Problem 32 - 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. \(y=x^{2}-8 x+18\)

Step-by-Step Solution

Answer

The given quadratic function is rewritten in the vertex form as \(f(x) = (x - 4)^{2} + 2\), with the vertex being (4, 2). To find the intercepts, we calculated that there are no real x-intercepts and the y-intercept is (0, 18). When graphing the function, plot the vertex at point (4, 2), the y-intercept at point (0, 18), and draw a parabola that opens upwards passing through these points. Make sure to label the vertex and y-intercept for clarity.

Step 1: Rewrite the quadratic function in vertex form

Given quadratic function: \(y = x^{2} - 8x + 18\) To rewrite the function in the form \(f(x) = a(x-h)^2 + k\), let's complete the square: \(y = (x^{2} - 8x) + 18\) Add and subtract the square of half of the coefficient of x (-4) inside the parentheses: \(y = (x^{2} - 8x + 16 - 16) + 18\) Simplify the expression: \(y = (x - 4)^{2} - 16 + 18\) \(y = (x - 4)^{2} + 2\) Now, the quadratic function is in the vertex form: \(f(x) = (x - 4)^{2} + 2\)

Step 2: Identify the vertex of the quadratic function

The vertex of a quadratic function in the form \(f(x) = a(x-h)^{2} + k\) is given by the point (h, k). In this case, we have: h = 4 k = 2 Thus, the vertex is (4, 2).

Step 3: Calculate the intercepts

1. X-intercepts: To find the x-intercepts, set y = 0 and solve for x: \(0 = (x - 4)^{2} + 2\) Subtract 2 from both sides and simplify: \((x - 4)^{2} = -2\) In this case, there are no real solutions for x, as the square of a real number cannot be negative. 2. Y-intercept: To find the y-intercept, we set x = 0: \(y = (0 - 4)^{2} + 2\) \(y = 16 + 2\) \(y = 18\) So, the y-intercept is (0, 18).

Step 4: Graph the quadratic function

To graph the function, first plot the vertex at point (4, 2). Next, plot the y-intercept at point (0, 18). Then, draw a parabola that opens upwards and goes through the vertex and the y-intercept. Please note that there are no x-intercepts in this case, as previously identified. Make sure to label important points on the graph – vertex and y-intercept – for clarity.

Key Concepts

Quadratic Functions

A quadratic function is a type of polynomial function that can be represented as \(y = ax^2 + bx + c\). Its characteristic curve is called a parabola. This curve can open upwards or downwards, depending on the coefficient \(a\).

• If \(a > 0\), the parabola opens upwards.
• If \(a < 0\), it opens downwards.

Understanding the basic properties of quadratic functions allows us to analyze and graph them effectively. In the given exercise, the quadratic function is \(y = x^2 - 8x + 18\), which fits the standard quadratic form. Completing the square is one method to convert it into vertex form to easily identify the graph's critical elements.

Vertex Form

The vertex form of a quadratic function is expressed as \(f(x) = a(x-h)^2 + k\), where \((h, k)\) represents the vertex of the parabola.Converting a quadratic function from its standard form to vertex form helps reveal its vertex directly, making it easier to understand its graphical behavior.
For the function \(y = x^2 - 8x + 18\), we completed the square to transform it into \((x - 4)^2 + 2\). Thus, the function in vertex form is \(f(x) = (x-4)^2 + 2\).
This tells us that:

• \(h = 4\)
• \(k = 2\)
• The vertex is at the point \((4, 2)\).

Intercepts

Intercepts are the points where the function intersects the x-axis and y-axis.

X-Intercepts

X-intercepts occur where \(y=0\). Solving \(0 = (x - 4)^2 + 2\), we find that there are no real solutions for \(x\) because the square cannot equal a negative number, indicating no real x-intercepts for this function.

Y-Intercept

Y-intercepts occur where \(x=0\). Substituting \(x = 0\) in \(y = (0 - 4)^2 + 2\), simplifies to \(y = 18\). Thus, the y-intercept is at \((0, 18)\).
The intercepts are valuable for understanding the graph's shape and position on a coordinate plane.

Graphing Parabolas

Graphing a parabola involves identifying key points such as the vertex and intercepts.

Plotting the Graph

Begin by plotting the vertex, \((4, 2)\). This provides a central reference for your graph. Since there are no x-intercepts, focus on the y-intercept at \((0, 18)\).
With these points:

• Vertex (4, 2)
• Y-intercept (0, 18)

Draw the parabola opening upwards, as indicated by the positive coefficient in the equation \((x-4)^2 + 2\). It should pass through these critical points, illustrating the upward "U" shape of the graph.
Label these points on your graph for clarity, providing a visual summary of the function's key features.