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Magazine Chapter 2: Problem 25
Do the graphs intersect in the given viewing rectangle? If they do, how many points of intersection are there? $$ y=6-4 x-x^{2}, y=3 x+18 ; \quad[-6,2] \text { by }[-5,20] $$
Short Answer
Expert verified
Yes, the graphs intersect at two points: \((-3, 15)\) and \((-4, 6)\).
Step by step solution
01
Write the Equations
The given equations are \( y = 6 - 4x - x^2 \) and \( y = 3x + 18 \). We want to find the intersection of these equations in the region \([-6, 2]\) for \(x\) and \([-5, 20]\) for \(y\).
02
Set the Equations Equal
To find the intersection points, set the equations equal to each other: \[ 6 - 4x - x^2 = 3x + 18 \].
03
Rearrange Terms to Form a Quadratic Equation
Rearrange the terms of the equation to form a standard quadratic equation: \(-x^2 - 7x - 12 = 0\).
04
Solve the Quadratic Equation
The quadratic equation \(-x^2 - 7x - 12 = 0\) can be solved using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \(a = -1\), \(b = -7\), \(c = -12\).
05
Calculate the Discriminant and Roots
Calculate the discriminant \(\Delta = b^2 - 4ac = (-7)^2 - 4\cdot(-1)\cdot(-12) = 49 - 48 = 1\). Since the discriminant is positive, there are two real roots. Applying the quadratic formula yields roots: \( x_1 = -3 \) and \( x_2 = -4 \).
06
Verify the Roots in the Viewing Rectangle
Both \(x_1 = -3\) and \(x_2 = -4\) are within the given \(x\) range \([-6, 2]\). Now, calculate the corresponding \(y\) values. For \(x = -3\), \( y = 15 \). For \(x = -4\), \( y = 6 \). Both \(y\) values are within the viewing range \([-5, 20]\).
07
Confirm Number of Intersection Points
The graphs intersect at two points within the given viewing rectangle: \((-3, 15)\) and \((-4, 6)\). Thus, there are two points of intersection.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a type of polynomial that is characterized by a degree of 2, which means the highest power of the variable is 2. It generally takes the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not equal to zero. Quadratic equations are crucial in studying parabolic shapes, which are simply the graphs formed by these equations.
In the given problem, the equation \( y = 6 - 4x - x^2 \) is a quadratic equation because it contains the term \( x^2 \). Understanding the structure of a quadratic equation will help in solving and graphing problems related to these parabolic shapes effectively.
In the given problem, the equation \( y = 6 - 4x - x^2 \) is a quadratic equation because it contains the term \( x^2 \). Understanding the structure of a quadratic equation will help in solving and graphing problems related to these parabolic shapes effectively.
Discriminant Calculation
The discriminant helps us determine the nature and number of roots for a quadratic equation. Calculated using \( \Delta = b^2 - 4ac \), the discriminant offers insight:
In our problem, the quadratic equation is \( -x^2 - 7x -12 = 0 \), where \( a = -1 \), \( b = -7 \), and \( c = -12 \). Calculating the discriminant gives us \( \Delta = 49 - 48 \), equalling 1, indicating two real roots.
- If \( \Delta > 0 \), two distinct real roots exist.
- If \( \Delta = 0 \), there is exactly one real root (also called a repeated root).
- If \( \Delta < 0 \), the roots are complex and not real.
In our problem, the quadratic equation is \( -x^2 - 7x -12 = 0 \), where \( a = -1 \), \( b = -7 \), and \( c = -12 \). Calculating the discriminant gives us \( \Delta = 49 - 48 \), equalling 1, indicating two real roots.
Quadratic Formula
The quadratic formula is a powerful tool that provides the solutions, or roots, of a quadratic equation \( ax^2 + bx + c = 0 \). Given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
it employs the calculated discriminant under the square root sign. By substituting the values of \( a = -1 \), \( b = -7 \), and \( c = -12 \), the roots are found to be \( x_1 = -3 \) and \( x_2 = -4 \).
This method is essential in finding the precise intersection points where two curves meet, as was done in this exercise.
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
it employs the calculated discriminant under the square root sign. By substituting the values of \( a = -1 \), \( b = -7 \), and \( c = -12 \), the roots are found to be \( x_1 = -3 \) and \( x_2 = -4 \).
This method is essential in finding the precise intersection points where two curves meet, as was done in this exercise.
Viewing Rectangle
A viewing rectangle is the window through which we observe the graphs of equations on a coordinate plane. It defines the limits within which the graphs are plotted and analyzed. For any set of equations, the viewing rectangle is crucial in determining the relevancy of intersection points or the visibility of graph behavior in specific regions.
In this problem, the interval for \( x \) is \([-6, 2]\), and the interval for \( y \) is \([-5, 20]\). Both intersection points \((-3, 15)\) and \((-4, 6)\) fall within this rectangle, confirming they are meaningful solutions in the given context.
In this problem, the interval for \( x \) is \([-6, 2]\), and the interval for \( y \) is \([-5, 20]\). Both intersection points \((-3, 15)\) and \((-4, 6)\) fall within this rectangle, confirming they are meaningful solutions in the given context.
Intersection Points
To find the intersection points of two graphs, we equate their equations and solve for the variable, typically \( x \). The resulting values indicate where on the \( x \)-axis both curves meet.
In the exercise, setting \( 6 - 4x - x^2 \) equal to \( 3x + 18 \) and solving for \( x \), we found \( x_1 = -3 \) and \( x_2 = -4 \). These \( x \)-values give us the \( y \)-coordinates by substituting back into either original equation. Thus, the intersection points are \((-3, 15)\) and \((-4, 6)\).
Finding intersection points is crucial in understanding how the behaviors of different functions interact and overlap in the given space.
In the exercise, setting \( 6 - 4x - x^2 \) equal to \( 3x + 18 \) and solving for \( x \), we found \( x_1 = -3 \) and \( x_2 = -4 \). These \( x \)-values give us the \( y \)-coordinates by substituting back into either original equation. Thus, the intersection points are \((-3, 15)\) and \((-4, 6)\).
Finding intersection points is crucial in understanding how the behaviors of different functions interact and overlap in the given space.
