Problem 49 - Precalculus with Limits

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. $$( 0,0 ) , ( 2 , - 2 ) , ( 4,0 )$$

Step-by-Step Solution

Answer

The equation of the parabola that passes through the points (0,0), (2,-2), and (4,0) is \(y = -x\).

Step 1: Set up the system of equations based on given points

The given points are (0,0), (2,-2) and (4,0). We substitute these points into the general equation of the parabola \( y = ax^2 + bx + c \) to create the following system of equations: 1) Substituting (0,0) into the equation: \( 0 = 0*a + 0*b + c => c = 0 \)2) Substituting (2,-2) into the equation: \( -2 = 4*a + 2*b + 0 \)3) Substituting (4,0) into the equation: \( 0 = 16*a + 4*b + 0 \)

Step 2: Solve the system of equations

We solve the system created in step 1 to find the values for a, b and c.1) From the first equation we know that c = 0.2) For the second equation: \( -2 = 4*a + 2*b => -1 = a + b. \)3) For the third equation: And \(0 = 16*a + 4*b => 0 = 4*a + b. \)Solving these simultaneous equations will give us the value of a and b.

Step 3: Find the values of a and b

To find the values of a and b, we can subtract the second equation from the third which gives: \(b = -1.\) Substitute b = -1 into the equation \( -1 = a + b \) would then give us the value of a = 0.

Step 4: Write the equation of the parabola

Substitute the obtained values of a, b, and c into the general equation of the parabola \(y = ax^2 + bx + c\), We get the equation of the parabola to be \(y = 0*x^2 + (-1)*x + 0 => y = -x \).

Key Concepts

System of Equations

When finding the equation of a parabola, a common method involves working with a system of equations. A system of equations is a set of two or more equations with the same variables. In our case, the variables are the coefficients of the parabola's equation, which are typically denoted as 'a', 'b', and 'c'.

To solve for the parabola that goes through the points \( (0,0), (2,-2), (4,0) \), we set up a system of equations by plugging these points into the general parabola equation, \( y = ax^2 + bx + c \). Each point will give us a separate equation, forming a system that, once solved, reveals the values of 'a', 'b', and 'c'. It's important to handle each equation with precision - a single error can lead us away from the correct parabolic curve.

After substituting the points, we end up with a system like this:

• \( 0 = a \cdot 0^2 + b \cdot 0 + c \)
• \( -2 = a \cdot 2^2 + b \cdot 2 + c \)
• \( 0 = a \cdot 4^2 + b \cdot 4 + c \)

This system is the cornerstone of finding our coefficients. Remember, solving a system often entails methods like substitution, elimination, or using matrices. For our example, we can use substitution since one of the equations directly gives us the value of 'c'.

Solving for Coefficients

With the system of equations set from the previous step, we can solve for the coefficients 'a', 'b', and 'c'. This involves manipulation of the system to isolate each coefficient. Often, coefficients 'a' and 'b' are of the most interest, since 'c' can be straightforwardly determined if one of the points has a zero x-value, as in our exercise.

Here are the steps the solution took:

• The first point \( (0,0) \) indicates that the parabola passes through the origin, simplifying our equation to \( c = 0 \).
• Next, we use the second point, \( (2,-2) \), to form an equation \( -2 = 4a + 2b \).
• The third point, \( (4,0) \), gives us \( 0 = 16a + 4b \).

Then, we can use the method of elimination or substitution to solve for 'a' and 'b'. For instance, if we subtract the second equation from the third one modified to have the same coefficient for 'b', we can find 'b' and subsequently 'a'. The order in which you tackle the coefficients can be flexible, but precision is crucial in each algebraic step to ensure accurate values.

Graphing Parabolas

Graphing parabolas is essential to visually verify the solution of a parabolic equation. Once we have the equation \( y = ax^2 + bx + c \), graphing it can help us see if it passes through the given points. Moreover, graphing provides an intuitive understanding of the curve's shape and direction.

In our specific exercise, after finding the coefficients, our parabola equation simplifies to \( y = -x \). While this is not a conventional parabolic shape - it's actually a straight line - it's still a special case of a parabola. When graphed, it should intersect the points \( (0,0), (2,-2), \) and \( (4,0) \). Graphing utilities or software can be handy for plotting these points and the resulting curve. It's important to keep in mind that while graphing utilities are helpful, a fundamental understanding of how to draw parabolas based on their equations, orientations, vertex positions, and axes of symmetry is a vital skill in mathematics.