Problem 47 - Precalculus Mathematics for Calculus

Question

The graph of a polynomial function is given. From the graph, find (a) the \(x\) - and \(y\) -intercepts, and (b) the coordinates of all local extrema. $$P(x)=-x^{2}+4 x$$

Step-by-Step Solution

Answer

x-intercepts: (0,0) and (4,0); y-intercept: (0,0); Local maximum at (2,4).

Step 1: Identify the Function

The polynomial function given is \(P(x)=-x^{2}+4x\). This is a quadratic function and can be expressed in the form \(P(x) = ax^2 + bx + c\) with \(a = -1\), \(b = 4\), and \(c = 0\).

Step 2: Find the x-intercepts

To find the x-intercepts, set \(P(x) = 0\).\[-x^{2} + 4x = 0\]Factor the equation:\[x(-x + 4) = 0\]Thus, \(x = 0\) or \(x = 4\).The x-intercepts are (0,0) and (4,0).

Step 3: Find the y-intercept

The y-intercept is the point where the graph crosses the y-axis. This occurs when \(x = 0\). Substitute \(x = 0\) into the function:\[P(0) = -(0)^2 + 4(0) = 0\]The y-intercept is at (0,0).

Step 4: Calculate the Vertex

The vertex of a parabola in standard form \(P(x) = ax^2 + bx + c\) can be calculated using the formula \(x = -\frac{b}{2a}\).Here, \(a = -1\) and \(b = 4\), so:\[x = -\frac{4}{2(-1)} = 2\]Substitute \(x = 2\) back into the function to find \(y\):\[P(2) = -(2)^2 + 4(2) = -4 + 8 = 4\]The vertex, which is a local maximum for this downward opening parabola, is at (2,4).

Step 5: List Local Extrema

Since the polynomial is a quadratic function opening downward, it has a single local maximum at the vertex. The local maximum occurs at (2,4).

Key Concepts

x-intercepts

The x-intercepts of a function are points where the graph crosses the x-axis. For a quadratic function like \( P(x) = -x^2 + 4x \), we find the x-intercepts by setting the function equal to zero. This is because at the x-intercept, the y-value is zero.
To solve \(-x^2 + 4x = 0\), you can factor out an \( x \):

• \( x(-x + 4) = 0 \)

This gives us two solutions: \( x = 0 \) and \( x = 4 \). So, the x-intercepts are the points (0,0) and (4,0). These points tell us where the parabola touches the x-axis. Identifying x-intercepts is crucial for understanding the roots of the function.

y-intercepts

The y-intercept of a graph is the point where it crosses the y-axis. This occurs when \( x = 0 \). For our quadratic function \( P(x) = -x^2 + 4x \), we can find the y-intercept by substituting \( x = 0 \) into the function.
So,

• \( P(0) = -(0)^2 + 4(0) = 0 \)

Thus, the y-intercept is at the origin (0,0). This means that our parabola touches both the x-axis and y-axis at this single point. Understanding the y-intercept helps when drawing the graph, as it gives a starting point on the y-axis.

local extrema

Local extrema in a function are points where the function reaches a local minimum or maximum. For a quadratic function like \( P(x) = -x^2 + 4x \), the parabola opens downwards since the leading coefficient \( a \) is negative. Therefore, this function has a local maximum.
A short guide to identify local extrema:

• Differentiate the function to find the critical points.
• Use the first or second derivative test to determine if these points are minima or maxima.

In this quadratic, it matches the vertex, which is found by other methods as shown in the next section.

vertex of a parabola

The vertex of a quadratic function is an important feature since it is the highest or lowest point on the graph, known as the parabola's peak. For a standard quadratic function in the form \( ax^2 + bx + c \), the vertex can be calculated using the formula:

• \( x = -\frac{b}{2a} \)

For \( P(x) = -x^2 + 4x \), with \( a = -1 \) and \( b = 4 \), we find:

• \( x = -\frac{4}{2(-1)} = 2 \)

Substitute \( x = 2 \) in the function to find \( y \):

• \( P(2) = -(2)^2 + 4(2) = 4 \)

Thus, the vertex is at (2,4). Since our parabola opens downward, this vertex represents a local maximum. The vertex aids in sketching the parabola accurately and understanding the graph’s behavior.