Problem 13 - Algebra 2

Question

Graph each equation. $$ y=\frac{1}{3}(x+6)^{2}+3 $$

Step-by-Step Solution

Answer

The graph is a parabola with a vertex at (-6, 3), opening upwards and wider than \(y=x^2\).

Step 1: Identify the Type of Equation

The equation is a quadratic equation in the form of a parabola, expressed as \(y = \frac{1}{3}(x+6)^2 + 3\). This is a vertex form of a quadratic equation, where \(a = \frac{1}{3}\), \(h = -6\), and \(k = 3\).

Step 2: Determine the Vertex

The vertex of the parabola can be directly extracted from the equation \(y = a(x-h)^2 + k\). For this equation, the vertex is \((-6, 3)\).

Step 3: Determine the Direction of Opening and Width

The parabola opens upwards because the coefficient \(a = \frac{1}{3}\) is positive. Since \(\left|a\right| < 1\), the parabola is wider than the standard parabola \(y = x^2\).

Step 4: Plot the Vertex on a Graph

Plot the vertex of the parabola at the point \((-6, 3)\) on the graph paper or coordinate plane.

Step 5: Identify the Axis of Symmetry

The axis of symmetry for this parabola is the vertical line that passes through the vertex. This line is \(x = -6\).

Step 6: Choose Additional Points

To accurately graph the parabola, choose additional x-values around the vertex, such as \(x = -7\) and \(x = -5\). Substitute these x-values into the equation to find corresponding y-values. For example, substituting \(x = -7\) gives \(y = \frac{1}{3}((-7+6)^2) + 3 = 4\). Similarly, substitute \(x = -5\) to find \(y\).

Step 7: Plot the Additional Points and Draw the Parabola

Plot these additional points \((-7, 4)\) and \((-5, 4)\) on the graph along with the vertex \((-6, 3)\). Use the symmetry property to plot more points if needed. Connect these points with a smooth curve to complete the parabola.

Key Concepts

Parabola

A quadratic equation typically manifests as a parabola when graphed on a coordinate plane. Parabolas have a distinctive 'U' shape and can open in either the upward or downward direction.
This orientation depends on the coefficient of the squared term in the equation:

• If the coefficient is positive, the parabola opens upwards.
• If negative, it opens downwards.

The parabola in our exercise is described by the equation: \[y = \frac{1}{3}(x+6)^2 + 3\] The upward opening is confirmed by the positive coefficient, \(\frac{1}{3}\). A noteworthy aspect of a parabola is that it is always symmetrical around a vertical line.

Vertex Form

The vertex form of a quadratic equation provides immediate insight into the properties of the corresponding parabola. It is expressed as:
\[y = a(x-h)^2 + k\] where:

• \(a\) determines the width and direction of the parabola.
• (\(h, k\)) is the vertex of the parabola.

For the equation in our exercise, \[y = \frac{1}{3}(x+6)^2 + 3,\] the vertex is directly given as \((-6, 3)\). This vertex indicates the point at which the parabola changes direction and is typically the highest or lowest point of the curve, depending on its direction of opening. The term \((x+6)^2\) indicates a horizontal shift of 6 units to the left from the origin.

Axis of Symmetry

The axis of symmetry is a crucial feature of a parabola, representing a vertical line that divides the parabola into two mirror-image halves. This line runs directly through the vertex of the parabola.
For any parabola in vertex form: \[y = a(x-h)^2 + k,\] the axis of symmetry is given by the equation \(x = h\). In this exercise, since \(h = -6\), our axis of symmetry is the line \(x = -6\). This means if you were to 'fold' the parabola along this line, both halves would match perfectly. The axis is an essential guide in accurately plotting further points to complete the graph of a parabola.

Graphing Quadratics

Graphing a quadratic function involves several steps to ensure accuracy and precision. First, identify and plot the vertex. This serves as a focal point for the rest of the graphing process.
Next, determine the parabola's direction of opening and width from the coefficient \(a\):

• For \(a = \frac{1}{3}\), the upward opening results in a wider-than-usual curve, compared to \(y = x^2\).

Determine the axis of symmetry, as it aids in verifying the accuracy of additional points. Afterward, choose and substitute 'x' values close to the vertex into the equation to find corresponding 'y' values. Points like \((-7, 4)\) and \((-5, 4)\) can be identified, plotted, and mirrored using the symmetry.
Finally, connect all these points with a smooth curve to complete and visualize the quadratic graph, ensuring each side mirrors the other as per the symmetry line.