Problem 49 - 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. $$ \begin{array}{|c|c|c|c|c|c|} \hline x & -2 & -1 & 0 & 1 & 2 \\ \hline y & -8 & -3 & 0 & 1 & 0 \\ \hline \end{array} $$

Step-by-Step Solution

Answer

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

Step 1: Understanding the Vertex

To identify the vertex of the quadratic function, look for the highest or lowest y-value in the table, which will typically correspond to the vertex. Here, the highest y-value of 1 occurs at x = 1; thus, the vertex is (1, 1).

Step 2: Finding the Axis of Symmetry

The axis of symmetry of a quadratic function can be determined by the x-coordinate of the vertex. Since the vertex is at (1, 1), the axis of symmetry is the vertical line x = 1.

Step 3: Formulating the General Form

The standard form of a quadratic function is given by: \[ f(x) = a(x - h)^2 + k \] where (h, k) is the vertex of the parabola. Substituting h = 1 and k = 1, we have: \[ f(x) = a(x - 1)^2 + 1 \]

Step 4: Determine the Coefficient a

To find the value of 'a', substitute another point from the table into the equation. Using the point (-2, -8), substitute x = -2 and y = -8:\[ -8 = a(-2 - 1)^2 + 1 \]This simplifies to:\[ -8 = 9a + 1 \]Solving for 'a':\[ 9a = -9 \]\[ a = -1 \]

Step 5: Finalizing the Equation

Now that we know 'a', substitute it back into the formula:\[ f(x) = -1(x - 1)^2 + 1 \]Simplify to express in standard form:\[ f(x) = -(x^2 - 2x + 1) + 1 \]\[ = -x^2 + 2x + 0 \]Thus, the quadratic function is:\[ f(x) = -x^2 + 2x \]

Key Concepts

Understanding the Vertex

In the context of quadratic functions, the vertex is a critical point that represents the peak or trough of the parabola. It can be thought of as the "turning point" where the graph changes direction. If you're looking at a table of values for a quadratic function, like the one provided in the exercise, the vertex is typically the point with the highest or lowest y-value, depending on whether the parabola opens upwards or downwards.

In our exercise, the highest y-value is 1, which occurs when the x-value is also 1. Therefore, the vertex of the given quadratic function is at \(1, 1\). The vertex provides us with essential information needed to write the equation for the quadratic function later.

Remember, for parabolas that open upwards, the vertex is the lowest point. Conversely, for those that open downwards, like in our example, it is the highest point.

Finding the Axis of Symmetry

The axis of symmetry in a quadratic function is a vertical line that divides the parabola into two mirror images. This line passes through the vertex of the parabola, making it one of the simplest ways to find the axis of symmetry is to use the x-coordinate of the vertex.

For our quadratic function, since the vertex is \(1, 1\), the axis of symmetry is the vertical line \(x = 1\). This means that if you were to fold the graph along this line, both sides of the parabola would match perfectly. The axis of symmetry is crucial because it helps in understanding the balance and structure of the parabola.

Standard Form of a Quadratic Equation

The standard form of a quadratic function is expressed as \[ f(x) = a(x - h)^2 + k \] where \(h, k\) represents the vertex. This form is particularly helpful because it clearly shows the vertex, enabling quick adjustments to the graph based on changes to the vertex's location.

• The variable \(a\) determines the direction and width of the parabola.
• When \(a\) is positive, the parabola opens upwards; when negative, it opens downwards.
• The magnitude of \(a\) affects how "wide" or "narrow" the parabola appears.

In our exercise, substituting the vertex \(1, 1\) into the formula, we arrive at \[ f(x) = a(x - 1)^2 + 1 \]. This sets the foundation for solving for the unknown coefficient \(a\).

Coefficient Determination

Determining the coefficient \(a\) is crucial, as it ultimately shapes the graph of the quadratic function, determining its direction and steepness. To find \(a\), we utilize other known points from the function's table of values by substituting them into the standard form equation.

Here, we use the point \((-2, -8)\). By substituting \(x = -2\) and \(y = -8\) into \[ f(x) = a(x - 1)^2 + 1 \], we perform simple algebraic steps:

• Replace \(y\) and \(x\) with \(-8\) and \(-2\), respectively, leading to the equation: \( -8 = a(-3)^2 + 1 \).
• Simplify to \( -8 = 9a + 1 \). Then solve for \(a\) by isolating it: \(9a = -9\), yielding \(a = -1\).

Thus, with \(a\) determined, you finalize the quadratic function equation as \[ f(x) = -1(x - 1)^2 + 1 \], and simplify it to \[ f(x) = -x^2 + 2x \]. This final form effectively captures the behavior depicted by the original table of values.