Problem 68 - 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. $$ y=-0.01 x^{2}-0.5 x+70 $$

Step-by-Step Solution

Answer

The tangent line is horizontal at the point (-25, 76.25).

Step 1: Understand What a Horizontal Tangent Line Means

A horizontal tangent line to a graph occurs where the derivative of the function equals zero. This is because the slope of a horizontal line is zero.

Step 2: Find the Derivative of the Function

The given function is \( y = -0.01x^2 - 0.5x + 70 \). To find the derivative, apply the power rule: \( y' = \frac{d}{dx}(-0.01x^2) + \frac{d}{dx}(-0.5x) + \frac{d}{dx}(70) \). The derivative is: \( y' = -0.02x - 0.5 \).

Step 3: Set the Derivative Equal to Zero

We set the derivative equal to zero and solve for \(x\): \( -0.02x - 0.5 = 0 \).

Step 4: Solve the Equation for x

Add 0.5 to both sides:\( -0.02x = 0.5 \).Now, divide both sides by -0.02:\( x = -\frac{0.5}{0.02} \).Simplify to find \( x = -25 \).

Step 5: Find the y-coordinate Corresponding to x

Plug \( x = -25 \) back into the original function \( y = -0.01x^2 - 0.5x + 70 \) to find \( y \):\( y = -0.01(-25)^2 - 0.5(-25) + 70 \).Calculate \( y = -0.01(625) + 12.5 + 70 \).\( y = -6.25 + 12.5 + 70 \).\( y = 76.25 \).

Step 6: State the Point

The point on the graph where the tangent line is horizontal is \( (-25, 76.25) \).

Key Concepts

Derivative Calculation

Calculating the derivative of a function is a key step in finding points where tangent lines are horizontal. Derivatives represent the slope of a function at any given point. For a tangent line to be horizontal, the slope, or derivative, must be zero. This means we're looking for where the rate of change of the function equals zero. To find the derivative of any function, we apply certain rules. One of the most common rules used in calculus is the power rule, which we will discuss later.
In this exercise, the goal was to find the derivative and then set it equal to zero to identify the x-values at which the function has a horizontal tangent line. Once we've identified these x-values, we can substitute them back into the original function to find the corresponding y-values and determine the points on the graph where horizontal tangents occur.

Quadratic Function

A quadratic function is a type of polynomial that involves terms up to the second power (squared terms). It generally takes the form \( ax^2 + bx + c \), where \( a\), \( b\), and \( c \) are constants. Quadratic functions create a parabolic shape when graphed.
In the given exercise, the specific quadratic function is "\( y = -0.01x^2 - 0.5x + 70 \)." Due to the presence of \( x^2\), it signals a quadratic function that opens upwards or downwards depending on the sign of the coefficient of \( x^2 \). Since the coefficient here is negative (\( -0.01 \)), our parabola opens downward. This particular orientation impacts where the horizontal tangent points can occur, since they will be on the curve’s maximum point.

Power Rule

The power rule for differentiation is one of the simplest yet powerful rules in calculus. It states that for any function \( y = x^n \), the derivative \( y' \) is \( nx^{n-1} \).
In this exercise, applying the power rule helps you find the derivative of \(-0.01x^2 - 0.5x + 70\). Here, the derivative of \(-0.01x^2\) using the power rule is \(-0.02x\), because we multiply by the exponent \( 2 \) and then reduce \( x^{2-1} \) to \( x^1 \).

• The derivative of \(-0.5x \) is simply \(-0.5\), as \( x^1 \) becomes \( x^0 \) (which is 1).
• Finally, the derivative of a constant, such as \( 70 \), is \( 0 \).

Combining these, the overall derivative function becomes \( y' = -0.02x - 0.5 \). By knowing this rule, you efficiently determine the slopes needed for further analysis.

Solving Equations

Solving equations is a fundamental aspect of mathematics and critical in calculus when finding where tangent lines are horizontal. Once the derivative is set to zero, an equation is formed that needs to be solved for the variable \( x \).
From our exercise, after finding the derivative \( y' = -0.02x - 0.5 \), we set it equal to zero to find critical points. So, you solve \( -0.02x - 0.5 = 0 \), isolating \( x \) by firstly adding \( 0.5 \) to both sides to get \(-0.02x = 0.5\).

• Next, divide both sides by \( -0.02 \), resulting in \( x = -\frac{0.5}{0.02} \).
• Calculating this gives \( x = -25 \).

Once \( x \) is found, substitution back into the original function provides the \( y \)-value, leading to the horizontal tangent point of \((-25, 76.25)\). Each step is essential to confirm precise solutions when dealing with more complex derivatives.