Problem 51
Question
Finding the Equation of a Parabola In Exercises \(49 - 54 ,\) 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. $$ ( 2,0 ) , ( 3 , - 1 ) , ( 4,0 ) $$
Step-by-Step Solution
Verified Answer
The equation of the parabola that passes through the given points is \(y = x^2 - 6x\).
1Step 1: Substituting Points into the Formula
Substituting each given point into the equation \(y = ax^2 + bx + c\) will give us a system of three equations that we can solve for a, b and c. The three equations we get are: \(0 = 4a + 2b + c\) (from point (2,0)), \(-1 = 9a + 3b + c\) (from point (3,-1)), \(0 = 16a + 4b + c\) (from point (4,0)).
2Step 2: Solving the System of Equations
We can solve these equations using substitution or elimination method. But the easiest way would be to perform pairwise subtractions of the equations. Subtract the first equation from the second, and the first equation from the third. This will eliminate 'c' and give us two new equations: \(-1 = 5a + b\) and \(0 = 12a + 2b\). On simplifying the second equation, we get \(0 = 6a + b\). Subtracting this new second equation from the first new equation, we get \(-1 = -a\), or \(a = 1\). Substituting a = 1 into our new second equation we get \(0 = 6 + b\) or \(b = -6\). Substituting a = 1 and b = -6 into our original first equation will give us \(c = 0\).
3Step 3: Forming the equation of parabola
We can form the equation of the parabola using the values a = 1, b = -6 and c = 0. It gives us \(y = x^2 - 6x\).
Key Concepts
Systems of EquationsQuadratic FunctionsGraphing Parabolas
Systems of Equations
Understanding systems of equations is essential when finding the equation of a parabola that passes through given points. A system of equations consists of two or more equations with a common set of variables. In our exercise, we had three points which led to three equations with variables a, b, and c representing the coefficients in the parabolic equation of the form
\(y = ax^2 + bx + c\).
Each of the given points generated a unique equation when substituted into the parabolic form. This resulted in a system of equations with the solutions for a, b, and c defining the specific parabola that crosses through those points. Solving a system can be approached by various methods such as substitution, elimination, matrix operations, or graphically. In the exercise, the elimination method was used effectively to minimize the variables step-by-step, making the system easier to solve.
\(y = ax^2 + bx + c\).
Each of the given points generated a unique equation when substituted into the parabolic form. This resulted in a system of equations with the solutions for a, b, and c defining the specific parabola that crosses through those points. Solving a system can be approached by various methods such as substitution, elimination, matrix operations, or graphically. In the exercise, the elimination method was used effectively to minimize the variables step-by-step, making the system easier to solve.
Quadratic Functions
Quadratic functions are mathematical expressions of the second degree, typically in the form of
\(y = ax^2 + bx + c\),
where a, b, and c are coefficients and \(a eq 0\). The graph of a quadratic function is a parabola, which can either open upwards or downwards based on the sign of coefficient 'a'.
The exercise demonstrates how the coefficients of a quadratic function determine its specific shape and position on the Cartesian plane. By altering these values, we adjust the parabola's width, direction, and location. The solution identifies these coefficients through the system of equations created by known points on the graph, reinforcing the relationship between a quadratic function's algebraic form and its geometric representation.
\(y = ax^2 + bx + c\),
where a, b, and c are coefficients and \(a eq 0\). The graph of a quadratic function is a parabola, which can either open upwards or downwards based on the sign of coefficient 'a'.
The exercise demonstrates how the coefficients of a quadratic function determine its specific shape and position on the Cartesian plane. By altering these values, we adjust the parabola's width, direction, and location. The solution identifies these coefficients through the system of equations created by known points on the graph, reinforcing the relationship between a quadratic function's algebraic form and its geometric representation.
Graphing Parabolas
Graphing parabolas is a visual way to comprehend the behavior of quadratic functions. A parabola is a symmetrical curve that has a vertex as its highest or lowest point. By plotting a parabola, one can observe its axis of symmetry, width, and direction of opening - all of which are determined by the quadratic equation's coefficients.
To sketch the parabola for the equation found in our exercise, \(y = x^2 - 6x\), we can start by identifying its vertex and then plot more points or apply the vertex form to understand its symmetry. Once some key points are plotted, the shape of the parabola becomes clear, allowing us to visualize the solution. Graphing utilities or software can provide a more precise graph and are excellent tools for verifying the accuracy of the algebraic solution.
To sketch the parabola for the equation found in our exercise, \(y = x^2 - 6x\), we can start by identifying its vertex and then plot more points or apply the vertex form to understand its symmetry. Once some key points are plotted, the shape of the parabola becomes clear, allowing us to visualize the solution. Graphing utilities or software can provide a more precise graph and are excellent tools for verifying the accuracy of the algebraic solution.
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