/*! 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 8 The graph of a function \(f\) pa... [FREE SOLUTION] | 91影视

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Chapter 1: Problem 8

The graph of a function \(f\) passes through the points \((0,1),\left(2, \frac{1}{3}\right)\) and \(\left(4, \frac{1}{5}\right)\). Find a quadratic function whose graph passes through these points.

Short Answer

Expert verified
The quadratic function that passes through the points (0,1), (2, 1/3) and (4, 1/5) is \(f(x) = 1\).

Step by step solution

01

Set Up System of Equations

Substitute the x and y values of each point into the general equation of a quadratic function \(f(x) = ax^2 + bx + c\). This gives three equations: \[1 = a*0^2 + b*0 + c\] \[\frac{1}{3} = a*2^2 + b*2 + c\] \[\frac{1}{5} = a*4^2 + b*4 + c\] Simplifying these, yields: \[1 = c\] \[\frac{1}{3} = 4a + 2b + 1\] \[\frac{1}{5} = 16a + 4b + 1\] From the first equation, we can observe that \(c = 1\)
02

Solve System of Equations

Subtract the first equation from the second and the third equation yielding: \[0 = 4a + 2b\] and \[0 = 16a + 4b\] which simplifies to \[0 = 2a + b\] and \[0 = 4a + b\] Now, we can subtract the first equation from the second yielding \(0 = 2a\), which simplifies to \(a = 0\). Substituting \(a = 0\) into the first equation yields \(b = 0\)
03

Formulate the Function

Substitute \(a = 0\), \(b = 0\), and \(c = 1\) into the general form of the quadratic equation \(f(x) = ax^2 + bx + c\). The resulting function is \(f(x) = 1\)

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Key Concepts

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

System of Equations
A system of equations is a collection of two or more equations with the same set of variables. Solving a system means finding the values of the variables that satisfy all equations simultaneously.
  • Each equation in the system represents a constraint on the variables.
  • In this exercise, we use the general quadratic formula, \(f(x) = ax^2 + bx + c\), to create a system of equations by substituting the given points into it.
  • The resulting system of equations helps us find the values of the coefficients \(a, b,\) and \(c\) that make the quadratic function pass through the given points.
We started with three equations, derived from the three given points.
The first equation, \(1 = c\), immediately helped us discover that \(c\) is 1.
The other two equations permitted us to solve for \(a\) and \(b\). This technique illustrates the power of systems of equations to solve real-world problems, like finding the best-fit curve in data analysis.
Quadratic Equation
A quadratic equation is a second-degree polynomial of the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(a eq 0\).
  • These functions graphically represent parabolas, which can open either up or down.
  • The quadratic in this exercise simplifies to a constant function, \(f(x) = 1\).
The quadratic function we derived, \(f(x) = 1\), is actually a special case of a quadratic where both \(a = 0\) and \(b = 0\), which implies the parabola is very flat.
This essentially makes it a horizontal line.
It visually demonstrates that not all sets of three points yield a typical parabolic curve, sometimes forming simpler lines instead.
Point-Slope Form
The point-slope form is a way of writing linear equations. It is given by \( y - y_1 = m(x - x_1) \), where \(m\) is the slope and \((x_1, y_1)\) is a point on the line.
  • This form aids in easily identifying a line's slope and its specific point.
  • While often associated with linear equations, its conceptual framework鈥攗sing specific points to develop equations鈥攖ranslates to quadratic functions as well.
For quadratics, while we don't use point-slope form directly, incorporating given points helps formulate these equations.
In a way, this process parallels deriving a line equation from a single point and a slope.
In the context of our quadratic function, each point helped build our system of equations. This step-by-step translation of points into equations showcases the adaptability of concepts like point-slope form across different types of algebraic formulas.

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Most popular questions from this chapter

Determine whether each statement is true or false. If a statement is true, give a reason or cite an appropriate statement from the text. If a statement is false, provide an example that shows the statement is not true in all cases or cite an appropriate statement from the text. (a) \(A 6 \times 3\) matrix has six rows. (b) Every matrix is row-equivalent to a matrix in row-echelon form. (c) If the row-echelon form of the augmented matrix of a system of linear equations contains the row \(\left[\begin{array}{lllll}1 & 0 & 0 & 0 & 0\end{array}\right]\) then the original system is inconsistent. (d) A homogeneous system of four linear equations in six variables has an infinite number of solutions.

The system below has one solution: \(x=1, y=-1,\) and \(z=2\) $$\begin{aligned} 4 x-2 y+5 z &=16 \\ x+y &=0 \\ -x-3 y+2 z &=6 \end{aligned}$$

In your own words, describe the difference between a matrix in row-echelon form and a matrix in reduced row-echelon form.

Consider the matrix \(A=\left[\begin{array}{rrr}2 & -1 & 3 \\ -4 & 2 & k \\ 4 & -2 & 6\end{array}\right]\) (a) If \(A\) is the augmented matrix of a system of linear equations, determine the number of equations and the number of variables. (b) If \(A\) is the augmented matrix of a system of linear equations, find the value(s) of \(k\) such that the system is consistent. (c) If \(A\) is the coefficient matrix of a homogeneous system of linear equations, determine the number of equations and the number of variables. (d) If \(A\) is the coefficient matrix of a homogeneous system of linear equations, find the value(s) of \(k\) such that the system is consistent.

Solve the homogeneous linear system corresponding to the coefficient matrix provided. $$\left[\begin{array}{cccc} 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]$$
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