Problem 28 - Applied Calculus

Question

Graph each function "by hand." [Note: Even if you have a graphing calculator, it is important to be able to sketch simple curves by finding a few important points.] $$ f(x)=3 x^{2}-6 x-9 $$

Step-by-Step Solution

Answer

Graph the parabola with vertex at (1, -12) and passing through (0, -9), (2, -9), and (-1, 0).

Step 1: Identify the Type of Function

The given function is a quadratic function of the form \( f(x) = ax^2 + bx + c \). Here, \( a = 3 \), \( b = -6 \), and \( c = -9 \). This indicates that the graph will be a parabola.

Step 2: Find the Vertex

The vertex of a parabola given by the function \( f(x) = ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). Substituting the given values, we calculate:\[ x = -\frac{-6}{2 \times 3} = 1 \]Next, plug \( x = 1 \) back into the function to find \( y \):\[ f(1) = 3(1)^2 - 6(1) - 9 = 3 - 6 - 9 = -12 \]Thus, the vertex is \( (1, -12) \).

Step 3: Identify the Axis of Symmetry

The axis of symmetry is a vertical line that passes through the vertex of the parabola. For our function, the axis of symmetry is \( x = 1 \).

Step 4: Find the Y-Intercept

The y-intercept occurs when \( x = 0 \). Substitute \( x = 0 \) into the function:\[ f(0) = 3(0)^2 - 6(0) - 9 = -9 \]Thus, the y-intercept is \( (0, -9) \).

Step 5: Find Additional Points

Select additional points on either side of the vertex to help graph the parabola. Let's calculate \( f(2) \) and \( f(-1) \):- \( f(2) = 3(2)^2 - 6(2) - 9 = 12 - 12 - 9 = -9 \) giving the point \( (2, -9) \).- \( f(-1) = 3(-1)^2 - 6(-1) - 9 = 3 + 6 - 9 = 0 \) giving the point \( (-1, 0) \).Graph these points: \((0, -9), (1, -12), (2, -9), (-1, 0)\).

Step 6: Sketch the Parabola

Using the vertex \((1, -12)\), the axis of symmetry \(x = 1\), the y-intercept \((0, -9)\), and the additional points \((2, -9)\) and \((-1, 0)\), sketch the parabolic curve. Ensure the curvature respects the symmetry about the axis and passes through the computed points.

Key Concepts

Parabola Vertex

When graphing a quadratic function like \( f(x) = 3x^2 - 6x - 9 \), the parabola's vertex is an essential feature. The vertex is the point where the parabola changes direction. It provides crucial information about the curve's lowest or highest point, depending on whether the parabola opens upwards or downwards.
To find the vertex of a quadratic function in the standard form \( f(x) = ax^2 + bx + c \), use the formula:

• The x-coordinate is given by \( x = -\frac{b}{2a} \).

In our example, substituting \( b = -6 \) and \( a = 3 \) gives \( x = 1 \). Then, plug \( x = 1 \) back into the function to find the y-coordinate:

• \( f(1) = 3(1)^2 - 6(1) - 9 = -12 \)

Thus, the vertex is at \((1, -12)\). This means the curve has its lowest point here because the parabola opens upwards (since \( a > 0 \)).
Understanding and finding the vertex allows you to sketch the parabola more accurately.

Axis of Symmetry

The axis of symmetry is another important characteristic of a parabola that helps in graphing quadratic functions. It is a vertical line that passes through the vertex, essentially splitting the parabola into two mirror images. This symmetry helps confirm that you've plotted the curve correctly.
For any quadratic function \( f(x) = ax^2 + bx + c \), the axis of symmetry can be found using the same x-coordinate as the vertex:

• This is given by \( x = -\frac{b}{2a} \).

In the given function, the axis of symmetry is \( x = 1 \).
The parabola is symmetric about this line, meaning that points on the parabola to the left of this line will have their counterparts to the right, equidistant from the axis. This property is what makes the parabolic shape so recognizable and consistent.

Y-Intercept

The y-intercept is the point where the graph of the quadratic function crosses the y-axis. This occurs when \( x = 0 \). For graphing, the y-intercept provides a starting point and another fixed location through which the graph must pass.
To find the y-intercept, substitute \( x = 0 \) in the function \( f(x) = ax^2 + bx + c \):

• In our example, calculating \( f(0) \) gives \( f(0) = 3(0)^2 - 6(0) - 9 = -9 \).

Thus, the y-intercept is \((0, -9)\).
This piece of information is particularly useful because it locks down one point on the graph that can be used alongside other key features like the vertex to sketch the curve of the parabola.

Quadratic Functions

Quadratic functions are polynomial functions of degree 2, expressed in the form \( f(x) = ax^2 + bx + c \), where \( a eq 0 \). The graph of a quadratic function is a parabola, which can open upward or downward:

• Upwards if \( a > 0 \), resembling a U-shape.
• Downwards if \( a < 0 \), resembling an upside-down U.

The quadratic function given, \( f(x) = 3x^2 - 6x - 9 \), has a positive \( a \) value of 3, indicating an upward-opening parabola.
Graphing a quadratic function involves locating several critical points, such as the vertex, axis of symmetry, and y-intercept, along with additional points to define the shape.
The beauty of quadratic functions lies in their symmetry and simplicity, yet they help model a wide range of real-world phenomena, from physics and engineering to economics and beyond. Understanding these core features makes sketching and interpreting quadratic graphs much easier.