Problem 29 - Precalculus

Question

Find the quadratic function passing through the points (-2,1),(1,4),(3,-2).

Step-by-Step Solution

Answer

The quadratic function is \( f(x) = -\frac{4}{5}x^2 -\frac{9}{5}x + \frac{1}{5} \).

Step 1: Formulate the general quadratic function

A quadratic function can be expressed generally as \( f(x) = ax^2 + bx + c \). Our objective is to find the coefficients \(a\), \(b\), and \(c\) such that the function passes through the given points.

Step 2: Apply the first point (-2, 1)

Substitute \(x = -2\) and \(f(x) = 1\) into the quadratic function. This gives us the equation \( 1 = a(-2)^2 + b(-2) + c \), simplifying to \( 4a - 2b + c = 1 \).

Step 3: Apply the second point (1, 4)

Using the point \((1, 4)\), substitute \(x = 1\) and \(f(x) = 4\) into the quadratic function. This results in the equation \( 4 = a(1)^2 + b(1) + c \), simplifying to \( a + b + c = 4 \).

Step 4: Apply the third point (3, -2)

For the point \((3, -2)\), substitute \(x = 3\) and \(f(x) = -2\) into the quadratic function. This produces the equation \( -2 = a(3)^2 + b(3) + c \), which simplifies to \( 9a + 3b + c = -2 \).

Step 5: Solve the system of equations

We now have a system of three linear equations: \(4a - 2b + c = 1\), \(a + b + c = 4\), and \(9a + 3b + c = -2\). Solve this system to find values of \(a\), \(b\), and \(c\). Begin by eliminating \(c\) from two pairs of equations.\Subtract the second equation from the first, yielding \(3a - 3b = -3\). Divide by 3 to simplify to \(a - b = -1\). Subtract the second equation from the third to get \(8a + 2b = -6\), then simplify to \(4a + b = -3\). Now solve the simplified equation system: \(a - b = -1\) and \(4a + b = -3\). Add the two equations to eliminate \(b\), resulting in \(5a = -4\), so \(a = -\frac{4}{5}\). Substitute \(a = -\frac{4}{5}\) into \(a - b = -1\) to find \(b = -\frac{9}{5}\). Finally, use \(a + b + c = 4\) to find \(c = \frac{1}{5}\).

Step 6: Write the quadratic function

With \(a = -\frac{4}{5}\), \(b = -\frac{9}{5}\), and \(c = \frac{1}{5}\), the quadratic function is \( f(x) = -\frac{4}{5}x^2 -\frac{9}{5}x + \frac{1}{5} \).

Key Concepts

Solving Systems of Equations

To find the quadratic function that passes through specific points, you first need to set up and solve a system of equations. This involves using the general quadratic equation, \(f(x) = ax^2 + bx + c\), and substituting each point into this equation to form multiple linear equations.
To illustrate, consider the points (-2, 1), (1, 4), and (3, -2). You substitute each point into the quadratic function to get three separate equations:

• First Point (-2, 1): \(4a - 2b + c = 1\)
• Second Point (1, 4): \(a + b + c = 4\)
• Third Point (3, -2): \(9a + 3b + c = -2\)

These three equations form a system that represents the relationships specified by the known points on the parabola. The next step is to solve these equations simultaneously. By using logical steps to eliminate one variable at a time, you can determine the values of \(a\), \(b\), and \(c\) that define your quadratic equation. It involves a bit of algebraic manipulation like adding or subtracting equations to isolate variables and simplify expressions. This solution process allows you to fit a curve precisely through given points, using the properties of parabolas.

Curve Fitting

Curve fitting is an essential concept in mathematics and data analysis. It involves finding a curve, often a line or parabola, that best fits a set of data points. The aim is to find an equation that models the data trend effectively. In the context of quadratic functions, curve fitting determines the coefficients \(a\), \(b\), and \(c\) such that the quadratic curve fits all given points exactly.
In our example of finding a quadratic function passing through (-2, 1), (1, 4), and (3, -2), the task was to determine the exact equation of the quadratic curve that fits these points. The approach involved using these points to derive a system of equations, which was then solved to find the coefficients for our quadratic equation.
Curve fitting is not only about being able to draw a curve through certain points. It's about understanding trends and making predictions. A well-fitted curve provides insights into the underlying structure of the data, making it a powerful tool in both statistics and real-world applications. Using mathematical techniques like those described ensures that the fitted curve accurately represents the selected data points.

Parabola

A parabola is a U-shaped curve that is symmetric around its vertex. It is the graph of a quadratic function of the form \(y = ax^2 + bx + c\). The coefficients \(a\), \(b\), and \(c\) determine the specific shape and position of the parabola.
In a quadratic equation where \(a > 0\), the parabola opens upwards, while if \(a < 0\), it opens downwards. The vertex is either a maximum or a minimum point, depending on the direction of opening.
For example, in solving the exercise to find the quadratic function through the given points, the resultant formula was \(f(x) = -\frac{4}{5}x^2 - \frac{9}{5}x + \frac{1}{5}\). Here, \(a = -\frac{4}{5}\) makes the parabola open downwards.

• Vertex: It's found by using the formula \(x = -\frac{b}{2a}\).
• Axis of symmetry: It's a vertical line through the vertex, given by \(x = -\frac{b}{2a}\).

Understanding parabolas is crucial for grasping the nature of quadratic functions as they apply to real-world phenomena, such as the path of projectiles, optimizing areas, or even in economics in profit forecasts. Hence, learning to work with parabolas by means of quadratic functions is a valuable mathematical skill.