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# 3.10: Transformational Form of y = x²

Difficulty Level: At Grade Created by: CK-12

Introduction

In this lesson you will learn to write the equation of the base quadratic function \begin{align*}y=x^2\end{align*} to reflect the transformations that the graph has undergone. The equation will include any of the transformations that have been applied to \begin{align*}y=x^2\end{align*}. These transformations will include a vertical reflection, a vertical stretch, a vertical translation and a horizontal translation

Objectives

The lesson objectives for The Transformation Form of \begin{align*}y=x^2\end{align*} are:

• Understanding the transformational form of the equation.
• Identifying the transformations of \begin{align*}y=x^2\end{align*} from the equation.
• Applying the transformations to create the equation.
• Drawing the graph by using the transformations of \begin{align*}y=x^2\end{align*}.

## Transformational Form of y = x²

Introduction

In this concept you will learn to apply transformations of the graph of \begin{align*}y=x^2\end{align*} to create an equation for the image graph. The equation will be written in transformational form. The transformational form of an equation is of the form \begin{align*}\boxed{} \ \frac{1}{a}(y-k)=(x-h)^2\end{align*} where

• \begin{align*}a\end{align*} is the vertical stretch \begin{align*}(VS)\end{align*}
• \begin{align*}k\end{align*} is the vertical translation \begin{align*}(VT)\end{align*}
• \begin{align*}h\end{align*} is the horizontal translation \begin{align*}(HT)\end{align*}
• \begin{align*}\boxed{}\end{align*} is the location of a negative sign which indicates a vertical reflection \begin{align*}(VR)\end{align*}

As the name indicates, you will be able to identify the transformations from the equation. The transformations can then be used to draw the image graph of \begin{align*}y=x^2\end{align*}.

Guidance

Another way to write the transformational form of \begin{align*}y=x^2\end{align*} is \begin{align*}\boxed{\boxed{} \ \frac{1}{VS}(y-VT)=(x-HT)^2}\end{align*}

The transformations that deal with the vertical components are the ‘\begin{align*}y\end{align*}’ side of the equation and those that deal with the horizontal components are on the ‘\begin{align*}x\end{align*}’ side of the equation. These apply to the \begin{align*}x-\end{align*} and \begin{align*}y-\end{align*}axes of the Cartesian grid.

Given the equation \begin{align*}\frac{1}{3} (y-2)=(x+4)^2\end{align*}, list the transformations of \begin{align*}y=x^2\end{align*}.

\begin{align*}& \boxed{\boxed{} \ \frac{1}{VS} (y-VT) = (x-HT)^2}\\ & \quad \ \ {\color{red}\updownarrow} \qquad \quad \ {\color{red}\updownarrow} \qquad \qquad \quad {\color{red}\updownarrow}\\ & \quad \ \ \frac{1}{3} \ (y \ - \ 2) \ = (x \ + \ 4)^2\end{align*}

The vertical stretch is 3. The vertical translation is 2. The negative sign did not change in the equation. It remained the same as it was in the general, transformational form of the equation. This indicates that the \begin{align*}VT\end{align*} is a positive value. The horizontal translation is negative 4. The negative sign given in the general, transformational form of the equation changed to a positive sign in the given equation. This change indicates that the horizontal translation was a negative value. There is no negative sign before \begin{align*}\frac{1}{3}\end{align*} so the image graph is not a vertical reflection.

Example A

Given the following function in transformational form, identify the transformations of \begin{align*}y=x^2\end{align*}.

\begin{align*}-2(y+1)=(x-2)^2\end{align*}

\begin{align*}VR\end{align*} – Is there a negative sign in front of \begin{align*}\frac{1}{VS}\end{align*}? YES. The image graph is a vertical reflection in the \begin{align*}x-\end{align*}axis.

\begin{align*}VS\end{align*} – Is there a number in front of ‘\begin{align*}y\end{align*}’ or the parenthesis containing the variable ‘\begin{align*}y\end{align*}’? YES. However, the number in front of the parenthesis containing the variable ‘\begin{align*}y\end{align*}’ is not in the form \begin{align*}\frac{1}{VS}\end{align*}. The vertical stretch factor is the reciprocal of the numerical coefficient of ‘\begin{align*}y\end{align*}’. Therefore, the vertical stretch of this function is \begin{align*}\frac{1}{2}\end{align*}.

\begin{align*}VT\end{align*} – Is there a number after the variable ‘\begin{align*}y\end{align*}’? YES. The value of this number is the opposite of the sign that appears in the equation. The vertical translation is -1.

\begin{align*}HT\end{align*} – Is there a number after the variable ‘\begin{align*}x\end{align*}’? YES. The value of this number is the opposite of the sign that appears in the equation. The horizontal translation is +2.

