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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. $$(0,3),(1,4),(2,3)$$

Short Answer

Expert verified
The equation of the parabola that passes through the points (0,3), (1,4) and (2,3) is \(y = x^2 + 3\).

Step by step solution

01

Formulate the equations

Substituting the coordinates from each point into the equation of the parabola \(y = ax^2 + bx + c\) gives us three equations. For (0,3), we have \(3 = a(0)^2 + b(0) + c\), simplifying to \(c = 3\). For (1,4), we have \(4 = a(1)^2 + b(1) + c\), which simplifies to \(4 = a + b + c\). For (2,3), we have \(3 = a(2)^2 + b(2) + c\), which simplifies to \(3 = 4a + 2b + c\).
02

Solve the equations

We now have a system of three equations: \(c = 3\), \(4 = a + b + c\), and \(3 = 4a + 2b + c\). Substituting \(c = 3\) into the other two gives \(4 = a + b + 3\) and \(3 = 4a + 2b + 3\). Solving these for a and b gives \(a = 1\) and \(b = 0\).
03

Formulate the equation of the parabola

Substituting a, b, and c into \(y = ax^2 + bx + c\) gives the equation of the parabola as \(y = x^2 + 3\).

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Key Concepts

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

Quadratic Functions
Quadratic functions are a type of polynomial function that's characterized by terms up to the second degree. The general form of a quadratic function is \( y = ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are coefficients where \( a eq 0 \) because otherwise, the equation would be linear, not quadratic.
Quadratic functions present unique properties in mathematics:
  • They graph into a U-shaped curve known as a parabola.
  • The "a" coefficient determines the direction of the parabola: if \( a > 0 \), it opens upwards, and if \( a < 0 \), it opens downwards.
  • The vertex form of the parabola can also be expressed to understand the turning point of the curve, but in the standard form, finding the vertex requires completing the square or using vertex formulas.
Quadratics appear naturally in many real-world contexts, such as in physics to describe projectile motion. Understanding the role of each coefficient helps in modeling and interpreting these scenarios correctly.
System of Equations
A system of equations consists of multiple equations that share the same set of unknowns and are solved simultaneously. In the context of quadratic functions, solving a system of equations involves finding the coefficients \( a \), \( b \), and \( c \) in the parabola's equation that fits through given points.
The process includes:
  • Substituting each given point into the general quadratic equation \( y = ax^2 + bx + c \).
  • Formulating individual equations from each substitution.
  • Using substitution or elimination methods to solve for the unknown coefficients.
In our example, we used a system consisting of three equations derived from the points (0,3), (1,4), and (2,3). After simplifying, substituting, and solving, we determined \( a = 1 \), \( b = 0 \), and \( c = 3 \), leading to the quadratic equation \( y = x^2 + 3 \). Mastering these skills is essential for tackling various mathematical and real-world problems.
Graphing Utilities
Graphing utilities are digital tools, such as graphing calculators or software, that allow us to visualize mathematical functions, such as quadratic equations. These tools are crucial for confirming algebraic solutions visually.
Using graphing utilities involves these steps:
  • Input the quadratic equation into the tool directly.
  • Plot given points to ensure the curve passes through them.
  • Adjust view settings if needed to get a clear view of the parabola.
For the exercise, once you have the quadratic equation \( y = x^2 + 3 \), entering it into a graphing utility will help confirm that it passes through points (0,3), (1,4), and (2,3). By providing a visual representation, graphing utilities strengthen understanding and reassure that the algebraic equation aligns with the given data points.

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