Problem 27 - 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. \(f(x)=x^{2}-2 x-3\)

Step-by-Step Solution

Answer

The function can be rewritten as \(f(x) = (x-1)^{2} - 4\), with a vertex at (1, -4). The axis of symmetry is \(x = 1\), the x-intercepts are at (-1, 0) and (3, 0), and the y-intercept is at (0, -3). By plotting these points and connecting them, the graph of the function is an upward-opening parabola.

Step 1: Identify and compare coefficients

Compare the given function with the general quadratic function \(f(x) = Ax^2 + Bx + C\) to identify the coefficients. Here, \(A = 1, B = -2, C = -3\).

Step 2: Complete the square

To complete the square, let's find the square of half of the linear coefficient to add and subtract it in the function: \[(\frac{B}{2A})^2 = (\frac{-2}{2\cdot1})^2 = (-1)^2 = 1\] Now we have to add and subtract this value inside the function: \(f(x) = x^{2} - 2x - 3 = (x^{2} - 2x + 1) -1 -3\)

Step 3: Rewrite the function in the form \(f(x) = a(x-h)^{2} + k\)

Now group the terms to get a perfect square trinomial: \(f(x) = (x^{2} - 2x + 1) - 4 = (x-1)^{2} - 4\) So, the function is now in the vertex form: \(f(x) = 1(x-1)^{2} - 4\), with \(a = 1, h = 1, k = -4\).

Step 4: Find the vertex

The vertex of the function is given by the point \((h, k) = (1, -4)\).

Step 5: Determine the axis of symmetry

The axis of symmetry is a vertical line passing through the x-coordinate of the vertex. So, the equation of the axis of symmetry is \(x = h = 1\).

Step 6: Calculate the intercepts

To find the x-intercepts, let \(f(x) = 0\): \((x-1)^{2} - 4 = 0 \Rightarrow (x-1)^{2} = 4 \Rightarrow x-1 = \pm2\) So, the x-intercepts are \(x = 1-2 = -1\) and \(x = 1+2 = 3\). Now, to find the y-intercept, let \(x = 0\): \(f(0) = (0-1)^{2} - 4 = 1 - 4 = -3\) So, the y-intercept is \(y = -3\).

Step 7: Sketch the graph of the function

1. Plot the vertex at the point \((1, -4)\). 2. Draw the axis of symmetry as the vertical line \(x = 1\). 3. Plot the x-intercepts at \((-1, 0)\) and \((3, 0)\). 4. Plot the y-intercept at \((0, -3)\). 5. Since \(a = 1 > 0\), the parabola opens upward. 6. Connect the points with a smooth curve to form the graph of the function. Now you have the graph of the function \(f(x) = (x-1)^{2} - 4\) including the intercepts.

Key Concepts

Quadratic Functions

Quadratic functions are a type of polynomial function where the highest degree of the variable is 2. They take the general form:

• \(f(x) = Ax^2 + Bx + C\)

Here, the constants \(A\), \(B\), and \(C\) are called coefficients. The coefficient \(A\) must be non-zero for the function to be quadratic. If \(A\) is positive, the parabola opens upward, and if negative, it opens downward. Understanding these basics helps when analyzing and graphing the function.
Quadratic functions create a U-shaped curve called a parabola when graphed. This shape is symmetric, meaning if you were to fold the graph along a vertical line passing through its vertex, both halves would match exactly.

Vertex Form

Converting a quadratic function to vertex form helps to easily identify the vertex of the parabola. The vertex form of a quadratic function is expressed as:

• \(f(x) = a(x-h)^2 + k\)

In this expression:

• \(a\) indicates the direction of the parabola. If \(a > 0\), it opens upward, and if \(a < 0\), it opens downward.
• \((h, k)\) gives the coordinates of the vertex, the highest or lowest point on the graph.

To convert from standard form to vertex form, we use a method called "completing the square". This involves adjusting the quadratic and linear terms to create a perfect square trinomial, simplifying the process of graphing the parabola and finding its characteristics.

Graphing Parabolas

When graphing parabolas, especially from the vertex form \(f(x) = a(x-h)^2 + k\), we focus on key characteristics:

• Vertex: The point \((h, k)\) where the parabola changes direction. This is the "tip" of the parabola.
• Axis of Symmetry: A vertical line through the vertex, given by \(x = h\). This line divides the parabola into two mirror-image halves.
• Direction: Determined by \(a\). If \(a > 0\), the parabola opens upwards; if \(a < 0\), it opens downwards.

By plotting the vertex and using the axis of symmetry, you can draw the basic shape of the parabola. Additional points, such as intercepts, help to finalize the graph's shape. Remember, a smooth curve should be drawn to connect all plotted points.

Intercepts

Intercepts are where the graph of a function crosses the axes. For parabolas, finding these points is crucial for accurate graphing:

• X-intercepts occur where the graph crosses the x-axis. Set \(f(x) = 0\) and solve for \(x\). For example, if \((x-h)^2 = k\), then \(x = h \pm \sqrt{k}\).
• Y-intercept is where the graph crosses the y-axis. Set \(x = 0\) in the function \(f(x)\) to find \(y\).

In our exercise, the x-intercepts were at \(-1\) and \(3\), and the y-intercept was at \(-3\). These points help shape the parabola on the graph, ensuring it reflects the function accurately across both axes.