Problem 19 - Algebra and Trigonometry

Question

Find the quadratic function \(y=a x^{2}+b x+c\) whose graph passes through the given points. $$ (-1,6),(1,4),(2,9) $$

Step-by-Step Solution

Answer

The quadratic function that passes through the points (-1,6), (1,4), and (2,9) is \(y = x^2 - x + 6\)

Step 1: Substitute the given points into the quadratic function's general form

When faced with the first point (-1,6), the output \(y\), is 6 when the input \(x\) is -1. We plug these values into the general quadratic equation to get: \(6=a(-1)^2 + b(-1) + c \). Repeating the same process for the other points gives: \(4=a(1)^2 + b(1) + c\) and \(9=a(2)^2 + b(2) + c\). Thus, our system of equations is: \[ \begin{align*} 6 = a - b + c,\ 4 = a + b + c,\ 9 = 4a + 2b + c. \end{align*} \]

Step 2: Solve the system of equations

Subtract the first equation from the second and from the third to get rid of \(c\) and \(b\). This will result in the following set of equations: \[ \begin{align*} -2 = 3a + b,\ 3 = 3a + b. \end{align*} \] As the two equations are equal, we can deduce that \(a = 1\). Substituting \(a = 1\) back to the first and second equations in step 1, we obtain \(b = -1\) and \(c = 6\) respectively.

Step 3: Formulate the final quadratic function

Now, substitute the values of \(a\), \(b\) and \(c\) back into the general form of the quadratic function to get the final function: \(y = x^2 - x + 6\).

Key Concepts

Graphing

Graphing a quadratic function helps visualize its behavior, such as its vertex, axis of symmetry, and whether it opens upwards or downwards. In the quadratic function form \(y = ax^2 + bx + c\), the sign of \(a\) dictates the direction. If \(a > 0\), it opens upwards, resembling a smiley face. If \(a < 0\), it opens downwards, resembling a frowny face. The vertex
– which represents either the maximum or minimum point – can be found using the formula \(-\frac{b}{2a}\).

• This gives the x-coordinate of the vertex.
• Plugging this into the function finds the y-coordinate.

Graphically representing a quadratic involves plotting several points and observing
their symmetry around the vertex.
Since we were given specific points that lie on the graph, it guides us in finding the expression of the function.
To efficiently graph a quadratic function, especially using points like \((-1,6), (1,4), (2,9)\),
we substitute these points into the function's standard form and adjust \(a\), \(b\), and \(c\) accordingly.

Systems of Equations

A system of equations is a collection of two or more equations with the same set of variables. Solving systems allows us to find solutions that satisfy all included equations. In this case, finding a quadratic function through specified points involves setting up a system with three equations. We use each point in the form \((x,y)\) and the structure \(y = ax^2 + bx + c\).
Thus, we have:

• \(6 = a - b + c\)
• \(4 = a + b + c\)
• \(9 = 4a + 2b + c\)

These represent a system of linear equations. Solving them yields values for \(a\), \(b\), and \(c\) that
allow us to replace variables with constants, forming our specific quadratic equation.
Determining these unknowns can be done using techniques like substitution or elimination maintaining mathematical consistency.

Substitution Method

The substitution method is a way of solving systems of equations by replacing one variable with an expression
or value derived from another equation. In our quadratic function exercise, we started with:

• \(6 = a - b + c\)
• \(4 = a + b + c\)
• \(9 = 4a + 2b + c\)

We manipulated these to eliminate variables step-by-step.
Subtracting appropriate equations from one another simplified the system,
leading to smaller, more direct equations like \(-2 = 3a + b\) and \(3 = 3a + b\).
Both equations help us identify that \(a = 1\). By substituting this back, we found \(b = -1\) and \(c = 6\).
Substitution allows for direct solutions by reducing the number of variables,
effectively breaking down the system until only constants and one variable remain.

Formulating Equations

Formulating equations is the process of setting mathematical expressions to represent real-world scenarios or given data. Here, we derived our three simultaneous equations from
inserted points into the quadratic function format.
Given points \((-1,6), (1,4), (2,9)\), each provides a unique equation:

• 6 for \((x,y) = (-1,6)\) becoming \(6 = a(-1)^2 + b(-1) + c\).
• 4 for \((x,y) = (1,4)\) becoming \(4 = a(1)^2 + b(1) + c\).
• 9 for \((x,y) = (2,9)\) becoming \(9 = a(2)^2 + b(2) + c\).

These formulated equations are the foundation for determining \(a\), \(b\), and \(c\).
Formulating effectively involves transforming given data into clear mathematical statements.
These statements help in breaking down complex problems into manageable parts.