Problem 61 - University Calculus: Early Transcendentals

Question

Find all points \((x, y)\) on the graph of \(f(x)=3 x^{2}-4 x\) with tangent lines parallel to the line \(y=8 x+5\).

Step-by-Step Solution

Answer

The point is (2, 4).

Step 1: Understand the Problem

We need to find points on the curve defined by a quadratic function \( f(x) = 3x^2 - 4x \) where the tangent line is parallel to another line given by the equation \( y = 8x + 5 \). Since parallel lines have the same slope, we need to find where the derivative of \( f(x) \) is equal to 8.

Step 2: Calculate the Derivative

Start by differentiating the function \( f(x) = 3x^2 - 4x \). Using the power rule, the derivative is \( f'(x) = 6x - 4 \). This derivative represents the slope of the tangent line at any point \( x \) on the curve.

Step 3: Set the Derivative Equal to Given Slope

Set the derivative \( f'(x) = 6x - 4 \) equal to 8, the slope of the given line, to find the points where the tangent is parallel. Solve the equation \( 6x - 4 = 8 \).

Step 4: Solve for x

Add 4 to both sides to get \( 6x = 12 \). Then, divide both sides by 6 to solve for \( x \). This gives \( x = 2 \).

Step 5: Find Corresponding y-Values

Substitute \( x = 2 \) back into the original function to find the y-coordinate. \( f(2) = 3(2)^2 - 4(2) = 12 - 8 = 4 \).

Step 6: Write Down the Solution

The point on the graph of \( f(x) = 3x^2 - 4x \) where the tangent line is parallel to \( y = 8x + 5 \) is \((2, 4)\).

Key Concepts

derivative of quadratic function

When dealing with any quadratic function, the derivative is essentially the tool you'll use to determine the slope of the tangent line at any given point on the function's curve. In the case of a standard quadratic function like \( f(x) = ax^2 + bx + c \), the derivative indicates how steep or flat the curve is at each point. Calculating the derivative is straightforward thanks to the power rule:

• For a term \( ax^n \), its derivative is \( nax^{n-1} \).
• The derivative of a constant, like \( c \), is zero.

Applying this to \( f(x) = 3x^2 - 4x \), we differentiate each term:

• \( (3x^2)' = 6x \) using \( n=2 \).
• \( (-4x)' = -4 \) using \( n=1 \).

The combined result is \( f'(x) = 6x - 4 \), which tells us the slope of the tangent line for any \( x \) on our graph. This derivative becomes incredibly useful for identifying keystone properties of the graph, such as where the lines have specific slopes.

tangent line parallelism

To understand how tangent line parallelism works, one must first recognize that two lines are parallel if they have the same slope. Therefore, in any problem involving finding points of parallel tangent lines, the slope of the tangent needs to match the slope of the other line.
In our example, you have a line \( y = 8x + 5 \) with a slope of 8. The task is to find locations on \( f(x) = 3x^2 - 4x \) where the tangents hold this same slope value.
The job of the derivative \( f'(x) = 6x-4 \), as explained, is to tell us about slopes. To find where and when the \( f(x) \) curve develops a slope of 8, you set \( 6x - 4 \) equal to 8, interpreting it as a direct requirement for parallelism. Solving \( 6x - 4 = 8 \), leads you to \( x = 2 \). This approach demonstrates the practical application of calculus to comparative slope analysis between a curve and another linear equation.

slope calculation

In calculus, calculating slopes is a central theme, especially those of tangent lines to curves like parabolas. Consider a curve expressed by \( f(x) = 3x^2 - 4x \). The rate of change or slope at any point along the curve is denoted by the derivative \( f'(x) = 6x - 4 \).
When calculating slope for a specific tangent line required to match a slope, such as the slope of 8 from a parallel line, the task becomes solving an equation derived from the derivative. Thus, we derive the need to solve \( 6x - 4 = 8 \).
Solving it involves:

• Adding 4 to both sides yields \( 6x = 12 \).
• Dividing by 6 gives \( x = 2 \).

Once \( x = 2 \) is established, the slope confirms the match with the given slope in the problem, guiding us to the quadratic function values needed, showing how derivative application connects directly to slope alignment solutions in calculus.