Example B

When the transformations of \begin{align*}y=x^2\end{align*} were determined from the given equation, the reciprocal of the numerical coefficient of ‘\begin{align*}y\end{align*}’ was the vertical stretch factor, and the opposite signs of the values that appeared for \begin{align*}VT\end{align*} and \begin{align*}HT\end{align*} were used to identify the vertical and horizontal translations. These same applications will be used to determine the equation in transformational form from a given list of transformations of \begin{align*}y=x^2\end{align*}.

Given the following transformations, determine the equation of the image of \begin{align*}y=x^2\end{align*} in transformational form.

\begin{align*}VR & \rightarrow NO\\ VS & \rightarrow 3\\ VT & \rightarrow 5\\ HT & \rightarrow -4\end{align*}

\begin{align*}\boxed{\boxed{} \ \frac{1}{VS} (y-VT)=(x-HT)^2}\end{align*}

\begin{align*}VR\end{align*} – The image is not reflected in the \begin{align*}x-\end{align*}axis. A negative sign is not required.

\begin{align*}VS\end{align*} – The reciprocal of 3 is \begin{align*}\frac{1}{3}\end{align*}.

\begin{align*}VT\end{align*} – The opposite of a +5 is a -5.

\begin{align*}HT\end{align*} – The opposite of a -4 is a +4.

The equation of the image of \begin{align*}y=x^2\end{align*} is \begin{align*}\boxed{{\color{red}\frac{1}{3}} (y {\color{red}-5})=(x {\color{red}+4})^2}\end{align*}

Example C

If a parabola opens downward, it is reflected in the \begin{align*}x-\end{align*}axis. To show this reflection in the equation, a negative sign is written before the vertical stretch factor in the equation.

From the vertex of the parabola, the points of \begin{align*}y=x^2\end{align*} are plotted left and right one and \begin{align*}\uparrow\end{align*} or \begin{align*}\downarrow\end{align*} one; left and right two and \begin{align*}\uparrow\end{align*} or \begin{align*}\downarrow\end{align*} four; left and right three and \begin{align*}\uparrow\end{align*} or \begin{align*}\downarrow\end{align*} nine. If the \begin{align*}y-\end{align*}values of 1, 4 or 9 are changed, the vertical stretch factor is the number by which the \begin{align*}y-\end{align*}values were multiplied to create the new values.

If the vertex of the parabola is not located at (0, 0), the new coordinates of the vertex will indicate the horizontal translation and the vertical translation of \begin{align*}y=x^2\end{align*}.

Using \begin{align*}y=x^2\end{align*} as the base function, identify the transformations that have occurred to produce the following image graph. Use these transformations to write the equation in transformational form.

\begin{align*}VR\end{align*} – NO The parabola does not open downward.

\begin{align*}VS\end{align*} – The \begin{align*}y-\end{align*}values of 1 and 4 are now up 3 and up 12. \begin{align*}VS = 3\end{align*}.

\begin{align*}VT\end{align*} – The \begin{align*}y-\end{align*}coordinate of the vertex is -5.

\begin{align*}HT\end{align*} – The \begin{align*}x-\end{align*}coordinate of the vertex is +3.

The equation is \begin{align*}\boxed{{\color{red}\frac{1}{3}} (y {\color{red}+5})=(x {\color{red}-3})^2}\end{align*}

Example D

When the equation of the base quadratic function is written in transformational form, the function can also be expressed in mapping notation form. This form describes how to obtain the image of a given graph by using the changes in the ordered pairs.

The standard base table of values for the base quadratic function \begin{align*}y=x^2\end{align*} is given by:

\begin{align*}& X \qquad -3 \qquad -2 \qquad -1 \qquad 0 \qquad 1 \qquad 2 \qquad 3\\ & Y \qquad \quad \ 9 \qquad \quad \ 4 \qquad \quad \ 1 \qquad 0 \qquad 1 \qquad 4 \qquad 9\end{align*}

When these ordered pairs are plotted, we get the base parabola. The mapping rule used to generate the image of a quadratic function is \begin{align*}(x,y) \rightarrow (x^\prime,y^\prime)\end{align*} where \begin{align*}(x^\prime,y^\prime)\end{align*} are the coordinates of the image graph. The resulting mapping rule from the equation

\begin{align*}\boxed{} \ \frac{1}{VS} (y-VT)=(x-HT)^2\end{align*} is \begin{align*}(x,y) \rightarrow (x+HT,VR \ \ VS \ \ y+VT)\end{align*}. A mapping rule details the transformations that were applied to the coordinates of the base function \begin{align*}y=x^2\end{align*}.

