Problem 73 - Calculus and its applications

Question

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=-\frac{1}{3} x^{3}+6 x^{2}-11 x-50 $$

Step-by-Step Solution

Answer

The tangent line is horizontal at \((1, -\frac{166}{3})\) and \((11, 0)\).

Step 1: Understanding the Problem

We need to find points on the graph of the function where the tangent line is horizontal. A tangent line is horizontal when its slope is zero. Therefore, we need to find where the derivative of the function is equal to zero.

Step 2: Compute the Derivative

The given function is \( y = -\frac{1}{3} x^{3} + 6x^{2} - 11x - 50 \). Let's find its derivative, \( \frac{dy}{dx} \). Differentiating term by term, we get: \[ \frac{dy}{dx} = -x^{2} + 12x - 11 \].

Step 3: Set Derivative to Zero and Solve

To find where the tangent line is horizontal, set the derivative equal to zero: \[ -x^{2} + 12x - 11 = 0 \]. This is a quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a = -1 \), \( b = 12 \), and \( c = -11 \).

Step 4: Use the Quadratic Formula

The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) helps solve for \( x \). Substitute \( a = -1 \), \( b = 12 \), and \( c = -11 \): \[ x = \frac{-12 \pm \sqrt{12^2 - 4(-1)(-11)}}{2(-1)} \]. Simplify this to find the solutions for \( x \).

Step 5: Simplify and Determine the Values of x

Calculate \( x = \frac{-12 \pm \sqrt{144 - 44}}{-2} \). \( x = \frac{-12 \pm \sqrt{100}}{-2} \). \( \sqrt{100} = 10 \), so the solutions are \( x = \frac{-12 + 10}{-2} = 1 \) and \( x = \frac{-12 - 10}{-2} = 11 \).

Step 6: Find y-values for x-values

Substitute \( x = 1 \) and \( x = 11 \) back into the original function to find the corresponding \( y \)-values. \( y(1) = -\frac{1}{3}(1)^3 + 6(1)^2 - 11(1) - 50 \) and \( y(11) = -\frac{1}{3}(11)^3 + 6(11)^2 - 11(11) - 50 \).

Step 7: Calculate Corresponding y-values

Calculate \( y(1) = -\frac{1}{3} + 6 - 11 - 50 = -55.333... \). Calculate \( y(11) = -\frac{1}{3}(1331) + 6(121) - 121 - 50 \). Simplify to find \( y(11) = 0 \).

Step 8: State the Points

The points on the graph where the tangent line is horizontal are \((1, -55.333)\) and \((11, 0)\).

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation of degree two. It typically takes the form \( ax^2 + bx + c = 0 \). Here, the coefficients \(a\), \(b\), and \(c\) are constants with \(a eq 0\). The quadratic equation can help us find the x-values where certain conditions are met in a function, like where the slope of a tangent line is zero, indicating a horizontal line.
To solve a quadratic equation like \(-x^2 + 12x - 11 = 0\), we use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
This formula requires a few steps:

• Calculate the discriminant \(b^2 - 4ac\). If it's positive, there are two real solutions; if zero, one solution; if negative, no real solutions.
• Insert \(a\), \(b\), and \(c\) into the formula to solve for \(x\).

By solving the quadratic equation in this exercise, we find specific x-values where the tangent is horizontal.

Derivative Calculation

To determine tangent lines on a graph, especially those that are horizontal, we use derivatives. The derivative of a function represents the slope of the tangent line at any point along the function.
For our function, \( y = -\frac{1}{3} x^{3} + 6 x^{2} - 11 x - 50 \), we differentiate term by term:

• The derivative of \(-\frac{1}{3}x^3\) is \(-x^2\).
• The derivative of \(6x^2\) is \(12x\).
• The derivative of \(-11x\) is \(-11\).
• Constants, like \(-50\), disappear when differentiated.

The final derivative is \(\frac{dy}{dx} = -x^2 + 12x - 11\). Setting this equal to zero allows us to find where the tangent to the curve is horizontal.

Graph Analysis

Graph analysis involves examining the features and behaviors of a graph to understand the function it represents. For horizontal tangents, we look at where the graph "flattens out."
These flatten out points align with where our derivative equals zero. By analyzing the quadratic derived from the derivative, \(-x^2 + 12x - 11 = 0\), we determine the x-values where the tangent line to the graph is horizontal.
When graphed, the function \( y = -\frac{1}{3} x^{3} + 6 x^{2} - 11 x - 50 \) would show horizontal tangents at specific x-values. This tells us the points where the slope of the function is neither going up or down.
Knowing these can help identify maximum or minimum points or turning points on the graph.

Function Evaluation

Function evaluation is a technique to find a specific y-value for given x-values. Once we've identified x-values for horizontal tangents, we substitute them back into the original function to find their corresponding y-values.
For example, after finding \(x = 1\) and \(x = 11\) as solutions from the quadratic equation, we evaluate these by substituting them into the original equation:

• \(y(1) = -\frac{1}{3}(1)^3 + 6(1)^2 - 11(1) - 50\)
• \(y(11) = -\frac{1}{3}(11)^3 + 6(11)^2 - 11(11) - 50\)

Calculating these gives the points \((1, -55.333)\) and \((11, 0)\). These are the precise points on the graph where tangent lines are horizontal.