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91Ó°ÊÓ

Find a possible expression for a quadratic function \(f(x)\) having the given zeros. There can be more than one correct answer. $$x=\frac{1}{2} \text { and } x=3$$

Short Answer

Expert verified
The equation of the quadratic function with roots \(x = \frac{1}{2}\) and \(x = 3\) can be \(f(x) = x^2 - \frac{7}{2}x + \frac{3}{2}\).

Step by step solution

01

Identify the roots

Our given roots are \(x = \frac{1}{2}\) and \(x = 3\).
02

Plug the roots into the quadratic formula

To find our quadratic function, we will simply substitute the roots h and k in the formula \(f(x) = a(x-h)(x-k)\). Let's use a=1 for simplicity. Substituting h as \(\frac{1}{2}\) and k as 3 into the formula, we have \(f(x) = (x - \frac{1}{2})(x - 3)\).
03

Simplify the expression

Multiplying out the expression we get \(f(x) = x^2 - \frac{7}{2}x + \frac{3}{2}\).

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Key Concepts

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

Quadratic Formula
Understanding the quadratic formula is essential for solving quadratic equations and finding the zeros of a quadratic function. The quadratic formula is expressed as:

\[\begin{equation}x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\end{equation}\]
This formula is derived from the standard form of a quadratic equation, which is:\[\begin{equation}ax^2 + bx + c = 0\end{equation}\]
Given a quadratic equation in standard form, you can use the quadratic formula to find the values of x that make the equation true—these are called the roots or zeros. The terms under the square root, known as the discriminant \[\begin{equation}D = b^2 - 4ac\end{equation}\]
, can tell us about the nature of the roots. A positive discriminant indicates two distinct real roots, a discriminant of zero indicates one real root (also called a repeated or double root), and a negative discriminant indicates two complex roots.
Roots of a Quadratic Equation
When we talk about the roots of a quadratic equation, we’re talking about the values of x that satisfy the equation \[\begin{equation}ax^2 + bx + c = 0 \end{equation}\]
In our example, the roots are \[\begin{equation}x = \frac{1}{2} \text{ and } x = 3\end{equation}\]
These roots are the 'x-intercepts' or 'zeros' of a quadratic function because they are the points where the graph of the function crosses the x-axis. Finding these is crucial for graphing the quadratic function, and they provide important information about the function's characteristics.
Simplifying Algebraic Expressions
Simplifying algebraic expressions is a fundamental skill in algebra. It involves combining like terms and using distributive properties to make expressions easier to work with. In our solution, we simplified the expression \[\begin{equation}(x - \frac{1}{2})(x - 3)\end{equation}\]
by expanding the multiplication and combining like terms to get:\[\begin{equation}f(x) = x^2 - \frac{7}{2}x + \frac{3}{2}\end{equation}\]
This simplification step is critical for interpreting the standard form of quadratic functions and for further calculations like evaluating the function for given values of x or finding the vertex form.
Finding a Quadratic Function
A quadratic function can be determined when the zeros are given, as these zeros are the roots or solutions of the quadratic equation. By knowing that a quadratic function in its factored form looks like \[\begin{equation}f(x) = a(x - h)(x - k)\end{equation}\],
where \[\begin{equation}h \end{equation}\]
and\[\begin{equation}k\end{equation}\]
are the zeros, we can use this form to write the function. We simply replace h and k with the given zeros and choose a suitable value for 'a' (often taken as 1 for simplicity). We then expand this to find the quadratic function in standard form. This conversion from factored form to standard form helps us understand the behavior of the function and is useful for graphing and analyzing the function's properties such as its vertex, axis of symmetry, and the direction it opens.

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Most popular questions from this chapter

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=(-2 x+5)^{2}$$

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

A carpenter wishes to make a rain gutter with a rectangular cross-section by bending up a flat piece of metal that is 18 feet long and 20 inches wide. The top of the gutter is open. $$2$$ (a) Write an expression for the cross-sectional area in terms of \(x,\) the length of metal that is bent upward. (b) How much metal has to be bent upward to maximize the cross-sectional area? What is the maximum cross-sectional area?

In Exercises \(87-96,\) find two functions \(f\) and \(g\) such that \(h(x)=(f \circ g)(x)=f(g(x)) .\) Answers may vary. $$h(x)=4(2 x+9)^{5}-(2 x+9)^{8}$$

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.)

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