Problem 47 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

The local maximum occurs at (4.00, 16.00).

Step 1: Identifying the Important Features of the Quadratic Function

The given function is a quadratic function of the form \( y = -x^2 + 8x \). Since the coefficient of \( x^2 \) is negative, the parabola opens downwards. We need the vertex to find the maximum point, as it provides the necessary information about the local extremum.

Step 2: Finding the Vertex of the Parabola

The vertex form of a parabola \( y = ax^2 + bx + c \) gives the x-coordinate of the vertex as \( x = -\frac{b}{2a} \). For our function, \( a = -1 \) and \( b = 8 \), therefore \( x = -\frac{8}{2(-1)} = 4 \). Substituting \( x = 4 \) back in the equation to find \( y \), we get \( y = -(4)^2 + 8(4) = 16 \). Thus, the vertex is at \((4, 16)\).

Step 3: Determining the Local Extrema

Since the quadratic function opens downwards, the vertex represents the maximum point of the parabola. Therefore, the coordinates of the local extremum for this function are the vertex coordinates \((4, 16)\).

Step 4: Graphing the Polynomial

To sketch the graph, plot the vertex at \((4, 16)\) and note the direction of opening (downwards). Calculate additional points, if necessary, for accuracy: for example, at \(x = 0\), \(y = 0\) and at \(x = 8\), \(y = 0\). The viewing rectangle \([-4, 12]\) for \(x\) and \([-50, 30]\) for \(y\) confirms that all critical graph features fit into these boundaries.

Key Concepts

Local Extrema

When identifying local extrema in a quadratic function like the one given, we are essentially looking for the highest or lowest point on the graph. In our function, \( y = -x^2 + 8x \), local extrema refer to the vertex of the parabola because it opens downwards due to the negative coefficient of \( x^2 \). This means there's a maximum point, but no minimum as the arms of the parabola extend infinitely downward.

To determine the local extrema, we compute the vertex's coordinates, calculated in the previous steps, which are \((4, 16)\). This point represents the highest point (maximum) on the parabola. Since the vertex coordinates give us the local maximum value, it concludes the search for local extrema in this downward-opening parabola.

In more complex situations, local extrema analysis might involve checking for both maximums and minimums, but for quadratic functions in standard form, the vertex typically represents one local extremum, either maximum or minimum.

Vertex of a Parabola

The vertex of a parabola is crucial in understanding its shape and orientation. For any quadratic function in the standard form \( y = ax^2 + bx + c \), the vertex can be found using the vertex formula, \( x = -\frac{b}{2a} \). In the exercise, the quadratic function given is \( y = -x^2 + 8x \), where \( a = -1 \) and \( b = 8 \).

By applying the formula, we find the x-coordinate of the vertex:

• \( x = -\frac{8}{2(-1)} = 4 \)

This x-value indicates the vertical line of symmetry for the parabola. Substituting back into the equation gives the y-coordinate:

• \( y = -(4)^2 + 8(4) = 16 \)

Thus, the vertex is at \((4, 16)\).

Understanding the vertex's location helps in drawing the graph accurately - it provides the highest point (or lowest, for upward-opening parabolas) and divides the parabola symmetrically.

Quadratic Function Analysis

Quadratic function analysis involves examining the properties and features of a quadratic equation to understand its graph and key characteristics. For the given function \( y = -x^2 + 8x \), the negative sign before the \( x^2 \) term indicates that the parabola opens downward. This suggests that the graph will have a definite maximum point, which coincides with its vertex.

To plot this parabola accurately, consider the following steps:

• First, calculate the vertex, which we already know is \((4, 16)\).
• Next, identify the x-intercepts by setting \( y = 0 \): \( 0 = -x^2 + 8x \), factoring gives \( x(x - 8) = 0 \), hence \( x = 0 \) and \( x = 8 \).

This tells us where the graph crosses the x-axis.

The y-intercept is found by setting \( x = 0 \), confirming \( y = 0 \). These intercepts, combined with the vertex, allow us to sketch the parabola accurately within the viewing rectangle given by the problem, \([-4, 12]\) by \([-50, 30]\). By ensuring the graph fits within these limits, we get better clarity on the parabola's behavior. Through this analysis, one quickly identifies key features like direction, symmetry, and intercepts of quadratic functions.