Finding The Derivative And Tangent Line Of A Quadratic Function F(x) = 4x^2 - 3x + 6

Jul 12, 2025 by ADMIN 85 views

In calculus, derivatives are a fundamental concept that allows us to analyze the rate at which a function's output changes with respect to its input. Geometrically, the derivative of a function at a specific point represents the slope of the tangent line to the function's graph at that point. This tangent line is a straight line that touches the curve at a single point, providing a linear approximation of the function's behavior near that point. Understanding derivatives and tangent lines is crucial in various fields, including physics, engineering, economics, and computer science, as they provide powerful tools for modeling and analyzing dynamic systems.

At its core, the derivative of a function, often denoted as f'(x), quantifies the instantaneous rate of change of the function f(x) at a particular value of x. This rate of change is determined by finding the limit of the difference quotient as the change in x approaches zero. The difference quotient represents the average rate of change of the function over a small interval, and the limit process allows us to zoom in on the instantaneous rate of change at a single point. The derivative f'(x) is itself a function that provides the slope of the tangent line to the graph of f(x) at any given point. This slope is a measure of the steepness and direction of the curve at that point, indicating whether the function is increasing, decreasing, or stationary. The derivative allows us to analyze the behavior of the function in detail, identifying critical points, intervals of increase and decrease, and concavity, which are essential for understanding the function's overall shape and characteristics.

The tangent line, as a geometric representation of the derivative, offers a visual and intuitive understanding of the function's behavior at a specific point. It is the straight line that best approximates the function near that point, sharing the same slope as the curve at that location. The equation of the tangent line can be determined using the point-slope form of a line, which requires the slope of the line (the derivative at the point) and a point on the line (the point of tangency). The tangent line provides a linear model of the function's behavior in the vicinity of the point, making it a valuable tool for approximating function values and analyzing local changes. In practical applications, tangent lines are used to estimate the behavior of complex systems, optimize processes, and solve problems involving rates of change. For example, in physics, tangent lines are used to determine the instantaneous velocity of an object, while in economics, they are used to analyze marginal cost and revenue.

Let's consider the specific problem of finding the derivative of the function $f(x) = 4x^2 - 3x + 6$ at $x = 4$ , denoted as $f'(4)$ , and then using this derivative to determine the equation of the tangent line to the parabola $y = 4x^2 - 3x + 6$ at the point $(4, 58)$ . This problem provides a concrete example of how to apply the concepts of derivatives and tangent lines to a quadratic function. Quadratic functions, characterized by their parabolic shape, are common in various mathematical and scientific contexts, making this example particularly relevant.

The problem involves two main steps: first, finding the derivative of the function, and second, using the derivative to find the equation of the tangent line. To find the derivative, we will apply the rules of differentiation, which provide a systematic way to calculate the derivative of a function. In this case, we will use the power rule and the sum/difference rule to differentiate the quadratic function. The power rule states that the derivative of $x^n$ is $nx^{n-1}$ , while the sum/difference rule states that the derivative of a sum or difference of functions is the sum or difference of their derivatives. Applying these rules, we can find the derivative of $f(x) = 4x^2 - 3x + 6$ , which will be another function that gives the slope of the tangent line at any point on the parabola.

Once we have the derivative, we can evaluate it at $x = 4$ to find the slope of the tangent line at the point $(4, 58)$ . This slope, $f'(4)$ , is a crucial piece of information for determining the equation of the tangent line. To find the equation of the tangent line, we will use the point-slope form of a line, which is given by $y - y_1 = m(x - x_1)$ , where $m$ is the slope of the line and $(x_1, y_1)$ is a point on the line. In this case, we know the slope $f'(4)$ and the point $(4, 58)$ , so we can plug these values into the point-slope form to obtain the equation of the tangent line. The final step is to rewrite the equation in the slope-intercept form, $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept. This form is convenient for graphing and analyzing the tangent line, and it allows us to easily identify the slope and y-intercept, which provide further insights into the line's behavior.

Find the derivative f'(x):

To find the derivative of the function $f(x) = 4x^2 - 3x + 6$ , we will apply the power rule and the sum/difference rule of differentiation. The power rule states that the derivative of $x^n$ is $nx^{n-1}$ , while the sum/difference rule states that the derivative of a sum or difference of functions is the sum or difference of their derivatives. Applying these rules, we get:

$f'(x) = rac{d}{dx}(4x^2 - 3x + 6) = rac{d}{dx}(4x^2) - rac{d}{dx}(3x) + rac{d}{dx}(6)$

Now, we apply the power rule to each term:

$rac{d}{dx}(4x^2) = 4 imes 2x^{2-1} = 8x$

$rac{d}{dx}(3x) = 3 imes 1x^{1-1} = 3$

$rac{d}{dx}(6) = 0$ (the derivative of a constant is zero)

Therefore, the derivative of the function is:

$f'(x) = 8x - 3$

Evaluate f'(4):

To find the slope of the tangent line at $x = 4$ , we need to evaluate the derivative at $x = 4$ :

$f'(4) = 8(4) - 3 = 32 - 3 = 29$

So, the slope of the tangent line at the point $(4, 58)$ is 29.

Find the equation of the tangent line:

Now that we have the slope of the tangent line, $m = 29$ , and a point on the line, $(4, 58)$ , we can use the point-slope form of a line to find the equation of the tangent line. The point-slope form is given by:

$y - y_1 = m(x - x_1)$

Plugging in the values $m = 29$ , $x_1 = 4$ , and $y_1 = 58$ , we get:

$y - 58 = 29(x - 4)$

Now, we can simplify the equation and rewrite it in the slope-intercept form, $y = mx + b$ :

$y - 58 = 29x - 116$

$y = 29x - 116 + 58$

$y = 29x - 58$

Therefore, the equation of the tangent line to the parabola $y = 4x^2 - 3x + 6$ at the point $(4, 58)$ is $y = 29x - 58$ .

Identify m and b:

The equation of the tangent line is in the form $y = mx + b$ , where $m$ is the slope and $b$ is the y-intercept. From the equation $y = 29x - 58$ , we can see that:

$m = 29$

$b = -58$

Therefore, the slope of the tangent line, $m$ , is 29, and the y-intercept, $b$ , is -58. The equation of the tangent line is $y = 29x - 58$ .

In conclusion, we have successfully found the derivative of the function $f(x) = 4x^2 - 3x + 6$ at $x = 4$ , which is $f'(4) = 29$ . This value represents the slope of the tangent line to the parabola at the point $(4, 58)$ . Using this slope and the point-slope form of a line, we determined the equation of the tangent line to be $y = 29x - 58$ . This process demonstrates the fundamental connection between derivatives and tangent lines, providing a powerful tool for analyzing the behavior of functions and their graphs. Understanding these concepts is essential for various applications in calculus and related fields.