Problem 30 - 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. \(h(x)=x^{2}-4 x+1\)

Step-by-Step Solution

Answer

The function \(h(x) = x^2 - 4x + 1\) can be rewritten in the form \(f(x) = a(x-h)^2 + k\) by completing the square. We find that \(f(x) = 1(x - 2)^2 - 3\). The x-intercepts are \(x = 2 + \sqrt{3}\) and \(x = 2 - \sqrt{3}\), and the y-intercept is (0, 1). The vertex of the parabola is at the point (2, -3) and the graph opens upwards.

Step 1: Identify the coefficients a, b, and c

For the given function \(h(x) = x^2 - 4x + 1\), we can see that a=1, b=-4, and c=1.

Step 2: Calculate the value of h and k

To complete the square, we need to find the values for h and k. We will use the following formulas: - \(h = \frac{-b}{2a}\) - \(k = h(x) - ah^2\) First, find the value of h: \[h = \frac{-(-4)}{2(1)}\] \[h = \frac{4}{2}\] \[h = 2\] Next, find the value of k by plugging h into the original function and subtracting from it a times h squared: \[k = h(x) - ah^2 = (1(2)^2 - 4(2) + 1) - (1)(2)^2\] \[k = (4 - 8 + 1) - 4 = -3\]

Step 3: Rewrite the function in the desired form

Now that we have the values of h and k, we can rewrite the given function h(x) in the form f(x) = a(x-h)^2 + k: \[f(x) = 1(x - 2)^2 - 3\]

Step 4: Find the x and y intercepts

To find the x-intercept(s), we set y (or f(x)) to 0 and solve for x: \[0 = 1(x - 2)^2 - 3\] \[3 = (x - 2)^2\] \[\sqrt{3} = x - 2; -\sqrt{3} = x - 2\] \[x = 2 + \sqrt{3} ; x = 2 - \sqrt{3}\] There are two x-intercepts: \(x = 2 + \sqrt{3}\) and \(x = 2 - \sqrt{3}\). To find the y-intercept, we set x to 0 and solve for y (or f(x)): \[f(0) = 1(0 - 2)^2 - 3\] \[f(0) = 1(-2)^2 - 3\] \[f(0) = 1(4) - 3\] \[f(0) = 4 - 3\] \[f(0) = 1\] The y-intercept is: (0, 1).

Step 5: Sketch the graph of the function

To sketch the graph of the function, plot the vertex (h, k) = (2, -3), the x-intercepts at \(x = 2 + \sqrt{3}\) and \(x = 2 - \sqrt{3}\), and the y-intercept at (0, 1). Since the coefficient a is positive, the parabola will open upwards. Connect the points with a smooth curve.

Key Concepts

Quadratic Functions

Quadratic functions are one of the most common types of polynomial equations you'll encounter in mathematics. These functions are written in the form \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants. The graph of a quadratic function is a U-shaped curve called a parabola. It's important to know that the parabola can open upwards or downwards, depending on whether the coefficient \(a\) is positive or negative.

• A positive \(a\) makes the parabola open upwards, resembling a "U" shape.
• A negative \(a\) causes it to open downwards, like an upside-down "U".

One compelling feature of quadratic functions is their symmetry. The line of symmetry passes through the lowest (minimum if \(a\) > 0) or highest point (maximum if \(a\) < 0) of the parabola, known as the vertex.

Vertex Form

The vertex form of a quadratic function offers a clear understanding of the function's graph, primarily focusing on its vertex. Converting a standard quadratic equation to the vertex form involves completing the square. The vertex form is expressed as \(f(x) = a(x-h)^2 + k\). Here's what the components mean:

• \(a\): This value affects the opening width and direction of the parabola.
• \(h\): The x-coordinate of the vertex. It shows where the parabola shifts horizontally.
• \(k\): The y-coordinate of the vertex. It represents the vertical shift.

The vertex, represented as \((h, k)\), is crucial as it indicates the peak or the lowest point of the parabola. When graphing, always start at the vertex as it gives the central shape and direction of the parabola.

X-Intercepts

The x-intercepts of a quadratic function are the points where the graph crosses the x-axis, meaning the value of \(y = 0\) at these points. To find them, you solve the equation \(ax^2 + bx + c = 0\). Here is how you do it for vertex form:

• Start with the equation in vertex form: \(f(x) = a(x-h)^2 + k\).
• Set \(f(x)\) to zero: \(0 = a(x-h)^2 + k\).
• Rearrange and solve for \(x\).

These solutions will give you the points where the quadratic graph crosses the x-axis. Often, a parabola will have two x-intercepts, but it could also have one or none, depending on how it sits relative to the axis. In our example \(h(x) = x^2 - 4x + 1\), after converting to vertex form, the x-intercepts were found to be at \(x = 2 + \sqrt{3}\) and \(x = 2 - \sqrt{3}\).

Y-Intercept

The y-intercept of a function is the point where the graph crosses the y-axis, meaning this is where \(x = 0\). Finding the y-intercept is straightforward in any form of the quadratic equation. Simply set \(x = 0\) in the equation and solve for \(y\):

• Substitute \(x = 0\) into \(f(x)\).
• Calculate \(f(0)\) to find the y-coordinate of the intercept.

For the function \(h(x) = x^2 - 4x + 1\), the y-intercept is \((0, 1)\). This means the graph will cross the y-axis at one "up" unit. Identifying the y-intercept helps when sketching graphs, providing a key point through which the parabola passes.