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# 3.9: Vertex Form of a Quadratic Function

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Practice Use Vertex Form of Quadratics
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Given the equation $y=3(x+4)^2+2$ , list the transformations of $y=x^2.$

### Guidance

The equation for a basic parabola with a vertex at $(0,0)$ is $y=x^2$ . You can apply transformations to the graph of $y=x^2$ 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 $y=a(x-h)^2+k$ where:

• $|a|$ is the vertical stretch factor. If $a$ is negative, there is a vertical reflection and the parabola will open downwards.
• $k$ is the vertical translation.
• $h$ 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 $y=x^2$ .

$y=-\frac{1}{2}(x-2)^2-1$

Solution:

• $a$ – Is $a$ negative? YES. The parabola will open downwards.
• $a$ – 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 $\frac{1}{2}$ .
• $k$ – 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 .
• $h$ – Is there a number after the variable ‘ $x$ ’? 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 $y=x^2$ in vertex form.

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

Solution:

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

The equation of the image of $y=x^2$ is $y=3(x+4)^2+5$ .

#### Example C

Using $y=x^2$ 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:

$a$ – The parabola does not open downward so $a$ will be positive.

$a$ – The $y$ -values of 1 and 4 are now up 3 and up 12. $a = 3$ .

$k$ – The $y-$ coordinate of the vertex is –5 so $k=-5$ .

$h$ – The $x-$ coordinate of the vertex is +3 so $h=3$ .

The equation is $y=3(x-3)^2-5$ .

#### Example D

In general, the mapping rule used to generate the image of a function is $(x,y) \rightarrow (x^\prime,y^\prime)$ where $(x^\prime,y^\prime)$ are the coordinates of the image graph. The resulting mapping rule from $y=x^2$ to the image $y=a(x-h)^2+k$ is $(x,y) \rightarrow (x+h,ay+k)$ . The mapping rule details the transformations that were applied to the coordinates of the base function $y=x^2$ .

Given the following quadratic equation, $y=2(x+3)^2+5$ 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.

$y=2(x+3)^2+5: \quad (x,y) \rightarrow (x-3,2y+5)$

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

#### Concept Problem Revisited

Given the equation $y=3(x+4)^2+2$ , list the transformations of $y=x^2$ .

$a=3$ so the vertical stretch is 3 . $k=2$ so the vertical translation is up 2 . $h=-4$ so the horizontal translation is left 4 .

### Vocabulary

Horizontal translation
The horizontal translation is the change in the base graph $y=x^2$ that shifts the graph right or left. It changes the $x-$ coordinate of the vertex.
Mapping Rule
The mapping rule defines the transformations that have occurred to a function. The mapping rule is $(x,y) \rightarrow (x^\prime,y^\prime)$ where $(x^\prime,y^\prime)$ are the coordinates of the image graph.
Transformation
A transformation is any change in the base graph $y=x^2$ . The transformations that apply to the parabola are a horizontal translation, a vertical translation, a vertical stretch and a vertical reflection.
Vertex form of $y = x^2$
The vertex form of $y = x^2$ is the form of the quadratic base function $y=x^2$ that shows the transformations of the image graph. The vertex form of the equation is $y=a(x-h)^2+k$ .
Vertical Reflection
The vertical reflection is the reflection of the image graph in the $x$ -axis. The graph opens downward and the $y$ -values are negative values.
Vertical Stretch
The vertical stretch is the change made to the base function $y=x^2$ 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 $y=x^2$ .
Vertical Translation
The vertical translation is the change in the base graph $y=x^2$ that shifts the graph up or down. It changes the $y-$ coordinate of the vertex.

### Guided Practice

1. Identify the transformations of $y=x^2$ for the quadratic function $-2(y+3)=(x-4)^2$

2. List the transformations of $y=x^2$ and graph the function $=-(x+5)^2+4$

3. Graph the function $y=2(x-2)^2+3$ using the mapping rule method.

1. Rewrite the equation in vertex form. $a$$a$ is negative so the parabola opens downwards.

$a$ – The vertical stretch of this function is $\frac{1}{2}$ .

$k$ – The vertical translation is -3 .

$h$ – The horizontal translation is +4 .

2.

$a & \rightarrow negative\\a & \rightarrow 1\\k & \rightarrow +4\\h & \rightarrow -5$

3. Mapping Rule $(x,y) \rightarrow (x+2,2y+3)$

Make a table of values:

$x \rightarrow$ $x+2$ $y \rightarrow$ $2y+3$
–3 –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

Identify the transformations of $y=x^2$ in each of the given functions:

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

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

Using the following mapping rules, write the equation, in vertex form, that represents the image of $y = x^2$ .

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

Apr 30, 2013

Jul 15, 2014