Problem 46 - College Algebra with Corequisite Support

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. $$ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & 5 & 2 & 1 & 2 & 5 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The quadratic function is \( y = x^2 + 1 \).

Step 1: Identify the Vertex

To find the vertex, observe the given y-values. The minimum y-value is at \(x = 0\) where \(y = 1\). Therefore, the vertex of the quadratic function is \((0, 1)\).

Step 2: Determine the Axis of Symmetry

The axis of symmetry for a quadratic function \(y = ax^2 + bx + c\) is given by the formula \(x = -\frac{b}{2a}\). Since the vertex is \((0, 1)\), the axis of symmetry is \(x = 0\).

Step 3: Choose the General Form of the Quadratic Equation

The vertex form of a quadratic equation is \(y = a(x - h)^2 + k\), where \((h, k)\) is the vertex. Substituting the vertex \((0, 1)\), we get the equation \(y = a(x - 0)^2 + 1\) or \(y = ax^2 + 1\).

Step 4: Use Additional Points to Determine 'a'

Use another point from the table, for instance \((-1, 2)\), to find \(a\). Substitute \(x = -1\) and \(y = 2\) into the equation: \(2 = a(-1)^2 + 1\). This simplifies to \(2 = a + 1\). Solving for \(a\), we find \(a = 1\).

Step 5: Write the Final Equation

Substitute \(a = 1\) back into the equation from Step 3. The general form of the quadratic function is \(y = x^2 + 1\).

Key Concepts

Vertex of Quadratic Function

The vertex of a quadratic function is a key point on its graph, often either the highest or lowest point. It's where the curve turns around. In other words, it's the point where the function reaches its minimum or maximum value.
To locate the vertex, we should first observe the y-values of the function. These values tell us about the behavior of the graph. For instance, when given a table of values, as in our exercise,

• Find the smallest or largest y-value depending on the nature of the function (depending if it opens upwards or downwards).
• In the given exercise, the table shows that the minimum y-value is 1 at x = 0.

Thus, the vertex of the quadratic function is the point (0, 1). In general, knowing the vertex helps in sketching the graph accurately and understanding the function's range and applications.

Axis of Symmetry

The axis of symmetry is a vertical line that divides the parabola into two mirror images. This line passes through the vertex of the quadratic function. The graph of a parabola is always symmetric relative to this vertical line.
To determine the axis of symmetry for a quadratic function, you can use the formula for one in the standard form:

• Use the formula: \( x = -\frac{b}{2a} \) for the equation \( y = ax^2 + bx + c \), which represents the axis.
• In our exercise, we found that the vertex is \((0, 1)\).
• Since the vertex x-coordinate is 0, the axis of symmetry is simply \( x = 0 \).

Recognizing the axis of symmetry is especially useful as it helps in graphing and understanding the behavior and properties of the quadratic function.

General Form of Quadratic Equation

The general form of a quadratic equation is the expression \( y = ax^2 + bx + c \). In this form,

• \(a\), \(b\), and \(c\) are constants, where \(a eq 0\).
• \(a\) determines the direction of the parabola (upwards if \(a > 0\) and downwards if \(a < 0\)).
• \(c\) provides the y-intercept, the point where the graph intersects the y-axis.

In the exercise, after identifying the vertex, we derived the equation in vertex form \( y = a(x - 0)^2 + 1 \) and found the value of \(a\) using additional points from the table. This gave us the general form \( y = x^2 + 1 \), confirming that the function is a simple parabola opening upwards.

Vertex Form of Quadratic Equation

The vertex form of a quadratic equation provides a straightforward way to understand and graph the parabola. It is expressed as \( y = a(x - h)^2 + k \), where

• \((h, k)\) is the vertex of the parabola, making it easy to graph.
• \(a\) controls the opening width and direction of the parabola (narrower or wider).

In the given exercise, we determined the vertex as \((0, 1)\), so the vertex form became \( y = a(x - 0)^2 + 1 \) or simply \( y = ax^2 + 1 \).
By testing another point, we found that \(a = 1\), leading to the vertex form \( y = x^2 + 1 \). This form is especially helpful as it simplifies finding the vertex, and it directly reflects in graphing the function more intuitively.