Problem 28 - Brief 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

The parabola has a vertex at (1, -12), y-intercept (0, -9), and x-intercepts (-1, 0) and (3, 0).

Step 1: Identify the type of function

The given function is \( f(x) = 3x^2 - 6x - 9 \). This is a quadratic function, which means its graph is a parabola. Since the coefficient of \( x^2 \) is positive, the parabola opens upward.

Step 2: Find the vertex

The vertex of a parabola in the form \( ax^2 + bx + c \) can be found using the formula \( x = -\frac{b}{2a} \). For our function, \( a = 3 \) and \( b = -6 \). Thus, \( x = -\frac{-6}{2(3)} = 1 \). Substitute \( x = 1 \) back into the function to find \( y \): \( f(1) = 3(1)^2 - 6(1) - 9 = -12 \). So, the vertex is \( (1, -12) \).

Step 3: Find the y-intercept

The y-intercept is the point where the graph crosses the y-axis, which 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 4: Find the x-intercepts

X-intercepts occur where \( f(x) = 0 \). Solve the equation \( 3x^2 - 6x - 9 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 3 \), \( b = -6 \), \( c = -9 \).Compute the discriminant: \( b^2 - 4ac = (-6)^2 - 4(3)(-9) = 36 + 108 = 144 \).So, \( x = \frac{6 \pm \sqrt{144}}{6} \) results in \( x = \frac{6 \pm 12}{6} \). This gives \( x = 3 \) and \( x = -1 \). The x-intercepts are \( (3, 0) \) and \( (-1, 0) \).

Step 5: Sketch the graph

Plot the points found: the vertex \( (1, -12) \), the y-intercept \( (0, -9) \), and the x-intercepts \( (3, 0) \) and \( (-1, 0) \). Since the parabola opens upwards and the vertex is at the lowest point, draw a U-shaped curve passing through these points. Ensure the curve is symmetric about the vertical line through the vertex.

Key Concepts

Vertex of a Parabola

When dealing with quadratic functions, identifying the vertex is crucial as it provides the minimum or maximum point of the parabola. The vertex can be found using the formula: \[ x = -\frac{b}{2a} \]Here, the values of \( a \) and \( b \) are derived from the quadratic function written in the form \( ax^2 + bx + c \). For our example, with \( f(x) = 3x^2 - 6x - 9 \), we determine that \( a = 3 \) and \( b = -6 \). Plugging these into the formula yields \( x = 1 \).

Once we have the \( x \)-coordinate of the vertex, we find the \( y \)-coordinate by substituting \( x \) back into the function. As we calculated: - \( f(1) = 3(1)^2 - 6(1) - 9 = -12 \)
Thus, the vertex is \( (1, -12) \). This point is the lowest part of the parabola since it opens upwards.

Quadratic Formula

The quadratic formula is a reliable tool for finding the x-intercepts of a quadratic function. It is helpful when factoring is not straightforward. The quadratic formula for solving \( ax^2 + bx + c = 0 \) is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The discriminant \( b^2 - 4ac \) determines the nature of the roots:- If positive, you have two distinct real roots.- If zero, there is one real root.- If negative, the roots are complex and not real.

In our function example \( 3x^2 - 6x - 9 = 0 \), compute: - \( a = 3 \), \( b = -6 \), \( c = -9 \). - Discriminant: \((-6)^2 - 4(3)(-9) = 144 \), which is positive, giving two real roots.
Therefore, \( x = \frac{6 \pm 12}{6} \), results in roots \( x = 3 \) and \( x = -1 \). These are the x-intercepts of the parabola.

Parabola Sketching

Sketching a parabola involves plotting key points and drawing a smooth curve through them. For a comprehensive graph, it's essential to identify:- The vertex, which in our case is \( (1, -12) \).- X-intercepts, where the graph crosses the x-axis, \( (3, 0) \) and \( (-1, 0) \).- Y-intercept, where the graph crosses the y-axis, \( (0, -9) \).

To sketch:1. Plot the vertex, which is the lowest (or highest) point.2. Mark the intercepts on their respective axes.3. Draw the parabola opening upwards due to the positive \( a \)-value (3 in this case).4. Ensure symmetry about the vertical line passing through the vertex (axis of symmetry).
This symmetry helps in accurately sketching the other half of the parabola.

X-intercepts and Y-intercepts

Interpreting the intercepts of a quadratic function is key to understanding its graph. The x-intercepts are points where the function's value is zero:- Determined using the quadratic formula, our function has x-intercepts at \( (3, 0) \) and \( (-1, 0) \).
For the y-intercept:- Evaluate the function at \( x = 0 \), resulting in \( f(0) = -9 \), giving the y-intercept \( (0, -9) \).

These intercepts provide crucial points to plot initially when sketching, as they define where the parabola crosses the axes. Understanding, calculating, and correctly marking these ensure a more precise graph.