Mathematics, 29.11.2020 06:40 kimlyn58p0wyn0

Describe the transformation of each function compared to its parent function. I need help please!

Describe the transformation of each function compared to its parent function.

I need help please!

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Answer from: cameron12502

Transformation involves changing the position and/or size of a function.

The transformations are:

Translate f(x), 4 units right, and 9 units up.Translate f(x), 4 units left, then compress f(x) vertically by 1/4Vertically stretch f(x) by 3/2

$\mathbf{1.\ f(x) = (x - 4)^2 +9}$

The above function is a square function, and the parent function is:

$\mathbf{f(x) = x^2}$

First, the function is translated right by 4 units (this is represented by subtracting 4 from x.

i.e.

$\mathbf{f(x) = (x - 4)^2}$

Next, the function is translated up by 9 units (this is represented by adding 9 to the value of y.

i.e.

$\mathbf{f(x) = (x - 4)^2 + 9}$

So, the transformation from the parent function is: Translate f(x), 4 units right, and 9 units up.

$\mathbf{2.\ f(x) = \frac 14(x + 4)^2}$

The above function is a square function, and the parent function is:

$\mathbf{f(x) = x^2}$

First, the function is translated left by 4 units (this is represented by adding 4 from x.

i.e.

$\mathbf{f(x) = (x + 4)^2}$

Next, the function is vertically compressed a factor of 1/4 (this is represented by multiplying the function by 1/4

i.e.

$\mathbf{f(x) = \frac 14(x + 4)^2}$

So, the transformation from the parent function is: Translate f(x), 4 units left, then compress f(x) vertically by 1/4

$\mathbf{3.\ f(x) = \frac 32x^2}$

The above function is a square function, and the parent function is:

$\mathbf{f(x) = x^2}$

The function is vertically stretched a factor of 3/2 (this is represented by multiplying the function by 3/2

i.e.

$\mathbf{f(x) = \frac 34 \times x^2}\\$

$\mathbf{f(x) = \frac 34 \timex x^2}$

So, the transformation from the parent function is: Vertically stretch f(x) by 3/2

Read more about transformations at:

link

Answer from: avision42

Step-by-step explanation:

In general, for all problems like this, remember the acronym VOHI:

vertical outside, horizontal inside.

1. The parent function here would be $f(x)=x^{2}$ . 4 has been subtracted from the inside, so it is a horizontal change. You might think that it is moving left 4, but it's actually moving right 4. There is a reason for this, but it's not necessary for you to know right now. All you need to remember is that when inside the parentheses, minus means moving right, and plus means moving left. You also have a 9 added to the outside. Vertical outside. This is a vertical change, meaning up and down. This function has been moved right 4 and up 9.

2. First, the parent function is again $f(x)=x^{2}$ . This time, we have $\frac{1}{4}$ on the outside of our function. This means it is a vertical transformation. When the value on the outside is > 1, it is a vertical stretch. When the value is between 0 and 1 (0<a<1), it is a vertical shrink. $\frac{1}{4}$ is between 0 and 1. So, this is a vertical shrink by a factor of $\frac{1}{4}$ . Next, there is a 6 on the inside of the parentheses, meaning a horizontal translation. Remember from problem 1, though, minus=right and plus=left. So, this function has been vertically shrunk by $\frac{1}{4}$ and moved left 6.

3. Again, parent function: $f(x)=x^{2}$ . This one just has $\frac{3}{2}$ on the outside. We know that it's on the outside because otherwise there would be parentheses around both $\frac{3}{2}$ and x. There is not, so it is outside. Outside, vertical. This is a vertical change. Looking back at problem two, when a value is >1, which this is, it is a vertical stretch. So this function has a vertical stretch by $\frac{3}{2}$ .

I hope this helps!

Answer from: Quest

the reason why your question keeps getting deleted because this is not a school related question : )

i will answer your question though.

i never tried church's

Answer from: Quest

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range: