/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 31 Express the polynomial \(f(x)\) ... [FREE SOLUTION] | 91Ó°ÊÓ

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Express the polynomial \(f(x)\) in the form \(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\). Find a quadratic function that has zeros -4 and 9 and a graph that passes through the point (3,5).

Short Answer

Expert verified
The quadratic function is \(f(x) = -\frac{5}{42}x^2 + \frac{25}{42}x + \frac{30}{7}\).

Step by step solution

01

Understand the zeros

The zeros of a quadratic function are the values of \(x\) where the function equals zero. For our function, the zeros are -4 and 9. This implies that the function can be written as \(f(x) = a(x + 4)(x - 9)\), where \(a\) is a constant.
02

Expand the function

Expand \((x + 4)(x - 9)\) to form a quadratic equation: \(x^2 - 9x + 4x - 36\). Simplifying this gives \(x^2 - 5x - 36\). Thus, the function becomes \(f(x) = a(x^2 - 5x - 36)\).
03

Use the point to solve for a

Since the function passes through the point \((3,5)\), we substitute \(x = 3\) and \(f(x) = 5\) into the equation to find \(a\). This gives us \(5 = a(3^2 - 5 \cdot 3 - 36)\).
04

Solve for a

Calculate \(3^2 - 5 \cdot 3 - 36 = 9 - 15 - 36 = -42\). Thus, \(5 = a(-42)\). Solving for \(a\), we get \(a = \frac{5}{-42} = -\frac{5}{42}\).
05

Write the final quadratic function

Substitute \(a\) back into the expanded form to get the quadratic equation. So, the quadratic function is \(f(x) = -\frac{5}{42}(x^2 - 5x - 36)\). Distributing \(a\), we have \(f(x) = -\frac{5}{42}x^2 + \frac{25}{42}x + \frac{180}{42}\). Simplifying, \(f(x) = -\frac{5}{42}x^2 + \frac{25}{42}x + \frac{30}{7}\).

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Key Concepts

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

Polynomials
Polynomials are mathematical expressions that consist of variables, coefficients, and exponents. The general form of a polynomial is expressed as:
  • \( a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0 \)
Here, \(a_n, a_{n-1}, \cdots, a_1, a_0\) are constants, and \(n\) is a non-negative integer that represents the highest degree of the polynomial.
The degree of a polynomial indicates the highest power of the variable present in the expression.
A quadratic function, for example, is a polynomial of degree 2, typically written in the form \(ax^2 + bx + c\). Polynomials can be used to model a wide range of real-world scenarios and are fundamental to algebra and calculus.
Knowing how to express functions in polynomial form is crucial for solving equations analytically and understanding the behavior of graphs.
Zeros of a Function
Zeros, also known as roots, of a function are the values of \(x\) for which the function outputs zero.
  • For a quadratic function in the form \(f(x) = ax^2 + bx + c\), the zeros occur where \(f(x) = 0\).
  • These points are important because they indicate where the graph of the function intersects the x-axis.
  • In this exercise's context, the zeros given are -4 and 9, which are the solutions to the equation \(a(x + 4)(x - 9) = 0\).
To find the quadratic function with these zeros, we used the formula \(f(x) = a(x + 4)(x - 9)\).
Understanding zeros is essential in solving quadratic equations, graphing quadratic functions, and analyzing the property of the graphs.
Quadratic Equations
Quadratic equations are a specific type of polynomial equation where the highest degree of the variable is 2. The standard form is:
  • \(ax^2 + bx + c = 0\)
These equations can have up to two real solutions, also known as the equation's roots or zeros.
To solve a quadratic equation, you can apply several methods, such as factoring, using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), and completing the square. In this exercise, we factored the quadratic functions by recognizing the zeros: -4 and 9.
Furthermore, we demonstrate using known values, like a point on the graph, to determine additional constants, such as the coefficient \(a\). With the point (3, 5) provided in the exercise, it helped us solve for \(a\) and consequently determine the exact form of the quadratic equation: \(-\frac{5}{42}x^2 + \frac{25}{42}x + \frac{30}{7}\).Quadratic equations are fundamental in calculus and physics, modeling various phenomena such as projectile motion and optimization problems.

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Most popular questions from this chapter

In a note that appeared in The Two-Year College Mathematics Journal [vol. \(12(1981), \text { pp. } 334-336],\) Professors Warren Page and Leo Chosid explain how the process of testing for rational roots can be shortened. In essence, their result is as follows. Suppose that we have a polynomial with integer coefficients and we are testing for a possible root \(p / q .\) Then, if a noninteger is generated at any point in the synthetic division process, \(p / q\) cannot be a root of the polynomial. For example, suppose we want to know whether \(4 / 3\) is a root of \(6 x^{4}-10 x^{3}+2 x^{2}-9 x+8=0\) The first few steps of the synthetic division are as follows. $$ \begin{array}{rrrrrr} 4 / 3 & 6 & -10 & 2 & -9 & 8 \\ \hline & & 8 & -8 / 3 & & \\ \hline & 6 & -2 & & & \\ \hline \end{array} $$ since the noninteger \(-8 / 3\) has been generated in the synthetic division process, the process can be stopped; \(4 / 3\) is not a root of the polynomial. Use this idea to shorten your work in testing to see whether the numbers \(3 / 4,1 / 8\). and \(-3 / 2\) are roots of the equation $$ 8 x^{5}-5 x^{4}+3 x^{2}-2 x-6=0 $$

Express the polynomial \(f(x)\) in the form \(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\). (a) Find a quadratic function that has a maximum value of 2 and that has -2 and 4 as zeros. (b) Use a graphing utility to check that your answer in part (a) appears to be correct.

Let \(f(x)=x^{3}+3 x^{2}-x-3\) (a) Factor \(f(x)\) by using the basic factoring techniques in Appendix B.4. (b) Sketch the graph of \(f(x)=x^{3}+3 x^{2}-x-3 .\) Note that -3 is a lower bound for the roots. (c) Show that the number -3 fails the lower bound test. This shows that a number may fail the lower bound test and yet be a lower bound. (We say that the lower bound test provides a sufficient but not a necessary condition for a lower bound.)

If \(p\) and \(q\) are prime numbers, show that the equation \(x^{3}+p x-p q=0\) has no rational roots.

Express the polynomial \(f(x)\) in the form \(a_{n} x^{n}+a_{n-1} x^{n-1}+\cdots+a_{1} x+a_{0}\). (a) Find a third-degree polynomial function that has zeros \(-5,2,\) and 3 and a graph that passes through the point (0,1) (b) Use a graphing utility to check that your answer in part (a) appears to be correct.

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