Chapter 10: Problem 18
Graph each equation. $$ x=y^{2}-14 y+25 $$
Short Answer
Expert verified
The graph is a parabola opening to the right at vertex (7, -24).
Step by step solution
01
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 \( x = a(y - h)^2 + k \).
02
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: \( y^2 - 14y + 49 - 49 + 25 \).
03
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 \( x = (y - 7)^2 - 24 \).
04
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.
05
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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 \( y \).
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.
The vertex form is crucial because it reveals:
- 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:
By completing the square, you can easily rewrite and solve quadratic equations and readily find the vertex, aiding greatly in graphing parabolas effectively.
- 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.