Problem 51 - Algebra and Trigonometry

Question

\(51-58\) Graph the polynomial in the given viewing rectangle. Find the coordinates of all local extrema. State each answer rounded to two decimal places. $$ y=-x^{2}+8 x, \quad[-4,12] \text { by }[-50,30] $$

Step-by-Step Solution

Answer

The local extremum is a maximum at (4, 16).

Step 1: Understand the Polynomial

We need to graph the polynomial \( y = -x^{2} + 8x \) and find its local extrema. The polynomial is a quadratic, which is a parabola that opens downwards because the coefficient of \( x^2 \) is negative.

Step 2: Determine the Vertex

For a quadratic equation \( y = ax^2 + bx + c \), the vertex \( x \)-coordinate can be found using \( x = -\frac{b}{2a} \). Here, \( a = -1 \) and \( b = 8 \). Thus, \( x = -\frac{8}{2(-1)} = 4 \). Substituting back, \( y = -(4)^2 + 8(4) = 16 \). So, the vertex or a local extremum is \( (4, 16) \).

Step 3: Set Viewing Rectangle

The viewing rectangle is specified as \([-4, 12]\) for the \( x \)-axis and \([-50, 30]\) for the \( y \)-axis. Make sure that the vertex \( (4, 16) \) is within this range, and it is.

Step 4: Confirm the Graph's Shape

Since the parabola opens downward, the vertex \( (4, 16) \) is a maximum point. Check endpoints if needed within \([-4, 12]\), but since it is a parabola, the endpoints (as it is downward opening) will be lower than the vertex.

Step 5: Identify Local Extrema

From Step 2, the local extremum is a single maximum point at the vertex \( (4, 16) \).

Step 6: Verify Calculations

Verify by calculating the value of \( y = -x^2 + 8x \) at some key points like \( x = -4, x = 0, x = 12 \). Ensure these values are within the viewing rectangle:- At \( x = -4 \), \( y = -(-4)^2 + 8(-4) = -48 \).- At \( x = 0 \), \( y = 0 \).- At \( x = 12 \), \( y = -12^2 + 8(12) = -48 \).These confirm the parabola shape within the viewing range.

Key Concepts

Parabola

A parabola is a u-shaped curve, which is the graph of a quadratic function. In the function form \(y = ax^2 + bx + c\), the parabola's orientation depends largely on the coefficient \(a\):

• If \(a > 0\), the parabola opens upwards, creating a minimum point at its vertex.
• If \(a < 0\), it opens downwards, forming a maximum point.

The graph of a parabola is symmetrical around its vertex, meaning each side mirrors the other. This trait is useful for predicting the graph's shape and behavior when analyzing quadratic functions. Parabolas frequently appear in algebra and calculus due to their straightforward properties and predictable shape.

Local Extrema

Local extrema refer to the points in a graph where it reaches either a local maximum or a local minimum. These are critical points that indicate where a function's graph changes direction:

• A local maximum is the highest point in a specific section of the graph.
• A local minimum is the lowest point in a particular part.

In quadratic functions, the local extremum aligns with the vertex. For a downward-opening parabola, like in our example, this extremum is a local maximum. It's vital to identify local extrema as they show points of slope 'zero', depicting the function's behavior not only at that point but also around it.

Vertex

The vertex of a parabola is a critical point, representing either the peak (maximum) or trough (minimum) of the parabola. To find the vertex in a quadratic equation \(y = ax^2 + bx + c\):1. Use the formula \(x = -\frac{b}{2a}\) to find the x-coordinate of the vertex.2. Substitute the x-coordinate back into the equation to find the corresponding y-coordinate.
For instance, in our function \(y = -x^2 + 8x\), the vertex is computed as follows:

• Calculate \(x = -\frac{8}{2(-1)} = 4\)
• Then, \(y = -(4)^2 + 8(4) = 16\)

So, the vertex at \((4, 16)\) is a point of maximum height, framing the parabola's striking feature in the graph.

Graphing Polynomials

Graphing polynomials like quadratic functions can be a straightforward process once you understand the components and transformations:

• Identify the polynomial's degree: Quadratic functions are second-degree polynomials, featuring a characteristic parabola.
• Calculate key points, such as the vertex and intercepts, to outline the curve's skeleton.
• Determine the direction of opening by checking the coefficient of the squared term.

When plotting these functions, ensure the graph is symmetric about the vertex and matches the general direction determined by \(a\). The graph of \(y = -x^2 + 8x\) should be plotted carefully within the given viewing rectangle, from \([-4, 12]\) horizontally and \([-50, 30]\) vertically to visualize the function's full scope. Make sure the plotted function fits entirely within this range to accurately depict its behavior.