Problem 20 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 13-26: Sketch, on the same coordinate plane, the graphs of \(f\) for the given values of \(c\). (Make use of symmetry, shifting, stretching, compressing, or reflecting.) $$ f(x)=-\frac{1}{2}(x-c)^{2} ; \quad c=-2,0,3 $$

Step-by-Step Solution

Answer

Graph three downward parabolas with vertices at (-2,0), (0,0), and (3,0).

Step 1: Identify the base function

The base function is a simple quadratic function given by \( f(x) = -\frac{1}{2}x^2 \). This is a downward-opening parabola centered at the origin (0,0) and vertically compressed by a factor of 0.5.

Step 2: Understand the transformation

The function provided, \( f(x) = -\frac{1}{2}(x-c)^2 \), represents a horizontal shift of the base quadratic function. The term \( c \) indicates a shift of \( c \) units to the right. Thus, the vertex of the parabola changes from (0,0) to (c,0).

Step 3: Sketch for \( c = -2 \)

Substitute \( c = -2 \) into the function: \( f(x) = -\frac{1}{2}(x + 2)^2 \). This shifts the parabola 2 units to the left, with the vertex now at (-2,0). Sketch this parabola.

Step 4: Sketch for \( c = 0 \)

With \( c = 0 \), the function is \( f(x) = -\frac{1}{2}x^2 \). This means no horizontal shift is applied, and the vertex remains at the origin, (0,0). Draw this parabola, ensuring it overlaps the y-axis.

Step 5: Sketch for \( c = 3 \)

Plug \( c = 3 \) into the function: \( f(x) = -\frac{1}{2}(x - 3)^2 \). This shifts the parabola 3 units to the right, moving the vertex to (3,0). Sketch this parabola.

Step 6: Finalize the graph

In the coordinate plane, superimpose all three parabolas. Ensure each parabola is clearly marked with its respective vertex at (-2,0), (0,0), and (3,0), demonstrating the horizontal shifts.

Key Concepts

Parabolas

Parabolas are the U-shaped graphs commonly associated with quadratic functions. In mathematical terms, a basic parabola is defined by the equation \(y = ax^2 + bx + c\). When \(a\) is positive, the parabola opens upward, and when \(a\) is negative, it opens downward. For instance, the function \(f(x) = -\frac{1}{2}x^2\) represents a parabola that opens downward. Parabolas have a symmetric shape, and their highest or lowest point, depending on whether they open up or down, is called the vertex. The axis of symmetry is a vertical line that passes through the vertex, dividing the parabola into two mirror-image halves. This symmetry can be used to easily sketch and transform the graph.

Horizontal Shifts

Horizontal shifts in quadratic functions involve moving the entire graph of the parabola left or right along the x-axis. This shift is determined by the value of \(c\) in the expression \(f(x) = a(x-c)^2\). Let's break it down:

• If \(c\) is positive, the graph shifts \(c\) units to the right.
• If \(c\) is negative, the graph shifts \(\abs{c}\) units to the left.

The vertex of the parabola, originally at the origin (0,0) for the base quadratic function, moves to the new position at \((c,0)\) due to this shift. In the given exercise, when \(c = -2\), the parabola shifts 2 units to the left, situating the vertex at (-2,0). When \(c = 0\), no shift occurs, and with \(c = 3\), the graph moves 3 units to the right.

Graph Transformations

Graph transformations allow us to alter the appearance of a graph in various ways: shifting, stretching, compressing, and reflecting. They are essential for understanding how changes to the function's equation affect its graph.In the given function \(f(x) = -\frac{1}{2}(x-c)^2\), the graph experiences transformations primarily due to the negative sign and the coefficient \(-\frac{1}{2}\). The negative sign reflects the parabola over the x-axis, making it open downwards instead of upwards. The fraction \(\frac{1}{2}\) compresses the graph vertically, causing it to be wider than a standard parabola. Understanding these transformations helps in visualizing the shape and position of the graph before actually sketching it on the coordinate plane.

Vertex Shifts

The vertex of a parabola is a crucial point where the curve changes direction. In the vertex form of a quadratic function, such as \(f(x) = a(x-c)^2 + k\), the coordinates \((c, k)\) dictate the vertex's position. When \(f(x) = -\frac{1}{2}(x-c)^2\), the term \((x-c)^2\) shifts the vertex along the horizontal axis.

• The horizontal change by \(c\) units moves the vertex to \((c,0)\) in our exercise because the constant \(k\) is zero.
• Value of \(c\) affects only the horizontal shift since there is no vertical translation.

In the provided exercises, as \(c\) varies, the vertex's location shifts but stays on the x-axis. For \(c = -2\), \(0\), and \(3\), the vertex points are \((-2,0)\), \((0,0)\), and \((3,0)\), respectively. By focusing on vertex shifts, we can correctly place and sketch each parabola on the graph.