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3 years ago

Would the scale factor 4.2 enlarge,reduce,or preserve

Mathematics

1 answer:

Juli2301 [7.4K]3 years ago

8 0

Hey there! I'm happy to help!

Let's look at what scale factors would be able to enlarge, reduce, and preserve first. We will use the point (1,1) as an example.

ENLARGE

When we talk about scale factors, both of our numbers of the ordered pair will be multiplied by that number. To make it bigger, we will multiply it by anything larger than a one, we could have a scale factor such as 2, 100, 579395. etc.!

Dilation of 2

(1,1)⇒(2,2)

REDUCE

To reduce our ordered pair, our scale factor must be a number greater than zero but less than one because it cannot switch quadrants, which is what would happen if we used negative numbers. Our reduced number must be greater than zero. We can use 1/2, 1/4, 1/100, 0.67 etc.

Dilation of 1/2

(1,1)⇒(0.5,0.5)

PRESERVE

What would we multiply an ordered pair by to make it stay the same? One of course! This is the only number that will keep it the same!

Dilation of 1

(1,1)⇒(1,1)

Now, let's think of any scale factor as X and see what rules it has to follow to be enlarging, reducing, or preserving.

ENLARGING SCALE FACTOR: X>1

REDUCING SCALE FACTOR: 0<X<1

PRESERVING SCALE FACTOR: X=1

So, where does 4.2 fall? Well, as you can see, 4.2 is greater than one, so this scale factor would make our point larger.

Therefore, 4.2 would enlarge.

I hope that this helps! Have a wonderful day!

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3 years ago

Which function is undefined for x = 0? y=3√x-2 y=√x-2 y=3√x+2 y=√x=2

Andre45 [30]

For this case, we must indicate which of the given functions is not defined for $x = 0$

By definition, we know that:

$f (x) = \sqrt {x}$ has a domain from 0 to infinity.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. For it to be defined, the term within the root must be positive.

Thus, we observe that:

$y = \sqrt {x-2}$ is not defined, the term inside the root is negative when $x = 0$ .

While $y = \sqrt {x + 2}$ if it is defined for $x = 0.$

$f(x)=\sqrt[3]{x}$ , your domain is given by all real numbers.

Adding or removing numbers to the variable within the root implies a translation of the function vertically or horizontally. In the same way, its domain will be given by the real numbers, independently of the sign of the term inside the root.

So, we have:

$y = \sqrt [3] {x-2}$ with x = 0: $y = \sqrt [3] {- 2}$ is defined.