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3 years ago

Given the function P(x)=(x-1)^2 write the new function with a Dilation of 2 On THE X

Mathematics

1 answer:

yaroslaw [1]3 years ago

3 0

Answer: New function is $y=2(x-1)^2.$

Step-by-step explanation:

A dilation is a transformation that creates an image using a scale factor that has the exactly same shape as the original, but have a different size.

The given function : $P(x)=(x-1)^2$

We know that any function y= f(x) will become y=k f(x) after dilation of k units on the x, where k= scale factor for dilation.

Here, the given scale factor = 2 i.e. k=2

Then, the new function will be :

$2P(x)=2(x-1)^2$

Hence, the new function is $y=2(x-1)^2.$

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2 years ago

which of the functions have a range of real numbers greater than or equal to 1 or less than or equal to-1

lilavasa [31]

The question is incomplete. Here is the complete question:

Which of the functions have a range of all real numbers greater than or equal to 1 or less than or equal to -1? check all that apply.

A. $y=\sec x$

B. $y= \tan x$

C. $y= \cot x$

D. $y= \csc x$

Answer:

A. $y=\sec x$

D. $y=\csc x$

Step-by-step explanation:

Given:

The range is greater than or equal to 1 or less than or equal to -1.

The given choices are:

Choice A: $y=\sec x$

We know that, the $\sec x=\frac{1}{\cos x}$

The range of $\cos x$ is from -1 to 1 given as [-1, 1]. So,

$|\cos x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\cos x|}\geq 1\\|\sec x|\geq 1$

Therefore, on removing the absolute sign, we rewrite the secant function as:

$\sec x\leq -1\ or\ \sec x\geq 1\\$

Therefore, the range of $y=\sec x$ is all real numbers greater than or equal to 1 or less than or equal to-1.

Choice B: $y= \tan x$

We know that, the range of tangent function is all real numbers. So, choice B is incorrect.

Choice C: $y= \cot x$

We know that, the range of cotangent function is all real numbers. So, choice C is incorrect.

Choice D: $y=\csc x$

We know that, the $\csc x=\frac{1}{\sin x}$

The range of $\sin x$ is from -1 to 1 given as [-1, 1]. So,

$|\sin x|\leq 1\\\textrm{Taking reciprocal, the inequality sign changes}\\\frac{1}{|\sin x|}\geq 1\\|\csc x|\geq 1$

Therefore, on removing the absolute sign, we rewrite the cosecant function as:

$\csc x\leq -1\ or\ \csc x\geq 1\\$

Therefore, the range of $y=\csc x$ is all real numbers greater than or equal to 1 or less than or equal to-1.

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