Problem 63 - A Graphical Approach to College Algebra

Question

Find the equation of the parabola with vertical axis that passes through the data points shown or specified. Check your answer. $$(0,1),(1,0),(2,-5)$$

Step-by-Step Solution

Answer

The equation is \(y = -2x^2 + x + 1\).

Step 1: Understand the Problem

We have three points: \((0, 1), (1, 0), (2, -5)\), and we need to find the equation of a parabola with a vertical axis that passes through them. The general form of a vertical parabola is \(y = ax^2 + bx + c\). We need to determine the coefficients \(a\), \(b\), and \(c\) such that the parabola passes through all three points.

Step 2: Create Equations Using the Points

Substitute each point into the equation \(y = ax^2 + bx + c\) to create a system of equations.- For \((0,1)\): \[1 = c\]- For \((1,0)\): \[0 = a(1)^2 + b(1) + c = a + b + c\]- For \((2,-5)\): \[-5 = a(2)^2 + b(2) + c = 4a + 2b + c\]

Step 3: Solve the System of Equations

Substitute \(c = 1\) from the first equation into the second and third equations:- \(0 = a + b + 1\) gives us \(a + b = -1\).- \(-5 = 4a + 2b + 1\) simplifies to \(4a + 2b = -6\).Now, solve the system:1. \(a + b = -1\)2. \(4a + 2b = -6\)From equation 1, \(b = -1 - a\).Substitute in equation 2:\[4a + 2(-1 - a) = -6\]\[4a - 2 - 2a = -6\]\[2a - 2 = -6\]\[2a = -4\]\[a = -2\]Now substitute \(a = -2\) back into \(b = -1 - a\):\[b = -1 - (-2) = 1\]

Step 4: Write the Final Equation

Remember \(c = 1\). Using \(a = -2\), \(b = 1\), and \(c = 1\), the equation of the parabola is:\[y = -2x^2 + x + 1\]

Step 5: Verify the Solution

Check that the equation \(y = -2x^2 + x + 1\) satisfies all three points:- For \((0, 1)\): \(-2(0)^2 + 0 + 1 = 1\). Matches.- For \((1, 0)\): \(-2(1)^2 + 1 + 1 = 0\). Matches.- For \((2, -5)\): \(-2(2)^2 + 2 + 1 = -5\). Matches.All points satisfy the equation, confirming our solution is correct.

Key Concepts

Understanding System of Equations

When it comes to finding the equation of a parabola that passes through certain points, we often rely on a system of equations. This is because a parabola's equation has three unknown coefficients —\(a\), \(b\), and \(c\)— in its general form, \(y = ax^2 + bx + c\). To find these coefficients, we need at least three unique points that the parabola passes through. Each point provides one equation in the system. For instance:

• Point \((0, 1)\) gives us the equation \(1 = c\).
• Point \((1, 0)\) gives \(0 = a + b + c\).
• Point \((2, -5)\) gives \(-5 = 4a + 2b + c\).

These equations together form a system that we can solve to find the values of \(a\), \(b\), and \(c\). The beauty of a system of equations is its ability to pinpoint exact values for these variables, given enough information.

Quadratic Function Details

The quadratic function is a very interesting and widely used mathematical concept. It represents a parabola in its graph form, and its standard equation is \(y = ax^2 + bx + c\). Here:

• The \(ax^2\) term determines the opening direction and the wideness of the parabola. If \(a\) is positive, the parabola opens upwards; if negative, it opens downwards.
• The \(bx\) term allows the parabola to shift left or right depending on its interaction with the \(a\) coefficient.
• The constant \(c\) shifts the parabola up or down, indicating the y-intercept of the parabola.

Understanding these components can help you graph the quadratic function effectively and anticipate how changes in \(a\), \(b\), or \(c\) affect the graph's shape and position.

Basic Steps in Solving Equations

Solving equations, like those formed in our system to find a parabola's equation, is primarily about isolating variables to find their values. Let's see how we approach solving such a system:

• Start with substituting known values: From \(1 = c\), we know \(c = 1\).
• Replace \(c\) in the other equations, reducing the number of unknowns.
• Solve the resulting simpler equations: \(0 = a + b + 1\) simplifies using \(c = 1\) to \(a + b = -1\).
• Keep substituting and eliminating: Use \(b = -1 - a\) in \(4a + 2b = -6\) to arrive at a single variable equation.
• Finally, find each variable: Solving \(2a = -4\) yields \(a = -2\), and substituting back gives \(b = 1\).

Through these systematic steps, solving equations becomes manageable. This methodical approach ensures precision and helps verify correctness by checking if these values satisfy all the initial equations.