Problem 24 - Calculus Single Variable

Question

Find the slope of the tangent line to the graph of the given function at the given point \(P\). $$ f(x)=-4 x^{2}+x+1 \quad P=(2,-13) $$

Step-by-Step Solution

Answer

The slope of the tangent line at point \( P=(2,-13) \) is \( -15 \).

Step 1: Differentiate the Function

To find the slope of the tangent line to the graph of the function at a given point, we first need to find the derivative of the function. The function is given as \( f(x) = -4x^2 + x + 1 \). The derivative, \( f'(x) \), represents the slope of the tangent line at any point \( x \). Calculating the derivative, we apply the power rule: \( \frac{d}{dx}[-4x^2] = -8x \), \( \frac{d}{dx}[x] = 1 \), and \( \frac{d}{dx}[1] = 0 \). Therefore, \( f'(x) = -8x + 1 \).

Step 2: Evaluate the Derivative at the Point

Next, we evaluate the derivative at the given point to find the slope of the tangent line at that point. The point is \( P=(2,-13) \), meaning we substitute \( x = 2 \) into the derivative: \( f'(2) = -8(2) + 1 = -16 + 1 = -15 \).

Step 3: Interpretation of the Result

The calculated derivative \( f'(2) = -15 \) indicates that the slope of the tangent line to the function \( f(x) \) at the point \( P=(2,-13) \) is \( -15 \). This means that as you move one unit to the right from this point, the value of \( f(x) \) decreases by 15 units.

Key Concepts

Power Rule for Derivatives

The power rule is a straightforward technique used for computing the derivative of a function that is a power of a variable. It states that if you have a function in the form of \( ax^n \), the derivative is found by multiplying the power \( n \) by the coefficient \( a \) and then subtracting one from the power.
For instance, when differentiating \(-4x^2\), we multiply \(-4\) by \(2\) and then reduce the power by one to get \(-8x^1 = -8x\).
This rule helps simplify the process, allowing us to focus on each term individually.

• The power rule ensures that we can efficiently calculate derivatives without complex steps.
• It is fundamental for finding slopes, especially when determining the behavior of polynomial functions.

Note that constants like \(1\) have a derivative of zero because they don’t change with \(x\). Using the power rule aids in easily obtaining each term’s derivative, such as in the exercise where \(f'(x) = -8x + 1\).

Understanding Tangent Lines

A tangent line is a straight line that touches a curve at just one point. This point is crucial as it tells us about the behavior of the function at that specific location.
In our exercise, we want to find out how steep the curve is at the point \(P=(2,-13)\). The slope of this tangent line is found using the derivative at that point.

• A tangent line reflects the instantaneous rate of change of the function.
• It shows how the function is increasing or decreasing at that exact spot.

The slope of the tangent can give us insight into whether the function is increasing or decreasing swiftly or gently. Knowing its slope helps us understand the nature of the curve around that point.
In real-world contexts, tangent lines can represent phenomena such as the speed of an object at a specific time.

Slope of a Function and Its Importance

The slope of a function at a given point defined by the derivative tells us how quickly the value of the function changes as the input changes. More precisely, it's the gradient or steepness of the graph at that specific point.
In the exercise, by evaluating the derivative \(f'(x) = -8x + 1\) at \(x = 2\), we obtain \(f'(2) = -15\). This slope of \(-15\) means that for every unit increase in \(x\), the function value decreases by 15 units.

• A positive slope suggests an upward trend, while a negative slope shows a downward trend.
• The higher the absolute value, the steeper the slope and the more rapid the change.

Slopes are useful in various fields such as physics for understanding motion or in economics to examine changes in costs or profits over time. Grasping how slopes work enhances our ability to interpret and predict functional behavior effectively.