Given the following quadratic equation, \begin{align*}\frac{1}{2}(y-5)=(x+3)^2\end{align*} write the mapping rule and create a table of values for the mapping rule.

The mapping rule for this function will tell exactly what changes were applied to the coordinates of the base quadratic function.

\begin{align*}\frac{1}{2}(y-5)=(x+3)^2 \quad (x,y) \rightarrow (x-3,2y+5)\end{align*}

These new coordinates of the image graph can be plotted to generate the graph.

Vocabulary

Horizontal translation
The horizontal translation is the change in the base graph \begin{align*}y=x^2\end{align*} that shifts the graph right or left. It changes the \begin{align*}x-\end{align*}coordinate of the vertex.
Mapping Rule
The mapping rule is another form used to express a quadratic function. The mapping rule defines the transformations that have occurred to the base quadratic function \begin{align*}y=x^2\end{align*}. The mapping rule is \begin{align*}(x,y) \rightarrow (x^\prime,y^\prime)\end{align*} where \begin{align*}(x^\prime,y^\prime)\end{align*} are the coordinates of the image graph.
Transformation
A transformation is any change in the base graph \begin{align*}y=x^2\end{align*}. The transformations that apply to the parabola are a horizontal translation \begin{align*}(HT)\end{align*}, a vertical translation \begin{align*}(VT)\end{align*}, a vertical stretch \begin{align*}(VS)\end{align*} and a vertical reflection \begin{align*}(VR)\end{align*}.
Transformational form of \begin{align*}y = x^2\end{align*}
The transformational form of \begin{align*}y = x^2\end{align*} is the form of the quadratic base function \begin{align*}y=x^2\end{align*} that shows the transformations of the image graph. The transformational form of the equation is \begin{align*}\boxed{\boxed{} \ \frac{1}{VS} (y-VT)=(x-HT)^2}\end{align*} A negative sign in \begin{align*}\boxed{}\end{align*} indicates a vertical reflection in the \begin{align*}x-\end{align*}axis. The reciprocal of the number before the parenthesis containing the variable ‘\begin{align*}y\end{align*}’ indicates the vertical stretch factor, and the opposite signed of the values for \begin{align*}VT\end{align*} and \begin{align*}HT\end{align*} indicate the coordinates of the vertex.
Vertical Reflection
The vertical reflection is the reflection of the image graph in the \begin{align*}x-\end{align*}axis. The graph opens downward and the \begin{align*}y-\end{align*}values are negative values.
Vertical Stretch
The vertical stretch is the change made to the base function \begin{align*}y=x^2\end{align*} by stretching (or compressing) the graph vertically. The vertical stretch will produce an image graph that appears narrower (or wider) then the original base graph of \begin{align*}y=x^2\end{align*}.
Vertical Translation
The vertical translation is the change in the base graph \begin{align*}y=x^2\end{align*} that shifts the graph up or down. It changes the \begin{align*}y-\end{align*}coordinate of the vertex.

Guided Practice

1. Identify the transformations of \begin{align*}y=x^2\end{align*} for the quadratic function \begin{align*}-2(y+3)=(x-4)^2\end{align*}

2. List the transformations of \begin{align*}y=x^2\end{align*} and graph the function \begin{align*}-(y-4)=(x+5)^2\end{align*}

3. Graph the function \begin{align*}\frac{1}{2}(y-3)=(x-2)^2\end{align*} using the mapping rule method.

Answers

1. \begin{align*}VR\end{align*} – There is a negative sign in front of \begin{align*}\frac{1}{VS}\end{align*}. The image graph is a vertical reflection in the \begin{align*}x-\end{align*}axis.

\begin{align*}VS\end{align*} – There is a number in front of ‘\begin{align*}y\end{align*}’ or the parenthesis containing the variable ‘\begin{align*}y\end{align*}’. The vertical stretch factor is the reciprocal of the numerical coefficient of ‘\begin{align*}y\end{align*}’. Therefore, the vertical stretch of this function is \begin{align*}\frac{1}{2}\end{align*}.

\begin{align*}VT\end{align*} – There is a number after the variable ‘\begin{align*}y\end{align*}’. The value of this number is the opposite of the sign that appears in the equation. The vertical translation is -3.

\begin{align*}HT\end{align*} – There is a number after the variable ‘\begin{align*}x\end{align*}’. The value of this number is the opposite of the sign that appears in the equation. The horizontal translation is +4.

