Question

Find the function \(y=a x^{2}+b x+c\) whose graph contains the points \((1,2),(-2,-7),\) and (2,-3).

Short answer

The function is \(y = -2x^2 + x + 3\).

Step-by-step solution

- Set up the system of equations

Given the quadratic function \(y = ax^2 + bx + c\) and the points (1, 2), (-2, -7), and (2, -3), substitute each point into the equation to create a system of equations.For (1, 2):\[2 = a(1)^2 + b(1) + c\]\[2 = a + b + c\]For (-2, -7):\[-7 = a(-2)^2 + b(-2) + c\]\[-7 = 4a - 2b + c\]For (2, -3):\[-3 = a(2)^2 + b(2) + c\]\[-3 = 4a + 2b + c\]

- Write the system of equations

The system of equations is:1. \(a + b + c = 2\)2. \(4a - 2b + c = -7\)3. \(4a + 2b + c = -3\)

- Eliminate one variable

Subtract equation 1 from equation 2:\[(4a - 2b + c) - (a + b + c) = -7 - 2\]\[3a - 3b = -9\]Simplify:\[a - b = -3\] (Equation 4)Subtract equation 1 from equation 3:\[(4a + 2b + c) - (a + b + c) = -3 - 2\]\[3a + b = -5\] (Equation 5)

- Solve the system

Add equation 4 and equation 5:\[(a - b) + (3a + b) = -3 + (-5)\]\[4a = -8\]\[a = -2\]Now, substitute \(a = -2\) back into equation 4:\[-2 - b = -3\]\[b = 1\]Next, substitute \(a = -2\) and \(b = 1\) back into equation 1:\[-2 + 1 + c = 2\]\[c = 3\]

- Write the final function

Substitute \(a = -2\), \(b = 1\), and \(c = 3\) back into the quadratic function form:\[y = -2x^2 + x + 3\]

Key concepts

system of equations

When we need to find a function that fits multiple points, we often rely on creating a system of equations.
A system of equations is a set of two or more equations with the same set of variables.
In this exercise, the system of equations is derived from substituting the given points into the general form of the quadratic function, which is \( y = ax^2 + bx + c \).
For each point like (1, 2), (-2, -7), and (2, -3), we substitute and get separate equations:

• (1,2): \( 2 = a(1)^2 + b(1) + c \) becomes \( 2 = a + b + c \)
• (-2,-7): \( -7 = a(-2)^2 + b(-2) + c \) becomes \( -7 = 4a - 2b + c \)
• (2,-3): \( -3 = a(2)^2 + b(2) + c \) becomes \( -3 = 4a + 2b + c \)

This gives us a system of three equations:

• 1. \( a + b + c = 2 \)
• 2. \( 4a - 2b + c = -7 \)
• 3. \( 4a + 2b + c = -3 \)

The challenge is solving this system to find the values of \(a\), \(b\), and \(c\).

solving quadratic equations

To solve for the quadratic function, we need to eliminate variables step by step.
First, subtract equation 1 from equations 2 and 3 to eliminate \(c\).
From equation 2:

• \( (4a - 2b + c) - (a + b + c) = -7 - 2 \)
• which simplifies to \( 3a - 3b = -9 \)
• dividing by 3 results in \( a - b = -3 \)

Then, from equation 3:

• \( (4a + 2b + c) - (a + b + c) = -3 - 2 \)
• which simplifies to \( 3a + b = -5 \)

Now, we have a simpler system of two equations:

• 4. \( a - b = -3 \)
• 5. \( 3a + b = -5 \)

Adding equations 4 and 5 to eliminate \(b\):

• \( (a - b) + (3a + b) = -3 + (-5) \)
• results in \( 4a = -8 \)
• giving \( a = -2 \)

Using \( a = -2 \) to find \( b \) from equation 4:

• \( -2 - b = -3 \)
• thus \( b = 1 \)

Finally, substituting \( a = -2 \) and \( b = 1 \) into equation 1 to find \( c \):

• \( -2 + 1 + c = 2 \)
• solving gives \( c = 3 \)

algebra

Algebra is the branch of mathematics dealing with symbols and the rules for manipulating those symbols.
In this exercise, we manipulated symbols representing the variables \(a\), \(b\), and \(c\) to find a quadratic function.
The process involved:

• 1. Setting up equations by substituting given points into the quadratic form \( y = ax^2 + bx + c \).
• 2. Simplifying and rearranging these equations to make it easier to solve for the unknowns.
• 3. Using algebraic methods like substitution and elimination to find the values of \(a\), \(b\), and \(c\).

These methods are fundamental in algebra and are used not only for quadratic equations but also for higher-order polynomials and other types of equations.
Understanding these techniques strengthens your ability to solve complex mathematical problems.