Problem 7 - The Calculus with Analytic Geometry

Question

Find the slope of the tangent line to the graph at the point \(\left(x_{1}, y_{1}\right) .\) Make a table of values of \(x, y\), and \(m\) at various points on the graph, and include in the table all points where the graph has a horizontal tangent. Draw a sketch of the graph. $$ y=4 x^{3}-13 x^{2}+4 x-3 $$

Step-by-Step Solution

Answer

Derivative: \(12x^{2} - 26x + 4\). Horizontal tangents at \(x = 2\) and \(x = \frac{1}{6}\).

Step 1 - Find the derivative of the function

To find the slope of the tangent line, compute the derivative of the function. Given the function: \[ y = 4x^{3} - 13x^{2} + 4x - 3 \] Differentiate it with respect to x:\[ \frac{dy}{dx} = \frac{d}{dx}(4x^{3}) - \frac{d}{dx}(13x^{2}) + \frac{d}{dx}(4x) - \frac{d}{dx}(-3) \] This results in:\[ \frac{dy}{dx} = 12x^{2} - 26x + 4 \]

Step 2 - Find the slope at the specific point

To find the slope at a specific point \(\left(x_{1}, y_{1}\right)\), substitute the x-coordinate of the point into the derivative. If \(x = x_{1}\), then:\[ m = 12x_{1}^{2} - 26x_{1} + 4 \]

Step 3 - Find points with horizontal tangents

Points where the graph has a horizontal tangent have a slope (derivative) of 0. Set the derivative equal to zero and solve for x:\[ 12x^{2} - 26x + 4 = 0 \] Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\), where \(a = 12\), \(b = -26\), and \(c = 4\):\[ x = \frac{26 \pm \sqrt{(-26)^{2} - 4 \cdot 12 \cdot 4}}{2 \cdot 12} \] \[ x = \frac{26 \pm \sqrt{676 - 192}}{24} \] \[ x = \frac{26 \pm \sqrt{484}}{24} \] \[ x = \frac{26 \pm 22}{24} \] Results in two solutions:\[ x = 2 \] \[ x = \frac{1}{6} \]

Step 4 - Make the table of values

Create a table with columns for x, y, and m. Calculate y and m for selected x-values, including the points found in Step 3 where the derivative is zero.For example:\( x = 2, y = 4(2)^{3} - 13(2)^{2} + 4(2) - 3 \)\( y = 4(8) - 13(4) + 8 - 3 \) \( y = 32 - 52 + 8 - 3 = -15 \)Thus, \( (x, y, m) = (2, -15, 0) \). Repeat for other values.

Step 5 - Draw a sketch of the graph

Plot the calculated points on a graph, including where the tangent is horizontal. Sketch the curve to show the nature of the function. Highlight the points with horizontal tangents.

Key Concepts

derivative of a polynomial function

The first step in finding the slope of a tangent line to a graph is to compute the derivative of the polynomial function. A derivative represents the rate of change of a function with respect to a variable. For the function \[ y = 4x^{3} - 13x^{2} + 4x - 3 \], the derivative is found using the power rule. This rule states that the derivative of \( ax^n \) is \( n \times ax^{n-1} \). So, for each term:

• Derivate of \( 4x^3 \) is \( 3 \times 4x^{2} = 12x^{2} \)
• Derivate of \( -13x^2 \) is \( 2 \times -13x = -26x \)
• Derivate of \( 4x \) is simply \( 4 \)
• Derivate of a constant \( -3 \) is \( 0 \)

Putting these together, we get:\[ \frac{dy}{dx} = 12x^{2} - 26x + 4 \]This new function \( \frac{dy}{dx} \), known as \( y' \), gives the slope of the tangent line to the curve at any point \( x \). This derivative is also called the first derivative of the function.

quadratic formula in calculus

When analyzing functions, we often need to find points where the slope is zero, indicating horizontal tangent lines. To do this, set the first derivative equal to zero and solve for \( x \). For our derivative function \[ 12x^{2} - 26x + 4 = 0 \], we use the quadratic formula to find the roots. The quadratic formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \(a\), \(b\), and \(c\) are coefficients from the quadratic equation. Here, \(a = 12\), \(b = -26\), and \(c = 4\). Plugging in these values, we solve for \( x \) as follows:\[ x = \frac{26 \pm \sqrt{676 - 192}}{24} \]\[ x = \frac{26 \pm \sqrt{484}}{24} \]\[ x = \frac{26 \pm 22}{24} \]This results in two solutions:\[ x = 2 \]\[ x = \frac{1}{6} \]These are the x-values where the tangent to the curve is horizontal.

horizontal tangent line analysis

After finding the x-values where the tangent line is horizontal, we calculate the corresponding \( y \)-values using the original function \[ y = 4x^{3} - 13x^{2} + 4x - 3 \]. For these x-values:

• When \( x = 2 \): \[ y = 4(2)^{3} - 13(2)^{2} + 4(2) - 3 \] \[ y = 32 - 52 + 8 - 3 = -15 \] Thus, the point is \( (2, -15) \)
• When \( x = \frac{1}{6} \): \[ y = 4\left(\frac{1}{6}\right)^{3} - 13\left(\frac{1}{6}\right)^{2} + 4\left(\frac{1}{6}\right) - 3 \] The calculation will yield another point. Let's complete this calculation: \[ y = \frac{4}{216} - \frac{13}{36} + \frac{2}{3} - 3 \] \[ y = \frac{4}{216} - \frac{78}{216} + \frac{144}{216} - \frac{648}{216} \] \[ y = \frac{4 - 78 + 144 - 648}{216} = \frac{-578}{216} \] Simplifying gives \( y \approx -2.68 \), so the point is \( \left( \frac{1}{6}, -2.68 \right) \)

With these points, you can create a table of values and plot them to visualize the graph. Points with a horizontal tangent line are where the derivative is zero, and these are critical for understanding the function's behavior.