Question

Find the function \(y=a x^{2}+b x+c\) whose graph contains the points \((1,-1),(3,-1),\) and (-2,14).

Short answer

The function is \( y = x^2 - 4x + 2\).

Step-by-step solution

Set up equations using given points

The function is given by the equation \( y = a x^2 + b x + c \). Plug the points \((1, -1)\), \((3, -1)\), and \((-2, 14)\) into this equation to form a system of equations. For point \((1, -1)\): \( -1 = a(1)^2 + b(1) + c \) which simplifies to \( a + b + c = -1 \). For point \((3, -1)\): \( -1 = a(3)^2 + b(3) + c \) which simplifies to \( 9a + 3b + c = -1 \).For point \((-2, 14)\): \( 14 = a(-2)^2 + b(-2) + c \) which simplifies to \( 4a - 2b + c = 14 \).

Write the system of equations

The system of equations derived from step 1 is as follows: 1. \( a + b + c = -1 \) 2. \( 9a + 3b + c = -1 \) 3. \( 4a - 2b + c = 14 \)

Subtract equations to eliminate c

To eliminate \(c\), subtract equation 1 from equation 2:\( (9a + 3b + c) - (a + b + c) \) which simplifies to \( 8a + 2b = 0 \) or \( 4a + b = 0 \). Next, subtract equation 1 from equation 3: \( (4a - 2b + c) - (a + b + c) \) which simplifies to \( 3a - 3b = 15 \) then \( a - b = 5 \).

Solve for a and b

We now have two simpler equations: \(4a + b = 0\) and \(a - b = 5\). Solve the first equation for \(b\): \( b = -4a \). Substitute \(b\) into the second equation: \( a - (-4a) = 5 \) which simplifies to \( 5a = 5\) thus \( a = 1\). Now substitute \(a\) back into equation \(4a + b = 0\): \( 4(1) + b = 0 \) which simplifies to \( b = -4 \).

Solve for c

Substitute \(a = 1\) and \(b = -4\) into p\(( a + b + c = -1 )\) which becomes \( 1 - 4 + c = -1 \). This simplifies to \( c = 2 \).

Write the final function

Combine the values for \(a\), \(b\), and \(c\): \[ y = 1x^2 - 4x + 2 \].

Key concepts

System of Equations

When dealing with multiple unknowns, a system of equations can be utilized to find consistent solutions for each variable. In this case, the quadratic function passes through specific points:

• (1, -1)
• (3, -1)
• (-2, 14)

Let's create equations from these points, remember from the standard quadratic equation: \( y = ax^2 + bx + c \). The main goal is to form and solve equations to find the unknowns \(a\), \(b\), and \(c\). This system lets us work with predictable relationships and seek out consistent variable values that satisfy all the given conditions.

Solving for Variables

Once the system of equations is set up, the next step is to solve for variables \(a\), \(b\), and \(c\). Here, we eliminate one variable at a time. For this specific exercise: We subtract equations to eliminate the variable \(c\): \( (9a + 3b + c) - (a + b + c) = 8a + 2b = 0 \) that further simplifies to \( 4a + b = 0 \). Likewise, we subtract another set to get \( a - b = 5 \). It’s clear now, the task is simplified to solving these new equations:

• \( 4a + b = 0 \)
• \( a - b = 5 \)

Solving these, we start with one equation to find a variable and substitute back to find others. This chain-solving strategy helps in reducing complex equations into more manageable forms.

Graphing Quadratic Equations

Quadratic functions are frequently represented by parabolas in a graph. Graphing these helps in visualizing the function's roots, vertex, and overall shape. Let’s consider our obtained solution: \( y = x^2 - 4x + 2 \). To graph this:

• Find the vertex
• Locate the axis of symmetry
• Identify x-intercepts (roots)
• Determine the y-intercept

The vertex can be derived using \( x = -\frac{b}{2a} \). Substituting \( a=1 \) and \( b=-4 \), we get \( x = 2 \). By inserting \( x=2 \) back into the equation, we calculate the vertex as \( y = -2 \). The parabola graph will curve upwards since the coefficient of \( x^2 \) is positive.

Parabola

A parabola is the graphical representation of a quadratic function. It’s crucial to understand its properties:

• The vertex is the highest or lowest point depending on if it opens upwards or downwards.
• The axis of symmetry is a vertical line passing through the vertex, splitting the parabola into two symmetrical parts.
• The direction of the opening is determined by the sign of the coefficient of \( x^2 \): positive opens upwards while negative opens downwards.

For our function \( y = x^2 - 4x + 2 \),

• The vertex is at (2, -2).
• The axis of symmetry is the line \( x = 2 \).
• The parabola opens upwards as the coefficient of \( x^2 \) is 1 (positive).

By graphing the parabola, you can analyze key features and present a clear visual understanding of quadratic functions.