Problem 48 - College Algebra

Question

For the following exercises, use the table of values that represent points on the graph of a quadratic function. By determining the vertex and axis of symmetry, find the general form of the equation of the quadratic function. \(\begin{array}{|c|c|c|c|c|c|}\hline x & {-2} & {-1} & {0} & {1} & {2} \\\ \hline y & {-2} & {1} & {2} & {1} & {-2} \\ \hline\end{array}\)

Step-by-Step Solution

Answer

The quadratic function is \(y = -x^2 + 2\).

Step 1: Identify Symmetric Points

Examine the given table to find pairs of points that are symmetric to each other. These points are useful for identifying the axis of symmetry. For example, \((-2, -2)\) and \((2, -2)\) are symmetric, as are \((-1, 1)\) and \((1, 1)\).

Step 2: Determine Axis of Symmetry

The axis of symmetry lies exactly in the middle of symmetric pairs. Between the points \(x = -2\) and \(x = 2\), the midpoint is \(x = 0\). Similarly, between \(x = -1\) and \(x = 1\), the axis is still \(x = 0\). Thus, the axis of symmetry is \(x = 0\).

Step 3: Locate the Vertex

Since the axis of symmetry is \(x = 0\), substitute \(x = 0\) into the table and find \(y = 2\), which gives the vertex \((0, 2)\).

Step 4: Write the Vertex Form Equation

The vertex form of a quadratic equation is \(y = a(x-h)^2 + k\), where \((h, k)\) is the vertex. Substituting \((0, 2)\), we have \(y = a(x-0)^2 + 2 = ax^2 + 2\).

Step 5: Determine the Coefficient 'a'

To find \(a\), use one of the other points from the table. Substituting \((1, 1)\) into the vertex form equation: \(1 = a(1)^2 + 2\). Solving this, we have \(1 = a + 2\), so \(a = -1\).

Step 6: Write the General Form Equation

Substitute \(a = -1\) into our vertex form equation: \(y = -x^2 + 2\), which is the general form of the quadratic equation.

Key Concepts

Vertex Form Equation

The vertex form of a quadratic equation is an incredibly useful way to express the equation of a parabola. It is written as \(y = a(x-h)^2 + k\), where \((h, k)\) represents the vertex of the parabola. This form is particularly handy because it directly reveals the vertex, which is the highest or lowest point on the graph.

• The coefficient \(a\) determines the direction and the width of the parabola. If \(a\) is positive, the parabola opens upward, and if \(a\) is negative, it opens downward.
• The vertex \((h, k)\) shifts the parabola along the coordinate plane.

For example, in the given problem, we found that the equation \(y = ax^2 + 2\) provided the vertex form with \(h = 0\) and \(k = 2\). The task was then to find \(a\) using other points on the graph to transition to the general form.

Axis of Symmetry

The axis of symmetry is a crucial part of understanding the structure of a parabola. It is a vertical line that runs through the vertex of the parabola, effectively dividing it into two mirror-image halves.
To find the axis of symmetry, look for points that mirror each other horizontally. In the exercise, the points \((-2, -2)\) and \((2, -2)\), as well as \((-1, 1)\) and \((1, 1)\), were symmetric. This symmetry led us to determine that the axis of symmetry is located at \(x = 0\).

• Once you have found the axis of symmetry, you can easily calculate the x-coordinate of the vertex, as it lies precisely on this line.
• The equation for the axis of symmetry in this case can be given simply as \(x = 0\).

Vertex of a Quadratic Function

The vertex of a quadratic function is the point that marks the peak or the trough of a parabola. It is the most significant point, since it indicates either the maximum or minimum value the quadratic function can attain.
In our example, after determining the axis of symmetry, we used it to find the vertex at \((0, 2)\). This was done by substituting the \(x\)-coordinate of the axis into the table of values.

• The vertex tells us whether the parabola opens upwards or downwards, depending on whether it is a maximum or minimum point.
• Knowing the vertex allows for a swift write-up of the vertex form equation, placing \((h, k)\) directly into the formula \(y = a(x - h)^2 + k\).

General Form of Quadratic Equation

The general form of a quadratic equation, \(ax^2 + bx + c\), is the standard version of the equation we often encounter. Though it doesn't immediately reveal the vertex, it fully defines the parabola by combining all its essential features.
In the context of our exercise, we used the vertex form \(y = ax^2 + 2\) and modified it to identify \(a = -1\) by substituting another point from the table into the equation. Reference points such as \((1, 1)\) helped us solve for \(a\) and complete the conversion to the general form which resulted in \(y = -x^2 + 2\).

• The general form displays the coefficients clearly, important for understanding transformations and intersections with the y-axis.
• Transforming from vertex to general form requires identifying coefficients through substitution and solving steps.