2.

\begin{align*}VR & \rightarrow YES\\ VS & \rightarrow 1\\ VT & \rightarrow +4\\ HT & \rightarrow -5\end{align*}

3. \begin{align*}\frac{1}{2}(y-3)=(x-2)^2\end{align*}

Mapping Rule \begin{align*}(x,y) \rightarrow (x+2,2y+3)\end{align*}

\begin{align*}VR & \rightarrow NO\\ VS & \rightarrow 2 \ - \ y- \text{values are multiplied by} \ 2\\ VT & \rightarrow +3- \ y- \text{values are also increased by} \ 3\\ HT & \rightarrow +2- \ x- \text{values are increased by} \ 2\end{align*}

Make a table of values:

\begin{align*}x \rightarrow x+2\end{align*} \begin{align*}y \rightarrow 2y+3\end{align*}
\begin{align*}-3\end{align*} -1 9 21
-2 0 4 11
-1 1 1 5
0 2 0 3
1 3 1 5
2 4 4 11
3 5 9 21

Draw the Graph

Summary

In this lesson you have learned to write the equation of a parabola in transformational form. This form of the equation shows any and all changes that have been made to the base quadratic \begin{align*}y=x^2\end{align*}. These changes transformations were then used to sketch the graph of the parabola.

You also learned to write a mapping rule that represented the transformations of \begin{align*}y=x^2\end{align*}. The mapping rule was used to create a table of values that reflected the changes of the base table of values of \begin{align*}y=x^2\end{align*}. The new table of values was then used to graph the quadratic function of the parabola.

You also learned to use the transformations of \begin{align*}y=x^2\end{align*} that were visible on the graph to write the equation of the quadratic function in transformational form. You had to remember to use the opposite sign for the vertical and horizontal translations as well as the reciprocal of the vertical stretch.

Problem Set

Complete the following table to identify the transformations of \begin{align*}y=x^2\end{align*} in each of the given functions:

Number \begin{align*}VR\end{align*} \begin{align*}VS\end{align*} \begin{align*}VT\end{align*} \begin{align*}HT\end{align*}
1.
2.
3.
4.
5.
1. \begin{align*}\frac{1}{4}(y+9)=(x-2)^2\end{align*}
2. \begin{align*}-6(y+7)=x^2\end{align*}
3. \begin{align*}-\frac{1}{3}(y+6)=(x-1)^2\end{align*}
4. \begin{align*}5(y-3)=(x+4)^2\end{align*}
5. \begin{align*}\frac{1}{5}y=(x+2)^2\end{align*}

Graph the following quadratic functions using the mapping rule method:

1. \begin{align*}\frac{1}{2}(y+5)=(x-4)^2\end{align*}
2. \begin{align*}-3(y-6)=(x-2)^2\end{align*}
3. \begin{align*}-\frac{1}{2}(y-7)=(x+3)^2\end{align*}
4. \begin{align*}-2(y-9)=(x+6)^2\end{align*}
5. \begin{align*}3y=(x-4)^2\end{align*}

Using the following mapping rules, write the equation, in transformational form, that represents the image of \begin{align*}y = x^2\end{align*}.

1. \begin{align*}(x,y) \rightarrow \left(x+1, -\frac{1}{2}y\right)\end{align*}
2. \begin{align*}(x,y) \rightarrow (x+6,2y-3)\end{align*}
3. \begin{align*}(x,y) \rightarrow \left(x-1, \frac{2}{3}y+2\right)\end{align*}
4. \begin{align*}(x,y) \rightarrow (x+3,3y+1)\end{align*}
5. \begin{align*}(x,y) \rightarrow \left(x-5,-\frac{1}{3}y-7\right)\end{align*}

Answers

Complete the following table...

Number \begin{align*}VR\end{align*} \begin{align*}VS\end{align*} \begin{align*}VT\end{align*} \begin{align*}HT\end{align*}
1. NO \begin{align*}4\end{align*} \begin{align*}-9\end{align*} \begin{align*}2\end{align*}
2. YES \begin{align*}\frac{1}{6}\end{align*} \begin{align*}-7\end{align*}
3. YES \begin{align*}3\end{align*} \begin{align*}-6\end{align*} \begin{align*}1\end{align*}
4. NO \begin{align*}\frac{1}{5}\end{align*} \begin{align*}3\end{align*} \begin{align*}-4\end{align*}
5. NO \begin{align*}5\end{align*} \begin{align*}-2\end{align*}

Graph the following quadratic...

