 3x^2 + 2x \). The derivative is found using the power rule. For \( 3x^2 \), the derivative is \( 6x \), and for \( 2x \), it's \( 2 \). Thus, the derivative \( f'(x) = 6x + 2 \).

Step 2: Evaluate the Derivative at Point P

Next, we evaluate the derivative at the given point \( P = (1,5) \) to find the slope of the tangent line at this point. Substitute \( x = 1 \) into the derivative, \( f'(1) = 6(1) + 2 = 6 + 2 = 8 \). So, the slope of the tangent line at point \( P \) is 8.

Step 3: Use Point-Slope Formula

Now we use the point-slope formula for a line, which is \( y - y_1 = m(x - x_1) \), where \( m \) is the slope and \( (x_1, y_1) \) is the point through which the line passes. Here, we have \( (x_1, y_1) = (1, 5) \) and \( m = 8 \). Thus, the equation becomes \( y - 5 = 8(x - 1) \).

Step 4: Convert to Slope-Intercept Form

To express the equation in slope-intercept form, \( y = mx + b \), solve the equation from Step 3. Start by distributing the 8: \( y - 5 = 8x - 8 \). Next, add 5 to both sides to isolate \( y \): \( y = 8x - 3 \). This is the slope-intercept form of the equation of the tangent line.

Key Concepts

Tangent Line

A tangent line is a straight line that touches a curve at one specific point, matching the curve's slope at that very point. Imagine zooming into the curve until it becomes straight; the tangent line is that straight line at that zoomed-in point.
For a function like our example, a parabolic function, drawing a tangent line at a point means identifying the direction or slope of the line that just grazes the curve. This line doesn't cross the curve at that small scale. It only touches the curve precisely at one point.
Understanding tangent lines becomes crucial in calculus as they help in modeling and understanding instantaneous rates of change. They remain fundamental in physics, economics, and engineering, as they estimate small changes and smooth out our understanding of more complex curves.

Derivative

The derivative of a function is a tool in calculus that helps us find the slope of the tangent line at any point on the function's curve. Think of it as a mathematical way to predict how a function behaves at an instant.
When we take the derivative of the function, such as in our example with the function \( f(x) = 3x^2 + 2x \), it helps us discover the formula for the slope at any point x. Using the power rule, the derivative of \( 3x^2 \) becomes \( 6x \), and for \( 2x \), it's \( 2 \). So, we get \( f'(x) = 6x + 2 \).
This derivative tells us how steep the graph is at any x value. The higher the derivative at x, the steeper the slope of the tangent line. In our example, the slope at \( x = 1 \) is \( 8 \), showing us that the tangent line rises quickly.

Slope-Intercept Form

The slope-intercept form is a straightforward way to express a linear equation, making it easy to read the slope and y-intercept directly from the equation. It follows the formula \( y = mx + b \), where \( m \) represents the slope, and \( b \) indicates the y-intercept.
For example, after determining the slope of the tangent line using the derivative and evaluating it at point P, we convert the equation into this form to make it easy to graph.
From our exercise, after using the point-slope form, we express the tangent line's equation as \( y = 8x - 3 \). Here, \( 8 \) is the slope, and \( -3 \) is the y-intercept. This format helps us sketch graphs and understand the linear relationship with ease.

Point-Slope Formula

The point-slope formula is extremely handy when we already know a line's slope and a point on the line. It helps us write out the equation of the line efficiently. The formula can be written as \( y - y_1 = m(x - x_1) \), where \( m \) represents the slope, and \( (x_1, y_1) \) is the known point.
In our exercise, with the slope at point P being \( 8 \) and using point \( (1, 5) \), we plug these into the formula as \( y - 5 = 8(x - 1) \). This form pinpoints exactly how the line is positioned relative to the origin.
While sometimes it's good to convert to the slope-intercept form for simplicity, the point-slope form remains extremely useful for quickly writing equations when working directly from given points and slopes.