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3 years ago

14

What is the transformations of f(x)=(x-2)^2-4

1 answer:

prisoha [69]3 years ago

4 0

This is how to solve the problem: "What is the transformations of f(x)=(x-2)^2-4"

Looking at the quadratic equation given,
f(x) = (x-2)^2 - 4
It's like seeing product of sum and difference of two squares.

(x-2)^2 - 4
[ (x-2) - 2 ] [ (x-2) + 2 ]
[ x - 2 - 2 ] [ x - 2 + 2 ]
[ x - 4 ] [ x ]
[ x^2 - 4x ]

So the final transformation of f(x) = (x-2)^2 -4 is f(x) = (x^2 - 4x).
In this way, we are able to show how a complicated equation to simple yet equal equation in quadratic form.

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True or false: the equation tan^2x+1=sec^2x

zhenek [66]

ANSWER

True

EXPLANATION

The given trigonometric equation is:

${ \tan}^{2} x + 1 = { \sec}^{2} x$

We take the LHS and simplify to arrive at the RHS.

${ \tan}^{2} x + 1 = \frac{{ \sin}^{2} x}{{ \cos}^{2} x} + 1$

Collect LCM on the right hand side to get;

${ \tan}^{2} x + 1 = \frac{{ \sin}^{2} x + {\cos}^{2} x}{{ \cos}^{2} x}$

This implies that

${ \tan}^{2} x + 1 = \frac{1}{{ \cos}^{2} x} .$

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Help me with this question

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3 years ago