1. \begin{align*}\frac{1}{2}(y+5)=(x-4)^2\end{align*} Mapping Rule: \begin{align*}(x,y) \rightarrow (x+4,2y-5)\end{align*}
\begin{align*}x \rightarrow x+4\end{align*} \begin{align*}y \rightarrow 2y-5\end{align*}
\begin{align*}-3\end{align*} 1 9 13
-2 2 4 3
-1 3 1 -3
0 4 0 -5
1 5 1 -3
2 6 4 3
3 7 9 13

1. \begin{align*}-\frac{1}{2}(y-7)=(x+3)^2\end{align*} Mapping Rule: \begin{align*}(x,y) \rightarrow (x -3, -2y+7)\end{align*}
\begin{align*}x \rightarrow x+4\end{align*} \begin{align*}y \rightarrow 2y-5\end{align*}
\begin{align*}-3\end{align*} -6 9 -11
-2 -5 4 -1
-1 -4 1 5
0 -3 0 7
1 -2 1 5
2 -1 4 -1
3 0 9 -11

1. \begin{align*}3y=(x-4)^2\end{align*} Mapping Rule: \begin{align*}(x,y) \rightarrow \left(x +4, \frac{1}{3}y \right)\end{align*}
\begin{align*}x \rightarrow x+4\end{align*} \begin{align*}y \rightarrow \frac{1}{3}y\end{align*}
\begin{align*}-3\end{align*} -6 9 3
-2 -5 4 \begin{align*}\frac{4}{3}\end{align*}
-1 -4 1 \begin{align*}\frac{1}{3}\end{align*}
0 -3 0 0
1 -2 1 \begin{align*}\frac{1}{3}\end{align*}
2 -1 4 \begin{align*}\frac{4}{3}\end{align*}
3 0 9 3

Using the following mapping rules...

1. \begin{align*}(x,y) \rightarrow \left(x+1, -\frac{1}{2}y\right)\end{align*}

\begin{align*}VR & \rightarrow YES\\ VS & \rightarrow \frac{1}{2}\\ VT & \rightarrow 0\\ HT & \rightarrow 1\end{align*}

Equation in transformational form \begin{align*}\boxed{-2y=(x-1)^2}\end{align*}

1. \begin{align*}(x,y) \rightarrow \left(x-1, \frac{2}{3}y+2 \right)\end{align*}

\begin{align*}VR & \rightarrow NO\\ VS & \rightarrow \frac{2}{3}\\ VT & \rightarrow 2\\ HT & \rightarrow -1\end{align*}

Equation in transformational form \begin{align*}\boxed{\frac{3}{2} (y-2)=(x+1)^2}\end{align*}

1. \begin{align*}(x,y) \rightarrow \left(x-5, - \frac{1}{3}y-7\right)\end{align*}

\begin{align*}VR & \rightarrow YES\\ VS & \rightarrow \frac{1}{3}\\ VT & \rightarrow -7\\ HT & \rightarrow -5\end{align*}

Equation in transformational form \begin{align*}\boxed{-3(y+7)=(x+5)^2}\end{align*}

## Summary

In this lesson you have learned how to graph the base quadratic function \begin{align*}y=x^2\end{align*} by creating a table of values. The points from the table were then plotted on the Cartesian grid to form a curved shape known as a parabola. You learned that a parabola has a turning point called a vertex, an axis of symmetry that is a vertical line that is parallel to the \begin{align*}y-\end{align*}axis and that passes through the vertex. You also discovered that each point on one side of the axis of symmetry had a corresponding point on the other side of the axis of symmetry. You also learned how to determine the domain and the range for a parabola and how to write these values using interval notation.

The next section of this lesson dealt with the changes or transformations that occurred to the image graph of \begin{align*}y=x^2\end{align*}. These transformations included a vertical reflection across the \begin{align*}x-\end{align*}axis, a vertical stretch that caused the \begin{align*}y-\end{align*}values of 1, 4 and 9 to either increase or to decrease, a vertical translation that caused the graph to move upward or downward and finally a horizontal translation that moved the graph to the left or the right. You learned to use these transformations to graph the image graph of \begin{align*}y=x^2\end{align*}. In addition, you learned to recognize the transformations as they appeared in a given graph.

The final section of this lesson dealt with representing the transformations in a quadratic function that was presented in transformational form. The transformations of \begin{align*}y=x^2\end{align*} were extracted from the equation and used to sketch the image graph of \begin{align*}y=x^2\end{align*}. You also learned how to express the quadratic function in the form of a mapping rule. The mapping rule method was then applied to graphing the image of \begin{align*}y=x^2\end{align*}.

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CK.MAT.ENG.SE.1.Algebra-I---Honors.3.10