Problem 84
Question
Find the equation of the parabola $$y=a x^{2}+b x+c$$ that passes through the points. To verify your result, use a graphing utility to plot the points and graph the parabola. $$(-2,9),(-1,0),(1,6)$$
Step-by-Step Solution
Verified Answer
The equation of the parabola that passes through the points (-2,9), (-1,0), and (1,6) is \(y = 2x^2 - x + 1\).
1Step 1: Substitute the given points into the parabola equation
Start by substituting the given points into the equation. For the point (-2,9), we get the equation \(9 = 4a -2b + c\). For the point (-1,0), we get \(0 = a - b + c\). And for the point (1,6), we get \(6 = a + b + c\). This results in the three equations: \[\begin{align*} 4a - 2b + c &= 9\\ a - b + c &= 0\\ a + b + c &= 6 \end{align*}\]
2Step 2: Solve the System of Equations
You will now solve these three equations simultaneously. Adding the last two equations should get rid of the variable b and results in \(2a + 2c = 6\), which simplifies further to \(a + c = 3\).Subtracting this from the first equation gives \(3a - 2b = 6\), which simplifies to \(a = 2\) and \(b = -1\). Finally, plugging \(a = 2\) into \(a + c = 3\) should give \(c = 1\)
3Step 3: Substitute Back to get the Parabola Equation
Substitute the calculated values of \(a = 2\), \(b = -1\), and \(c = 1\) back into the standard equation \(y = ax^2 + bx + c\) to give the equation of the parabola that passes through the three points as \(y = 2x^2 - x + 1\)
Key Concepts
System of EquationsQuadratic FunctionsGraphing Utility
System of Equations
When we are given multiple algebraic equations and we need to find the values of the variables that make all the equations true simultaneously, we are dealing with a system of equations. In the context of finding the equation of a parabola that passes through given points, we set up a system where each equation corresponds to a point on the parabola.
By substituting the coordinates of the points into the standard form of the quadratic equation, we obtain a system with three equations. Solving this system allows us to find the coefficients of the parabola, which are critical for describing its shape and position. There are various methods to solve systems of equations, including substitution, elimination, and using matrices. In our textbook example, the elimination method is used to simplify and reduce the equations, leading to finding the values of the coefficients a, b, and c that define the unique parabola through the given points.
By substituting the coordinates of the points into the standard form of the quadratic equation, we obtain a system with three equations. Solving this system allows us to find the coefficients of the parabola, which are critical for describing its shape and position. There are various methods to solve systems of equations, including substitution, elimination, and using matrices. In our textbook example, the elimination method is used to simplify and reduce the equations, leading to finding the values of the coefficients a, b, and c that define the unique parabola through the given points.
Quadratic Functions
Quadratic functions are a type of polynomial function with the highest degree of 2 and are represented by the equation \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is nonzero. The graph of a quadratic function is a parabola, which can open upwards as a cup or downwards as a cap, depending on the sign of the coefficient \( a \).
The attributes of a parabola include the vertex, the axis of symmetry, the direction of opening, the maximum or minimum value, and the intercepts. The process of finding the equation of our parabola involved using three points that lie on it which, when solved, gave us \( a = 2 \), \( b = -1 \) and \( c = 1 \) for our parabola's equation: \( y = 2x^2 - x + 1 \). This specific equation tells us that our parabola opens upward due to the positive \( a \) coefficient, crosses the y-axis at \( c = 1 \) and has a vertex that can also be calculated using these coefficients.
The attributes of a parabola include the vertex, the axis of symmetry, the direction of opening, the maximum or minimum value, and the intercepts. The process of finding the equation of our parabola involved using three points that lie on it which, when solved, gave us \( a = 2 \), \( b = -1 \) and \( c = 1 \) for our parabola's equation: \( y = 2x^2 - x + 1 \). This specific equation tells us that our parabola opens upward due to the positive \( a \) coefficient, crosses the y-axis at \( c = 1 \) and has a vertex that can also be calculated using these coefficients.
Graphing Utility
A graphing utility is an essential tool in mathematics, particularly when dealing with functions and their graphical representations. It is often used to visualize the shape and key points of a graph. In the educational context, graphing utilities serve as a means to confirm the accuracy of algebraic solutions.
After deriving the equation \( y = 2x^2 - x + 1 \) algebraically, a graphing utility helps us to plot this function and observe whether the graph indeed passes through the points \( (-2,9), (-1,0), and (1,6) \). This visual confirmation strengthens our understanding of both the algebraic process and the behavior of quadratic functions. Students are encouraged to use graphing utilities not only to check their work but to also explore the effects of changing coefficients \( a \) , \( b \), and \( c \) on the shape and location of the parabola.
After deriving the equation \( y = 2x^2 - x + 1 \) algebraically, a graphing utility helps us to plot this function and observe whether the graph indeed passes through the points \( (-2,9), (-1,0), and (1,6) \). This visual confirmation strengthens our understanding of both the algebraic process and the behavior of quadratic functions. Students are encouraged to use graphing utilities not only to check their work but to also explore the effects of changing coefficients \( a \) , \( b \), and \( c \) on the shape and location of the parabola.
Other exercises in this chapter
Problem 83
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