Problem 25 - Algebra for College Students

Question

Graph each quadratic function, and state its domain and range. $$y=-\frac{1}{3} x^{2}+5$$

Step-by-Step Solution

Answer

Domain: \( (-\infty, \infty) \); Range: \( (-\infty, 5] \)

Step 1: Identify the quadratic function form

The given quadratic function is in the standard form: \[ y = ax^2 + bx + c \] Here, it is: \[ y = -\frac{1}{3}x^2 + 5 \]. Notice that \(a = -\frac{1}{3}\), \(b = 0\), and \(c = 5\).

Step 2: Determine the vertex

Since the function is in the form \(y = ax^2 + c\) (no linear term), the vertex form simplifies to \((0, c)\). The vertex is thus at \((0, 5)\).

Step 3: Find the direction of the parabola

The sign of \(a\) determines the direction of the parabola. Since \(a = -\frac{1}{3}\), which is negative, the parabola opens downwards.

Step 4: Plot the vertex and axis of symmetry

Plot the vertex \((0, 5)\). The axis of symmetry is the vertical line \(x = 0\).

Step 5: Determine additional points

Choose values of \(x\) to find additional points: For \(x = 3\), \(y = -\frac{1}{3}(3)^2 + 5 = -3 + 5 = 2\). Point: (3, 2). For \(x = -3\), \(y = -\frac{1}{3}(-3)^2 + 5 = -3 + 5 = 2\). Point: (-3, 2). These points help define the shape of the parabola.

Step 6: Sketch the parabola

Plot the additional points and sketch the parabola using the vertex \((0, 5)\) and points \((3, 2)\) and \((-3, 2)\).

Step 7: State the domain and range

For any quadratic function, the domain includes all real numbers: \( \text{Domain} = (-\infty, \infty) \). Given that the parabola opens downwards with its vertex at (0,5), the range includes all values of \(y\) less than or equal to 5: \( \text{Range} = (-\infty, 5] \).

Key Concepts

Parabola

A parabola is a U-shaped curve that can open upwards or downwards. Parabolas are the graphs of quadratic functions, meaning they follow the equation \(y = ax^2 + bx + c\).
The direction in which a parabola opens depends on the coefficient \(a\) in front of the \(x^2\) term.
If \(a\) is positive, the parabola opens upwards. If \(a\) is negative, the parabola opens downwards.
In our specific function, \(y = -\frac{1}{3}x^2 + 5\), \(a = -\frac{1}{3}\), which is negative. Therefore, this parabola opens downwards.
To plot a parabola, you need to identify several key points: the vertex, the axis of symmetry, and additional points on either side of the vertex.

Vertex

The vertex of a parabola is its highest or lowest point, depending on whether it opens downwards or upwards, respectively. It's a crucial point because it represents the maximum or minimum value of the quadratic function.
For the quadratic in standard form, \(y = ax^2 + bx + c\), the vertex can be determined using the formula for the x-coordinate of the vertex:\[ x = -\frac{b}{2a} \]
However, since our given function is \(y = -\frac{1}{3}x^2 + 5\) (with \(b = 0\)), this formula simplifies, and the vertex is simply at (0, c).
In our case, the vertex is at (0, 5), representing the highest point of our downward-opening parabola.

Domain and Range

The domain and range of a quadratic function describe the possible values of \(x\) and \(y\), respectively.

• The domain refers to all the possible x-values. Quadratic functions are defined for all real numbers, so the domain is always \( -\infty < x < \infty \).
• The range represents all possible y-values. For a parabola that opens downwards, the y-values will be less than or equal to the y-coordinate of the vertex.

For the function \(y = -\frac{1}{3}x^2 + 5\), the highest y-value is at the vertex (0, 5), and it decreases as you move away from the vertex. Therefore, the range is \( -\infty < y \leq 5 \).

Standard Form

A quadratic function in standard form is written as \(y = ax^2 + bx + c\). This form makes it easy to identify the coefficients \(a\), \(b\), and \(c\), which in turn help us determine the direction of the parabola and the vertex.

• \(a\): Determines the direction of the parabola. If \(a\) is positive, it opens upwards. If \(a\) is negative, it opens downwards.
• \(b\): Affects the location of the vertex along the x-axis.
• \(c\): Corresponds to the y-coordinate of the vertex when \(b = 0\), and shows where the parabola intersects the y-axis.

In our example, \(y = -\frac{1}{3}x^2 + 5\):

• \(a = -\frac{1}{3}\) (downwards-opening parabola)
• \(b = 0\) (vertex at the y-axis)
• \(c = 5\) (vertex at y = 5)

Axis of Symmetry

The axis of symmetry of a parabola is a vertical line that divides the parabola into two mirror-image halves. This line always passes through the vertex.

• For a parabola in standard form \(y = ax^2 + bx + c\), the equation of the axis of symmetry is given by \(x = -\frac{b}{2a}\).

For our specific function \(y = -\frac{1}{3}x^2 + 5\), since \(b = 0\), the axis of symmetry simplifies to \(x = 0\).
This means the parabola is symmetrical about the y-axis.
Plotting the axis of symmetry helps in accurately sketching the parabola, ensuring that points on the parabola are evenly distributed on both sides of this line.