Problem 18 - Algebra 2

Question

Graph each equation. $$ x=y^{2}-14 y+25 $$

Step-by-Step Solution

Answer

The graph is a parabola opening to the right at vertex (7, -24).

Step 1: Rewrite the equation

We start with the equation \( x = y^2 - 14y + 25 \). Notice that this is a quadratic equation in terms of \( y \), which can be expressed in the form \( y = ax^2 + bx + c \). However, we want this in terms of \( y \). First, we group the \( y \) terms together to complete the square. Our aim is to rewrite the equation as \( x = a(y - h)^2 + k \).

Step 2: Complete the square

To complete the square on the \( y \) terms in \( y^2 - 14y + 25 \), take half of the coefficient of \( y \), which is \( 14 \), then square it. Half of \( 14 \) is 7, and \( 7^2 = 49 \). Insert and subtract this inside the expression: \( y^2 - 14y + 49 - 49 + 25 \).

Step 3: Simplify to vertex form

The expression \( y^2 - 14y + 49 \) is now a perfect square and can be expressed as \( (y - 7)^2 \). After simplifying, the equation becomes \( x = (y - 7)^2 - 24 \) since \( 49 - 24 = 24 \). So the equation in vertex form is \( x = (y - 7)^2 - 24 \).

Step 4: Identify key features

The equation is now in the form \( x = a(y - h)^2 + k \), where \( a = 1 \), \( h = 7 \), and \( k = -24 \). The vertex of this parabola is at \( (h, k) = (7, -24) \), and since \( a = 1 \), the parabola opens to the right.

Step 5: Graph the equation

Plot the vertex at the point \( (7, -24) \). The direction of opening is to the right, as determined from the coefficient \( a = 1 \), which is positive. To get more points, choose values of \( y \) around the vertex, solve for corresponding \( x \) values, and plot these points. Finally, sketch a U-shaped curve passing through these points that opens to the right.

Key Concepts

Quadratic Equation

A quadratic equation is a type of polynomial equation characterized by the highest power of the variable being squared. While traditionally expressed in the form \( ax^2 + bx + c = 0 \) where \( a, b, \) and \( c \) are constants and \( x \) is the variable, quadratic equations can also take different forms. In the given exercise, the equation \( x = y^2 - 14y + 25 \) is a quadratic equation in terms of the variable \( y \).

• The typical quadratic form is altered here since the variable of interest is \( y \) rather than \( x \).
• In this setting, \( y^2 \) is the highest power of \( y \), confirming that it's a quadratic in \( y \).

Recognizing the structure of quadratic equations is key to effectively manipulating them into useful forms, such as the vertex form, to easily identify their properties and graph them.

Vertex Form

The vertex form of a quadratic equation is especially useful for graphing and understanding the key features of a parabola. For a quadratic equation involving \( y \), as in \( x = a(y - h)^2 + k \), the vertex form highlights the vertex, \((h, k)\), of the parabola directly.

• In our exercise, once we transformed \( x = y^2 - 14y + 25 \) into \( x = (y - 7)^2 - 24 \), it became clear that \( h = 7 \) and \( k = -24 \).
• The vertex \((7, -24)\) represents the point where the parabola changes direction.

The vertex form is crucial because it reveals:

• The direction in which the parabola opens, determined by the value of \( a \). Here, \( a = 1 \), indicating the parabola opens to the right.
• The vertex gives the minimum or maximum point of the parabola, dependent on its orientation.

Completing the Square

Completing the square is a technique used to transform a quadratic equation into vertex form. It involves adjusting and rearranging the equation to create a perfect square trinomial. In the exercise, this process was applied to the expression \( y^2 - 14y + 25 \).Here’s how to complete the square:

• Take the coefficient of the linear term (\( y \)), halve it, and square the result. For \( -14y \), it is \( (\frac{-14}{2})^2 = 49 \).
• Add and subtract this squared number inside the equation: \( y^2 - 14y + 49 - 49 + 25 \).
• Reorganize it to form a perfect square trinomial: \( (y - 7)^2 \).

By completing the square, you can easily rewrite and solve quadratic equations and readily find the vertex, aiding greatly in graphing parabolas effectively.