Problem 52 - Calculus Early Transcendentals

Question

Give a graph of the polynomial and label the coordinates of the intercepts, stationary points, and inflection points. Check your work with a graphing utility. $$ p(x)=1+8 x-x^{2} $$

Step-by-Step Solution

Answer

y-intercept at (0, 1); x-intercepts at approx (-0.13, 0) and (8.13, 0); vertex at (4, 17); no inflection point.

Step 1: Identify the Type of Polynomial

The given function is \( p(x) = 1 + 8x - x^2 \), which is a quadratic polynomial because it is of degree 2. Quadratic functions generally have a parabolic shape.

Step 2: Determine the Y-Intercept

The y-intercept is found by evaluating the polynomial at \( x = 0 \). Substitute \( x = 0 \) into the polynomial to obtain: \( p(0) = 1 + 8(0) - (0)^2 = 1 \). Hence, the y-intercept is \( (0, 1) \).

Step 3: Find the X-Intercepts

Solve the equation \( 1 + 8x - x^2 = 0 \) to find the x-intercepts. This can be rewritten as \( -x^2 + 8x + 1 = 0 \). Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = -1 \), \( b = 8 \), and \( c = 1 \), we find the x-intercepts: \( x = \frac{-8 \pm \sqrt{8^2 - 4(-1)(1)}}{2(-1)} \). Simplified, \( x = \frac{-8 \pm \sqrt{64 + 4}}{-2} = \frac{-8 \pm \sqrt{68}}{-2} \). Calculating, \( x \approx -0.13 \) and \( x \approx 8.13 \) are the x-intercepts.

Step 4: Identify the Vertex (Stationary Point)

For a quadratic function in the form \( ax^2 + bx + c \), the x-coordinate of the vertex can be found using \( -\frac{b}{2a} \). Here, \( a = -1 \) and \( b = 8 \), giving \( x = -\frac{8}{2(-1)} = 4 \). Substitute \( x = 4 \) into \( p(x) \) to find the y-coordinate: \( p(4) = 1 + 8(4) - (4)^2 = 17 \). The vertex is at \( (4, 17) \), a maximum point.

Step 5: Check for Inflection Points

A quadratic function does not have inflection points since its second derivative, a constant \( -2 \), does not change sign. Therefore, no inflection point exists for this polynomial.

Step 6: Graph the Polynomial

Plot the function on a graph. The parabola opens downwards (due to the negative coefficient of \( x^2 \)). Label the y-intercept \( (0, 1) \), the x-intercepts approximately at \( (-0.13, 0) \) and \( (8.13, 0) \), and the vertex at \( (4, 17) \).

Step 7: Validate Using a Graphing Utility

Using a graphing calculator or software, input \( p(x) = 1 + 8x - x^2 \) and check that all calculated intercepts and the vertex match the plot shown by the tool. Adjust calculations if discrepancies are found.

Key Concepts

Polynomial Graphing

Understanding polynomial graphing is crucial when dealing with quadratic functions like \( p(x) = 1 + 8x - x^2 \). This particular function represents a quadratic polynomial because it has a degree of 2. The general form of a quadratic equation is \( ax^2 + bx + c \). Quadratic functions usually produce graph shapes known as parabolas. These parabolas can open upwards or downwards, depending upon the sign of the coefficient of \( x^2 \).
In our example, the function has a negative \( x^2 \) coefficient \(-1\), indicating that the graph opens downwards. When graphing, it's essential to identify critical points like the intercepts and vertex to accurately outline the parabola's shape. A useful step is to use graphing calculators or software to validate your work and gain a clearer visual understanding of the behavior of the polynomial.

Intercepts

Intercepts refer to the points where the graph crosses the axes. For the quadratic function \( p(x) = 1 + 8x - x^2 \), you identify intercepts by finding where the graph intersects the x-axis and y-axis.

• Y-Intercept: This is found when \( x = 0 \). Substitute \( x = 0 \) into the polynomial resulting in \( p(0) = 1 \). Therefore, the y-intercept is at \( (0, 1) \).
• X-Intercepts: These occur where \( y = 0 \). For \( 1 + 8x - x^2 = 0 \), solve this equation using the quadratic formula. The x-intercepts are \( x \approx -0.13 \) and \( x \approx 8.13 \). These points are \( (-0.13, 0) \) and \( (8.13, 0) \).

By pinpointing these intercepts, you lay the foundation for sketching the polynomial's graph accurately and understanding its intersection behavior with the axes.

Stationary Points

Stationary points in a quadratic function, like \( p(x) = 1 + 8x - x^2 \), are crucial in graphing as they denote the vertex.

• The x-coordinate of the vertex is calculated using the formula \(-\frac{b}{2a}\). With \( a=-1 \) and \( b=8 \), the calculation yields \( x = 4 \).
• To find the exact vertex point, substitute \( x = 4 \) back into the original polynomial to get \( p(4) = 17 \). Hence, the vertex or stationary point is \( (4, 17) \), a maximum since the parabola opens downwards.

Recognizing the vertex gives insight into the point of maximum height the parabola achieves and is integral in drawing the graph. This is useful in various applications such as optimization problems.

Inflection Points

Inflection points, often discussed in polynomial calculus, are where the curve changes concavity. However, for quadratic functions such as \( p(x) = 1 + 8x - x^2 \), inflection points do not exist.
This lack is because the second derivative of the quadratic function, which determines concavity changes, is a constant value. For this polynomial, the second derivative is \(-2\). Because it doesn’t change sign, there's no shift from concave up to down or vice versa. In more complex functions, inflection points mark the transition but remember, quadratic graphs maintain a consistent concavity.
Understanding that quadratic functions lack inflection points helps focus analysis strictly on the intercepts and stationary points, simplifying both computation and graph analysis.