Question

Multiple Choice If the line \(L\) fangent to the graph of a function \(f\) at the point \((2,5)\) passes through the point \((-1,-3),\) what is the slope of \(L_{2}\) . \(D$$\begin{array}{lllll}{\text { (A) }-3 / 8} & {\text { (B) } 3 / 8} & {\text { (C) }-8 / 3} & {\text { (D) } 8 / 3} & {\text { (E) undefined }}\end{array}\)

Short answer

The slope of the line L is 8/3(D).

Step-by-step solution

Substitute the Values

Substitute the given values into the slope formula. So, \(m = (-3 - 5) / (-1 - 2)\).

Simplify the Numerator and the Denominator

Now simplify the numerator and the denominator. The numerator -3 - 5 becomes -8 and the denominator -1 - 2 becomes -3.

Calculate the Slope

Now, divide the numerator by the denominator to get the slope. So, \(m = -8 / -3 = 8/3\). This means the slope of line L is 8/3.

Key concepts

Understanding the Slope Formula

The slope formula is fundamental in understanding the rate of change between two points on a line. In practical terms, it's what tells us how steep a line is. The slope, denoted as m, is calculated by the difference in the y-coordinates divided by the difference in the x-coordinates of two distinct points on that line.

The formula is written as:\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Using the slope formula is like finding the steepness of a hill in real life—positive slopes go uphill and negative slopes go downhill. A horizontal line has a slope of 0, as there is no change in y, while a vertical line's slope is undefined because the change in x would be 0, which causes a division by zero.

To apply this concept to the exercise given, we take the points (2,5) and (-1,-3). Substituting these into the slope formula, we get:\[ m = \frac{-3 - 5}{-1 - 2} \]This calculation proceeds through numerical simplification to find the grade of the line L.

Tangent Line Calculus

Tangent line calculus is a bit more complex than simply finding the slope between two points. When dealing with curves, the tangent line is a straight line that touches the curve at exactly one point and its slope at that point is the derivative of the function at that point.

For example, if you have a function f(x) and you want to find the slope of the tangent line at a specific point a, you would find f'(a), which is the derivative of f evaluated at x=a. This is precisely the slope of the tangent line at that point on the function.

In our multiple choice exercise, L is such a tangent line to a function f at the point (2,5), and we're calculating the slope as if it passed through another point (-1,-3). The concept of derivatives is not explicitly needed for this particular numerical slope calculation because the slope is determined by the two points that the tangent line is passing through.

Numerical Simplification Techniques

Numerical simplification is the process of reducing complexity in a mathematical expression to reach its simplest form. This often involves combining like terms, simplifying fractions, or applying arithmetic operations.

In the context of our problem, simplification is key to finding the correct slope. Starting with the raw substitution from the points gives us a messy fraction that we need to simplify. By subtracting the y-coordinates and the x-coordinates,\[ (-3 - 5) \text{ and } (-1 - 2), \]we get the numerator of -8 and a denominator of -3. We then divide these to get the slope of line L, which simplifies to:\[ m = \frac{-8}{-3} = \frac{8}{3} \]
Understanding numerical simplification can help students avoid errors and ensure accuracy in their calculations, especially when dealing with negative numbers or subtraction, as seen in this exercise.