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Chapter 6: Problem 85

Find the points of intersection of the graphs of the finctinns \(f(r)=r^{2}+3 r+7\) and \(p(r)=-2 r+3\)

Short Answer

Expert verified
The points of intersection are ( -1, 5) and (3, -3).

Step by step solution

01

- Set the functions equal to each other

To find the points of intersection, set the functions equal to each other:
02

- Rearrange the equation

Rearrange the equation to form a standard quadratic equation:
03

- Solve the quadratic equation

Use the quadratic formula to find the values of r: Use the quadratic formula: where which is simplified to: Substituting into the quadratic formula, we get: Therefore, we have: Simplifying, we get: Therefore: Thus, the solutions for r are:
04

- Solve for the intersection points

Substitute the values of r into the linear equation p(r) = -2r +3 to find the corresponding y-values: For r = -1: For r = 3: Thus, the points of intersection are: and

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Key Concepts

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

Quadratic Equations
A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. The highest power of \( x \) is 2, which makes it a quadratic equation.

Quadratic equations can be solved by various methods, such as:
  • Factoring,
  • Completing the square,
  • Using the quadratic formula.
The solutions to a quadratic equation are where the graph of the quadratic function intersects the x-axis. These points are called the roots of the equation.

In this exercise, the quadratic equation is formed by setting the quadratic function \( f(r) = r^2 + 3r + 7 \) equal to the linear function \( p(r) = -2r + 3 \).
Linear Equations
A linear equation is an equation of the form \( y = mx + b \), where \( m \) is the slope, and \( b \) is the y-intercept. The graph of a linear equation is a straight line.

Linear equations can be easily solved by isolating the variable. In this exercise, the linear equation is \( p(r) = -2r + 3 \).

To find the points of intersection, we need to set the linear equation equal to the quadratic equation and solve for the variable \( r \). This process will reveal where the two graphs intersect.
Quadratic Formula
The quadratic formula is a powerful tool for solving quadratic equations. It is given by:
\( r = \frac{ -b \pm \sqrt{ b^2 - 4ac } }{ 2a } \)

Here, \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation \( ax^2 + bx + c = 0 \).

The term under the square root, \( b^2 - 4ac \), is called the discriminant. It determines the nature of the roots:
  • If \( b^2 - 4ac > 0 \), the equation has two distinct real roots.
  • If \( b^2 - 4ac = 0 \), the equation has one real root (a repeated root).
  • If \( b^2 - 4ac < 0 \), the equation has no real roots (the roots are complex numbers).
In this exercise, after setting \( f(r) = r^2 + 3r + 7 \) equal to \( p(r) = -2r + 3 \), we rearrange it to form a standard quadratic equation and use the quadratic formula to find the values of \( r \).
Points of Intersection
The points of intersection of two functions are the values of \( x \) (or \( r \) in this case) where the two graphs meet, and the corresponding y-values.

To find these points, we follow these steps:
  • Set the two functions equal to each other.
  • Rearrange the equation to form a standard quadratic equation.
  • Solve the quadratic equation using a suitable method, such as the quadratic formula.
  • Substitute the values of the variable back into one of the original functions to find the corresponding y-values.
In this exercise:

1. We set \( r^2 + 3r + 7 = -2r + 3 \).
2. Rearrange to form \( r^2 + 5r + 4 = 0 \).
3. Use the quadratic formula to find the values of \( r \), which are \( r = -1 \) and \( r = -4 \).
4. Substitute these values into the linear equation \( p(r) = -2r + 3 \) to find the y-values.

The points of intersection are: \( r = -1, y = 5 \) and \( r = 3, y = -3 \).

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Most popular questions from this chapter

Analyzing Interest Rates on a Mortgage Colleen and Bill have just purchased a house for 650,000, with the seller holding a second mortgage of 100,000 . They promise to pay the seller 100,000 plus all accrued interest 5 years from now. The seller offers them three interest options on the second mortgage: (a) Simple interest at 6 % per annum (b) 5.5 % interest compounded monthly (c) $5.25 % interest compounded continuously Which option is best? That is, which results in paying the least interest on the loan?

Express y as a function of \(x .\) The constant \(C\) is a positive number. \(\ln y=\ln (x+C)\)

Express y as a function of \(x .\) The constant \(C\) is a positive number. \(2 \ln y=-\frac{1}{2} \ln x+\frac{1}{3} \ln \left(x^{2}+1\right)+\ln C\)

Solve each equation. $$ \ln e^{x}=5 $$

Solve each equation. $$ \log _{5}\left(x^{2}+x+4\right)=2 $$
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