Problem 54 - 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. $$ ( - 2 , - 3 ) , ( - 1,0 ) , \left( \frac { 1 } { 2 } , - 3 \right) $$

Step-by-Step Solution

Answer

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

Step 1: Substituting the given points into the equation

We start by substituting the coordinates of the three points given into the equation \(y = ax^2 + bx + c\) to get three equations \n1. Substituting (-2,-3), we get: -3 = 4a - 2b + c2. Substituting (-1,0), we get: 0 = a - b + c3. Substituting (0.5,-3), we get: -3 = 0.25a + 0.5b + c\nSo, we have the following system of equations:\n4a - 2b + c = -3 \na - b + c = 0 \n0.25a + 0.5b + c = -3

Step 2: Solve the system of equations

Now we solve this system of equations. This can be done using various methods, such as substitution, elimination or matrix method. Here's how we can do it using the elimination method:\nTo make it easier, we can multiply the third equation by 4 to get rid of decimal numbers, which gives: a + 2b + 4c = -12 \nNext we'll subtract the second equation from the first and from the new third equation which gives 3a - b = -3 and -b + 3c = -12.\nNow we see that the value of b can be obtained from the second equation and is b = -3 - 3a.\nThen, substituting b's value into the third equation gives us c = 4.\nFinally, substituting c = 4 into the second equation gives us a = -1 and b = 2.

Step 3: Form the equation of the parabola

Substitute the values of a, b, and c into the equation \(y = ax^2 + bx + c\) to find the equation of the parabola which is \(y = -x^2 + 2x + 4\)

Key Concepts

Quadratic Equations

Quadratic equations are fundamental in mathematics and are often used to describe parabolas. The general form of a quadratic equation is \(y = ax^2 + bx + c\). This represents a parabola on a graph. The "a" term determines the direction and width of the parabola. When "a" is positive, the parabola opens upwards; when negative, it opens downwards. The larger the absolute value of "a," the narrower and steeper the parabola. The "b" term affects the position of the vertex but not its symmetry. Finally, the "c" term represents the y-intercept, which is where the parabola crosses the y-axis.

To find the specific quadratic equation of a parabola given several points, like in the exercise, we plug these points into the general form. This translates to creating a system of equations that we must solve to find the specific values of a, b, and c. Once we determine these values, we can fully describe the exact position and shape of the parabola on the graph.

Systems of Equations

A system of equations consists of multiple equations that are all satisfied by the same set of values. In the given exercise, we derived a system from plugging the provided points into the quadratic equation. Solving this system helps us find the specific coefficients (a, b, c) of the parabola's equation.

Each equation in the system corresponds to a point on the parabola. Therefore, solving the system correctly ensures that the resulting parabola passes through all given points. Sometimes systems may have one solution, no solution, or infinitely many solutions. In geometrical problems like this, we seek one specific set of solutions that works for our curve.
Using techniques like elimination or substitution, we often aim to simplify and solve these equations step-by-step. Successfully solving the system gives us the exact formula for our parabola. Understanding and manipulating systems of equations is crucial for achieving accurate and meaningful results in complex problems.

Elimination Method

The elimination method is an efficient algebraic technique to solve systems of equations. It involves adding or subtracting equations to cancel out one or more variables. This helps us to solve for the remaining unknowns.

In the exercise, we used the elimination method after setting up a system of three equations from the parabola's points. By manipulating the equations, we eliminate variables to simplify and solve the system. For instance, you often need to add, subtract, or multiply each equation to line up similar terms for cancellation.
The beauty of elimination lies in its simplicity and power to reduce complex problems. Here, it enabled us to find values for a, b, and c efficiently. When used correctly, the elimination method swiftly solves equations that might otherwise appear challenging.
Mastering this method is invaluable for anyone looking to solve linear and quadratic systems efficiently.