Problem 48 - 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. $$ \begin{array}{cccccc} x & -2 & -1 & 0 & 1 & 2 \\ \hline y & -2 & 1 & 2 & 1 & -2 \end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Vertex

The vertex of a quadratic function is the point on the graph where the function reaches its maximum or minimum value. From the table, observe that the values of \(y\) are symmetric and the maximum value is \(2\) at \(x = 0\). Therefore, the vertex of the quadratic function is the point \((0, 2)\).

Step 2: Determine the Axis of Symmetry

The axis of symmetry of a quadratic function in standard form \(y = ax^2 + bx + c\) is determined by the vertex. Since the vertex is at \((0, 2)\), the axis of symmetry is \(x = 0\). This line divides the parabola into two mirror images.

Step 3: Use Symmetry to Identify Points

Using the axis of symmetry \(x = 0\), note that the values of \(y\) for \(x = -2\) and \(x = 2\) are both \(-2\), demonstrating symmetry. Similarly, \(y = 1\) for both \(x = -1\) and \(x = 1\). This confirms the quadratic nature of the function.

Step 4: Write the Quadratic Equation

Since the vertex is \((0, 2)\), use the vertex form of a quadratic equation, which is \(y = a(x - h)^2 + k\), where \(h = 0\) and \(k = 2\). Thus, the equation simplifies to \(y = ax^2 + 2\).

Step 5: Find the Value of 'a'

Select any point from the table other than the vertex to find \(a\). Using \((-1, 1)\), substitute into the equation: \(1 = a(-1)^2 + 2\). Simplify to get \(1 = a + 2\). Thus, \(a = 1 - 2 = -1\).

Step 6: Verify the Equation

Now our equation is \(y = -x^2 + 2\). Confirm this equation works with the other points in the table: for \(x = 1\), \(y = -(1)^2 + 2 = 1\); for \(x = 2\), \(y = -(2)^2 + 2 = -2\). The equation is verified.

Key Concepts

Vertex

The vertex of a quadratic function is a special point that helps to identify the overall shape of its graph, and it is either the highest or lowest point on the curve, depending on whether the parabola opens upwards or downwards. In the given exercise, using the table of values, the vertex was found at the point \((0, 2)\). This means that the function reaches its maximum value at this point, indicating the parabola opens downwards.

• The vertex form of a quadratic function: \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex.
• The vertex gives us crucial information about the parabola's orientation and maximum/minimum.
• In this exercise, \(h = 0\) and \(k = 2\), so the vertex form becomes \(y = a(x - 0)^2 + 2\).

Understanding the location of the vertex helps in graphing the quadratic function and provides valuable insights when transforming the function to different forms.

Axis of Symmetry

The axis of symmetry of a quadratic function is a vertical line that passes through the vertex, dividing the parabola into two mirror-image halves. It is a key feature because every parabola is symmetric with respect to this line. In the given problem, the axis of symmetry is derived directly from the vertex, which is \(x = 0\). This axis is crucial when predicting the behavior of the function on either side of the vertex.

• The formula to find the axis of symmetry from the standard form \(y = ax^2 + bx + c\) is \(x = -\frac{b}{2a}\).
• This axis ensures that the function's graph is balanced, with equal values for \(y\) on both sides at equal distances from the vertex.
• In the table, symmetry is displayed as \(y\) values repeat for opposite \(x\) values equidistant from the axis.

This understanding of symmetry helps in tackling both practical problems and theoretical explorations of quadratic functions.

Standard Form

The standard form of a quadratic function is expressed as \(y = ax^2 + bx + c\). This form is quite versatile as it can represent a wide range of parabolic shapes, but it doesn't directly reveal the vertex or axis of symmetry like the vertex form does.

• This form is useful for quickly identifying the degree and leading coefficient of the polynomial.
• Converting to standard form allows for easy application of algebraic techniques like factoring or completing the square.
• In the given exercise, the vertex and symmetry insights were indirectly used to validate the function’s equation \(y = -x^2 + 2\).

To discern deeper insights, the standard form can often be converted to other forms, such as the vertex form, to facilitate different applications or interpretations.

Vertex Form

The vertex form of a quadratic equation, \(y = a(x-h)^2 + k\), highlights the vertex' coordinates \((h, k)\). This form is exceptionally helpful for quickly sketching the graph, as it directly indicates both the vertex's position and the parabola's direction.

• This form makes it easy to decide whether the parabola opens upwards (\(a > 0\)) or downwards (\(a < 0\)).
• Knowing \(a\), you can determine how "steep" or "wide" the parabola is.
• In this exercise, it was converted from \(y = ax^2 + 2\) using the vertex at \((0,2)\) into \(y = -1(x-0)^2 + 2\), showing a vertex at \((0, 2)\), and confirming the parabola's downward opening with \(a = -1\).

The vertex form simplifies many tasks, including graphing, identifying vertex and symmetry, and translating between different quadratic equations forms.