Problem 21 - Intermediate Algebra

Question

The graph of each equation is a parabola. Find the vertex of the parabola and then graph it. $$x=-2 y^{2}+4 y+6$$

Step-by-Step Solution

Answer

The vertex of the parabola is at (8, 1).

Step 1: Identify the form of the equation

The given equation is in the form of a quadratic expression with respect to \( y \): \( x = -2y^2 + 4y + 6 \). This implies the parabola opens horizontally, either to the left or right.

Step 2: Convert to standard form

The standard form of a horizontally opening parabola is \( (y - k)^2 = 4p(x - h) \). First, rearrange the equation: \( x = -2(y^2 - 2y) + 6 \). Next, complete the square on the \( y \)-terms.

Step 3: Complete the square

Take the expression \( y^2 - 2y \), find \( \left(\frac{-2}{2}\right)^2 = 1 \), and add and subtract 1 inside the parentheses: \( x = -2(y^2 - 2y + 1 - 1) + 6 = -2((y-1)^2 - 1) + 6 \). Simplify to get: \( x = -2(y-1)^2 + 2 + 6 = -2(y-1)^2 + 8 \).

Step 4: Identify the vertex

The equation \( x = -2(y-1)^2 + 8 \) is in the form \( x = a(y-k)^2 + h \), where \( h = 8 \) and \( k = 1 \). Therefore, the vertex of the parabola, which is \( (h, k) \), is \( (8, 1) \).

Step 5: Graph the parabola

Plot the vertex \((8, 1)\) on the coordinate system. Since \( a = -2 \) in \( x = -2(y-1)^2 + 8 \), the parabola opens to the left due to the negative coefficient. The parabola is symmetric about the line \( y = 1 \). Sketch the graph using the vertex and direction of opening.

Key Concepts

Quadratic Expression

A quadratic expression is a polynomial where the highest degree of a variable is squared. In its standard form with respect to a variable, such as the variable \( y \), it takes the form \( ax^2 + bx + c \). Here, \( a \), \( b \), and \( c \) are constants, and \( a \) cannot be zero.
This expression can represent various shapes, but most commonly it describes a parabola.

• The sign of \( a \) determines whether the parabola opens upwards (positive) or downwards (negative) if concerning \( x \).
• If we consider a horizontal orientation like in our original exercise, \( x \) becomes an expression in terms of \( y \).
• The orientation changes and the signs help us determine if it opens to the left or the right.

Completing a quadratic expression is a key step towards understanding the parabola's vertex, opening direction, and other graph features.

Complete the Square

Completing the square is a method used to convert a quadratic expression into a perfect square trinomial form. This simplifies solving quadratics and finding vertices.To complete the square, focus on the quadratic and linear terms, such as the \( y^2 \) and \( -2y \) terms in \( y^2 - 2y \). Here are the steps involved:

• Take the coefficient (\( b \)) of the linear term (\( y \)) and halve it, then square it. For \( y^2 - 2y \), take \( -2/2 = -1 \), and \((-1)^2 = 1\).
• Add and subtract this square inside the expression: \( x = -2(y^2 - 2y + 1 - 1) + 6 \).
• Simplify this into a squared binomial: \( x = -2((y-1)^2 - 1) + 6 \).
• This form reveals the vertex, aiding graphing and solving tasks.

Completing the square is crucial for transitioning into the standard form of a parabola.

Standard Form of Parabola

The standard form of a parabola is vital for easy graph interpretation. For parabolas opening horizontally, like in our task, it takes the form \( (y - k)^2 = 4p(x - h) \). This form directly shows:

• The vertex of the parabola as \((h, k)\).
• The value \( p \) that indicates how "wide" or "tight" the parabola opens.

In our specific conversion to \( x = -2(y-1)^2 + 8 \), it becomes clear:

• Given \((y - 1)^2 = \left(-\frac{1}{2}\right)(x - 8)\) simplifies first.
• The vertex \((8, 1)\) is extracted easily because \( h = 8 \) and \( k =1 \).
• The negative coefficient signals it opens to the left.

Standard form simplifies graphing by accommodating these attributes in recognizable patterns.

Graphing Parabolas

Graphing parabolas becomes much more straightforward once you understand the vertex and orientation from the quadratic's form. Here's how to proceed:

• First, plot the vertex, the pivot point of the parabola's symmetry. For our equation, the vertex is \((8, 1)\).
• Observe the direction: A negative coefficient in this form indicates the parabola opens leftward, as ours does with \( a = -2 \).
• The parabola's symmetry helps in sketching: use the vertex as a center line (\( y = 1 \) here).
• Estimate other points by choosing \( y \) values and solving for \( x \) to get a better curve shape.

Graphing becomes a blend of understanding the algebraic form and visually plotting distinctive features.