Problem 14 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 13-22: (a) Use the quadratic formula to find the zeros of \(f\). (b) Find the maximum or minimum value of \(f(x)\). (c) Sketch the graph of \(f\). $$ f(x)=-x^{2}-6 x $$

Step-by-Step Solution

Answer

Zeros are \( x = 0 \) and \( x = -6 \); maximum value is 9 at \( x = -3 \).

Step 1: Identify The Quadratic Equation

The given function is \( f(x) = -x^2 - 6x \). This is a quadratic function in the form \( ax^2 + bx + c \). Here, \( a = -1 \), \( b = -6 \), and \( c = 0 \).

Step 2: Apply the Quadratic Formula

The quadratic formula for finding the zeros is \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Substitute \( a = -1 \), \( b = -6 \), \( c = 0 \) into the formula: \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(0)}}{2(-1)} \] Simplifying, this becomes \[ x = \frac{6 \pm \sqrt{36}}{-2} \] which further simplifies to \[ x = \frac{6 \pm 6}{-2} \].

Step 3: Calculate the Zeros

From \( x = \frac{6 + 6}{-2} \) and \( x = \frac{6 - 6}{-2} \), we get the zeros: \( x = 0 \) and \( x = -6 \).

Step 4: Determine Maximum Or Minimum Value

A quadratic function in the form \( ax^2 + bx + c \) has its vertex at \( x = -\frac{b}{2a} \). Substitute \( b = -6 \) and \( a = -1 \) into the vertex formula to get \[ x = -\frac{-6}{2(-1)} = 3 \]. Evaluate \( f(x) \) at \( x = 3 \): \[ f(-3) = -(-3)^2 - 6(-3) = -9 + 18 = 9 \]. This is a maximum value since \( a < 0 \).

Step 5: Sketch The Graph Of f(x)

To sketch the graph, note the following: The parabola opens downward (as \( a = -1 < 0 \)), its vertex is at (-3, 9), and it crosses the x-axis at \( x = 0 \) and \( x = -6 \). Draw a downward-opening curve through these points.

Key Concepts

Quadratic Formula

The quadratic formula is a powerful tool used to find the zeros, or roots, of a quadratic equation of the form \(ax^2 + bx + c = 0\). The formula is given by:

• \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula allows you to solve any quadratic equation by substituting the values of \(a\), \(b\), and \(c\) into it. In the context of our exercise, function \(f(x) = -x^2 - 6x\) corresponds to \(a = -1\), \(b = -6\), and \(c = 0\). Plugging these values into the formula, we find:

• \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(-1)(0)}}{2(-1)} \]
• This simplifies to \[ x = \frac{6 \pm 6}{-2} \]

By solving for \(x\), we determine the zeros of the function as \(x = 0\) and \(x = -6\). These zeros are the x-values where the parabola crosses the x-axis.

Zeros of a Function

Zeros of a function are the points where the graph of the function crosses the x-axis. For quadratic functions, these are found by setting the equation to zero and solving for \(x\). In this case, for the quadratic function \(f(x) = -x^2 - 6x\), applying the quadratic formula reveals that the zeros, or roots, are \(x = 0\) and \(x = -6\).These points have important implications:

• They mark the intersections of the graph with the x-axis.
• They help in determining the symmetry and shape of the parabola.

Finding zeros is a fundamental step in analyzing and graphing quadratic functions, as they not only locate the roots but also help in understanding other significant features of the graph.

Vertex of a Parabola

The vertex of a parabola is a key feature that provides important information about the graph. For a quadratic equation in the form \(ax^2 + bx + c\), the x-coordinate of the vertex can be found using the formula:

• \[ x = -\frac{b}{2a} \]

For the function \(f(x) = -x^2 - 6x\), with \(a = -1\) and \(b = -6\), the x-coordinate of the vertex is:

• \[ x = -\frac{-6}{2(-1)} = 3 \]

Substituting this value back into the function gives the y-coordinate:

• \[ f(-3) = -(-3)^2 - 6(-3) = -9 + 18 = 9 \]

Thus, the vertex of the parabola is located at \((-3, 9)\). Since the leading coefficient \(a = -1\) is negative, the parabola opens downward, making this point a maximum. The vertex helps in constructing the graph as it represents the peak point of the parabola.

Graphing Parabolas

Graphing a quadratic function involves plotting the vertex and zeros along with additional points if needed to create a detailed sketch. For the function \(f(x) = -x^2 - 6x\), follow these basic steps:

• Identify the vertex: We've calculated the vertex as \((-3, 9)\).
• Locate the zeros: We found zeros at \(x = 0\) and \(x = -6\). These are the x-intercepts.
• Determine the direction: The parabola opens downward because the leading coefficient \(a = -1\) is negative.

To sketch the graph:

• Plot the vertex at \((-3, 9)\).
• Mark the x-intercepts at \(0\) and \(-6\).
• Draw a smooth curve passing through these points that opens downwards, forming an inverted "U" shape.

Parabolas are symmetric; notice the symmetry about the vertical line \(x = -3\), which passes through the vertex. This symmetry helps in ensuring that the parabola's curve is accurately mirrored on both sides of the vertex.