Problem 33 - Beginning and Intermediate Algebra

Question

Rewrite function in the form \(f(x)=a(x-h)^{2}+k\) by completing the square. Then, graph the function. Include the intercepts. \(h(x)=-x^{2}-4 x+5\)

Step-by-Step Solution

Answer

\(h(x) = -(x - (-2))^2 + 9\) X-intercepts: \(x = 1\) and \(x = 3\) Y-intercept: \(h(0) = 5\) The graph is a downward opening parabola with vertex at \((-2, 9)\), x-intercepts at \(x = 1\) and \(x = 3\), and a y-intercept at \((0, 5)\).

Step 1: Identify a, b, and c

In the given quadratic function \(h(x) = -x^2 - 4x + 5\), we can identify the coefficients as follows: \(a = -1\), \(b = -4,\), \(c = 5.\)

Step 2: Complete the square for the quadratic and linear terms

To complete the square, we'll need to find a new constant term that will be added and subtracted inside the expression. We can do this using the following formula: \[\frac{b^2}{4a}.\] Using the values we identified in Step 1, we can plug them in and obtain: \[\frac{(-4)^2}{4 \cdot (-1)} = \frac{16}{-4} = -4.\] Now, rewrite the given function by adding and subtracting \(-4\) inside the expression, then adjusting the constant term accordingly: \[h(x) = -(x^2 + 4x - 4) + 5 + 4.\]

Step 3: Simplify the expression

After completing the square, rewrite the function in its simplified form as: \[h(x) = -(x^2 + 4x - 4) + 9.\] Now, factor the expression inside the parentheses: \[h(x) = -(x + 2)^2 + 9.\] Thus, the function in the form \(f(x) = a(x - h)^2 + k\) is as follows: \[h(x) = -(x - (-2))^2 + 9.\]

Step 4: Find the intercepts

To find the x-intercepts, set \(h(x)\) equal to 0: \[-(x - (-2))^2 + 9 = 0.\] To find the y-intercept, set \(x\) equal to 0: \[h(0) = -(0 - (-2))^2 + 9.\] Solve for the intercepts: X-intercepts: \(x = 1\) and \(x = 3\) Y-intercept: \(h(0) = 5\)

Step 5: Graph the function

Using the vertex form of the function, the intercepts, and the vertex (which is \((-2, 9)\)), graph the function: 1. Plot the vertex, \((-2, 9)\). 2. Plot the x-intercepts, \(x = 1\) and \(x = 3\). 3. Plot the y-intercept, \((0, 5)\). 4. Since the coefficient 'a' is negative, the parabola opens downward. The function graph should show a downward opening parabola with vertex at \((-2, 9)\), x-intercepts at \(x = 1\) and \(x = 3\), and a y-intercept at \((0, 5)\).

Key Concepts

Completing the Square

Completing the square is a method used to convert a quadratic function into a more manageable form, making it easier to identify key features, such as the vertex and intercepts. The process involves rearranging the quadratic equation into a perfect square trinomial, essentially a binomial squared.

Here's how it works: given a quadratic expression like \(ax^2 + bx + c\), you first isolate the quadratic and linear terms. In our exercise, with \(h(x) = -x^2 - 4x + 5\), the first step involved identifying the coefficients: \(a = -1\), \(b = -4\), and \(c = 5\).

Next, to complete the square, calculate the value \(\left(\frac{b}{2}\right)^2\) and adjust the equation by adding and subtracting this value. For our equation, the completed square step was:

• Calculate \(\left(\frac{-4}{2}\right)^2 = 4\),
• Add and subtract 4, giving us \(-((x^2 + 4x + 4) - 4)\),
• Re-write it as \(-(x + 2)^2 + 9\).

This rearrangement turns the quadratic equation into the vertex form, highlighting the vertex of the parabola.

Vertex Form

The vertex form of a quadratic equation is useful because it directly reveals the vertex of the parabola, making it easier to plot and understand how the function behaves graphically. In vertex form, the equation is expressed as \(f(x) = a(x-h)^2 + k\), where \((h, k)\) is the vertex of the parabola.

For our function \(h(x) = -(x + 2)^2 + 9\), you can easily see that the vertex is \((-2, 9)\). This form tells us that:

• \(h = -2\), which shifts the parabola 2 units to the left,
• \(k = 9\), which shifts it 9 units up,
• The coefficient \(a = -1\) indicates the parabola opens downwards because it is negative.

Understanding the vertex form simplifies both graphing the function and solving problems related to its vertex.

Graphing Quadratic Functions

Graphing quadratic functions is straightforward once you have the vertex form, as it serves as a guide to plotting its key features: the vertex, direction of opening, and intercepts. For \(h(x) = -(x + 2)^2 + 9\), follow these steps to graph it:

Start with the vertex. Here, the vertex is \((-2, 9)\). This point will be the highest point on the graph since the parabola opens downward.

Next, plot the intercepts, which in the exercise were found as follows:

• X-intercepts, where the function crosses the x-axis, were identified as \(x = 1\) and \(x = 3\).
• The y-intercept, where the function crosses the y-axis, is \(5\) which is found by setting \(x = 0\).

Finally, using the coefficient \(a = -1\), understand that the parabola opens downward, which affects how the entire shape stretches across the graph vertically.

Intercepts of Quadratic Equations

Intercepts give crucial information on how a quadratic function interacts with the axes. Understanding both x and y-intercepts makes it easier to fully model the function's behavior visually.

For x-intercepts, set the quadratic function equal to zero, \(f(x) = 0\), and solve for \(x\). In the provided exercise, it required solving \(-((x + 2)^2 - 9) = 0\), resulting in \(x = 1\) and \(x = 3\). These points show where the graph crosses the x-axis.

For the y-intercept, simply substitute \(x = 0\) in the function. This gives point \(h(0) = 5\), showing where the parabola intersects the y-axis.

Understanding these intercepts helps in sketching the graph correctly, ensuring it matches the calculated points and reflecting the parabola's symmetrical nature about its vertex.