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Chapter 2: Problem 34

Write each quadratic function in the form \(f(x)=a(x-h)^{2}+k\) by completing the square. Also find the vertex of the associated parabola and determine whether it is a maximum or minimum point. $$h(x)=x^{2}-4 x+6$$

Short Answer

Expert verified
The quadratic function in the form \(f(x)=a(x-h)^{2}+k\) is \(h(x) = (x - 2)^{2} + 2\), where the vertex of the parabola is \((2, 2)\) which is a minimum point.

Step by step solution

01

Rewrite in the Vertex Form

To rewrite the function \(h(x) = x^{2} - 4x + 6\) in the form \(f(x)=a(x-h)^{2}+k\), we need to complete the square. To do so, we place the constant term (6) to the right side of the equation. Hence, \(h(x) - 6 = x^{2} - 4x\). Next, we take half the coefficient of x, square it, and add it to both sides of the equation. Half of -4 is -2, and \((-2)^2 = 4\). Therefore, by adding 4 to both sides we obtain, \(h(x) - 6 + 4 = (x^{2} - 4x + 4)\) which simplifies to \(h(x) - 2 = (x - 2)^2\). Bringing back the constants on one side, we rewrite the equation in the vertex form as \(h(x) = (x - 2)^2 + 2\).
02

Find the Vertex

The vertex of a function in the form \(f(x) = a(x - h)^{2} + k\) is \((h , k)\). Thus, for the function \(h(x) = (x - 2)^{2} + 2\), the vertex is \((2, 2)\).
03

Determine the Minimum or Maximum

The vertex will be a minimum or maximum point depending on the sign of the coefficient 'a' in the form \(f(x) = a(x - h)^{2} + k\). If 'a' is greater than 0, then the parabola opens upwards and so the vertex is a minimum point. In the case of the function \(h(x) = (x - 2)^{2} + 2\), 'a' is 1 which is greater than 0. As such, the vertex is a minimum point.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quadratic Function
A quadratic function is a type of polynomial function that is characterized by the highest exponent being two. Written in the standard form, it looks like:
  • \( f(x) = ax^2 + bx + c \)
where \( a, b, \) and \( c \) are constants, and \( x \) is the variable. This form displays the typical parabola shape of a quadratic function on a graph.
The quadratic function is central to various topics in mathematics, including algebra and calculus, because of its ability to model real-world phenomena such as projectile motion, area computations, and more.
Vertex Form
The vertex form of a quadratic function provides a clearer depiction of its graph. In this form, the function is expressed as:
  • \( f(x) = a(x-h)^2 + k \)
Here, \( (h, k) \) represents the vertex of the parabola. The parameter \( a \) influences the width and direction of the parabola.
Completing the square is a method employed to transition a standard quadratic equation into vertex form. This process involves adjusting the equation to reveal the perfect square trinomial, which directly relates to the vertex of the parabola. Once in vertex form, determining the vertex and understanding the parabola's behavior becomes swift and intuitive.
Parabola Vertex
The vertex of a parabola is a very significant point. It is the spot where the parabola changes direction and represents either the maximum or minimum point of the function. In vertex form, the vertex can be effortlessly identified as \( (h, k) \).
  • The x-coordinate \( h \) indicates the parabola鈥檚 line of symmetry.
  • The y-coordinate \( k \) shows the parabola鈥檚 highest or lowest point.
By identifying the vertex, we gain more insights into the function's graph. This allows us to easily sketch the graph of the quadratic function or solve optimization problems related to the function.
Maximum and Minimum Points
Understanding maximum and minimum points is vital when analyzing quadratic functions. Depending on the value of \( a \) in the vertex form \( f(x) = a(x-h)^2 + k \), the orientation of the parabola changes:
  • If \( a > 0 \), the parabola opens upwards, and the graph's lowest point, or minimum, is at the vertex \( (h, k) \).
  • If \( a < 0 \), the parabola opens downwards, and the vertex becomes the highest point or maximum.
In optimization contexts, these points can represent crucial values, such as the lowest cost, shortest time, or maximum score. Recognizing whether your quadratic function provides a maximum or minimum helps in problem-solving and decision-making scenarios.

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Most popular questions from this chapter

Let \(n(t)\) represent the number of students attending a review session each week, starting with the first week of school. Let \(p(t)\) represent the number of tutors scheduled to work during the review session each week. Interpret the amount \(\frac{n(t)}{p(t)}\)

For function of the form \(f(x)=\) \(a x^{2}+b x+c,\) find the discriminant, \(b^{2}-4 a c,\) and use it to determine the number of \(x\)-intercepts of the graph of \(f .\) Also determine the number of real solutions of the equation \(f(x)=0.\) $$f(x)=-x^{2}+4 x-4$$

In Exercises \(67-86,\) find expressions for \((f \circ g)(x)\) and \((g \circ f)(x)\) Give the domains of \(f \circ g\) and \(g \circ f\). $$f(x)=|x| ; g(x)=\frac{2 x}{x-1}$$

When designing buildings, engineers must pay careful attention to how different factors affect the load a structure can bear. The following table gives the load in terms of the weight of concrete that can be borne when threaded rod anchors of various diameters are used to form joints. $$\begin{array}{cc} \text { Diameter (in.) } & \text { Load (1b) } \\ \hline 0.3750 & 2105 \\ 0.5000 & 3750 \\ 0.6250 & 5875 \\ 0.7500 & 8460 \\ 0.8750 & 11,500 \end{array}$$ (a) Examine the table and explain why the relationship between the diameter and the load is not linear. (b) The function $$f(x)=14,926 x^{2}+148 x-51$$ gives the load (in pounds of concrete) that can be borne when rod anchors of diameter \(x\) (in inches) are employed. Use this function to determine the load for an anchor with a diameter of 0.8 inch. (c) since the rods are drilled into the concrete, the manufacturer's specifications sheet gives the load in terms of the diameter of the drill bit. This diameter is always 0.125 inch larger than the diameter of the anchor. Write the function in part (b) in terms of the diameter of the drill bit. The loads for the drill bits will be the same as the loads for the corresponding anchors. (Hint: Examine the table of values and see if you can present the table in terms of the diameter of the drill bit.)

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=\sqrt[3]{4 x^{2}-1}$$
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