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# 3.9: Vertex Form of a Quadratic Function

Difficulty Level: At Grade Created by: CK-12

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

### Guidance

The equation for a basic parabola with a vertex at \begin{align*}(0,0)\end{align*} is \begin{align*}y=x^2\end{align*}. You can apply transformations to the graph of \begin{align*}y=x^2\end{align*} to create a new graph with a corresponding new equation. This new equation can be written in vertex form. The vertex form of a quadratic function is \begin{align*} y=a(x-h)^2+k\end{align*} where:

• \begin{align*}|a|\end{align*} is the vertical stretch factor. If \begin{align*}a\end{align*} is negative, there is a vertical reflection and the parabola will open downwards.
• \begin{align*}k\end{align*} is the vertical translation.
• \begin{align*}h\end{align*} is the horizontal translation.

Given the equation of a parabola in vertex form, you should be able to sketch its graph by performing transformations on the basic parabola. This process is shown in the examples.

#### Example A

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

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

Solution:

• \begin{align*}a\end{align*} – Is \begin{align*}a\end{align*} negative? YES. The parabola will open downwards.
• \begin{align*}a\end{align*} – Is there a number in front of the squared portion of the equation? YES. The vertical stretch factor is the absolute value of this number. Therefore, the vertical stretch of this function is \begin{align*}\frac{1}{2}\end{align*}.
• \begin{align*}k\end{align*} – Is there a number after the squared portion of the equation? YES. The value of this number is the vertical translation. The vertical translation is -1.
• \begin{align*}h\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

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

• Vertical stretch by a factor of 3
• Vertical translation up 5 units
• Horizontal translation left 4 units

Solution:

• \begin{align*}a\end{align*} – The image is not reflected in the \begin{align*}x\end{align*}-axis. A negative sign is not required.
• \begin{align*}a\end{align*} – The vertical stretch is 3, so \begin{align*}a=3\end{align*}.
• \begin{align*}k\end{align*} – The vertical translation is 5 units up, so \begin{align*}k=5\end{align*}.
• \begin{align*}h\end{align*} – The horizontal translation is 4 units left so \begin{align*}h=-4\end{align*}.

The equation of the image of \begin{align*}y=x^2\end{align*} is \begin{align*}y=3(x+4)^2+5\end{align*}.

#### Example C

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 vertex form.

Solution:

\begin{align*}a\end{align*} – The parabola does not open downward so \begin{align*}a \end{align*} will be positive.

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

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

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

The equation is \begin{align*}y=3(x-3)^2+5\end{align*}.

#### Example D

When the equation of the basic quadratic function is written in vertex 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*}y=a(x-h)^2+k\end{align*} is \begin{align*}(x,y) \rightarrow (x+h,ay+k)\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*}y=2(x+3)^2+5\end{align*} write the mapping rule and create a table of values for the mapping rule.

Solution:

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

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

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

#### Concept Problem Revisited

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

\begin{align*}a=3\end{align*} so the vertical stretch is 3. \begin{align*}k=2\end{align*} so the vertical translation is up 2. \begin{align*}h=-4\end{align*} so the horizontal translation is left 4.

### 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, a vertical translation, a vertical stretch and a vertical reflection.
Vertex form of \begin{align*}y = x^2\end{align*}
The vertex 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 vertex form of the equation is \begin{align*} y=a(x-h)^2+k\end{align*}.
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*}=-(x+5)^2+4\end{align*}

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

1. \begin{align*}a\end{align*}\begin{align*}a\end{align*} is negative so the parabola opens downwards.

\begin{align*}a\end{align*} – The vertical stretch of this function is \begin{align*}\frac{1}{2}\end{align*}.

\begin{align*}k\end{align*} – The vertical translation is -3.

\begin{align*}h\end{align*} – The horizontal translation is +4.

2.

\begin{align*}a & \rightarrow negative\\ a & \rightarrow 1\\ k & \rightarrow +4\\ h & \rightarrow -5\end{align*}

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

### Practice

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

Graph the following quadratic functions using the mapping rule method:

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

Using the following mapping rules, write the equation, in vertex 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*}

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Dec 19, 2012
Apr 29, 2014

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